Solutions to
Topology
Chapter 3 - Connectedness and Compactnes
James Munkres
Solutions by positrón0802
https://positron0802.wordpress.com
1 January 2021
Contents
3 Connectedness and Compactness
23 Connected Spaces . . . . . . . . . . . .
24 Connected Subspaces of the Real Line .
25 Components and Local Connectedness
26 Compact Spaces . . . . . . . . . . . . .
27 Compact Subspaces of the Real Line . .
28 Limit Point Compactness . . . . . . . .
29 Local Compactness . . . . . . . . . . .
Supplementary Exercises: Nets . . . . . . . .
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3
Connectedness and Compactness
23
Connected Spaces
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1
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Exercise 23.1. If (𝑋, 𝒯 0) is connected, then (𝑋, 𝒯) is connected. The converse is not true in
general.
Ð
Ð
Exercise 23.2. Suppose that 𝑛 𝐴𝑛 = 𝐵 ∪ 𝐶, where 𝐵 and 𝐶 are disjoint open subsets of 𝑛 𝐴𝑛 .
Since 𝐴1 is connected and a subset of 𝐵 ∪ 𝐶, by Lemma 23.2 it lies entirely within either 𝐵 or
𝐶. Without any loss of generality, we may assume 𝐴1 ⊂ 𝐵. Note that given 𝑛, if 𝐴𝑛 ⊂ 𝐵 then
𝐴𝑛+1 ⊂ 𝐵, for if 𝐴𝑛+1 ⊂ 𝐶 then 𝐴𝑛 ∩ 𝐴𝑛+1 ⊂ 𝐵 ∩ 𝐶 = ∅, in contradiction with the assumption. By
Ð
Ð
induction, 𝐴𝑛 ⊂ 𝐵 for all 𝑛 ∈ Z+, so that 𝑛 𝐴𝑛 ⊂ 𝐵. It follows that 𝑛 𝐴𝑛 is connected.
1
23
Connected Spaces
Exercise 23.3. For each 𝛼 we have 𝐴 ∩ 𝐴𝛼 ≠ ∅, so each 𝐴 ∪ 𝐴𝛼 is connected by Theorem 23.3.
In turn {𝐴 ∪ 𝐴𝛼 }𝛼 is a collection of connected spaces that have a point in common (namely any
Ð
Ð
point in 𝐴), so 𝛼 (𝐴 ∪ 𝐴𝛼 ) = 𝐴 ∪ ( 𝛼 𝐴𝛼 ) is connected by Theorem 23.3.
Exercise 23.4. Suppose that 𝐴 is a non-empty subset of 𝑋 that is both open and closed, i.e., 𝐴
and 𝑋 \ 𝐴 are finite or all of 𝑋 . Since 𝐴 is non-empty, 𝑋 \ 𝐴 is finite. Thus 𝐴 cannot be finite as
𝑋 \ 𝐴 is infinite, so 𝐴 is all of 𝑋 . Therefore 𝑋 is connected.
Exercise 23.5. Let 𝑋 have the discrete topology and let 𝑌 be a subspace of 𝑋 containing at least
two different points. Let 𝑝 ∈ 𝑌 . Then {𝑝} and 𝑌 \ {𝑝} are non-empty disjoint open sets in 𝑌 whose
union is 𝑌 , so 𝑌 is not connected. It follows that a discrete space 𝑋 is totally disconnected.
The converse does not hold: 𝑋 = Q (with the standard topology) is totally disconnected
(Example 4), but its topology is not the discrete topology.
Exercise 23.6. Suppose that 𝐶 ∩Bd 𝐴 = 𝐶 ∩𝐴∩𝑋 − 𝐴 = ∅. Then 𝐶 ∩𝐴 and 𝐶 ∩ (𝑋 \𝐴) are a pair of
disjoint non-empty sets whose union is all of 𝐶, neither of which contains a limit point of the other.
Indeed, if 𝐶 ∩(𝑋 −𝐴) contains a limit point 𝑥 of 𝐶 ∩𝐴, then 𝑥 ∈ 𝐶 ∩(𝑋 −𝐴)∩𝐴 0 ⊂ 𝐶 ∩𝐴∩𝑋 − 𝐴 = ∅,
a contradiction, and similarly 𝐶 ∩ 𝐴 does not contain a limit point of 𝐶 ∩ (𝑋 − 𝐴). Then 𝐶 ∩ 𝐴 and
𝐶 ∩ (𝑋 − 𝐴) constitute a separation of 𝐶, contradicting the fact that 𝐶 is connected (Lemma 23.1).
Exercise 23.7. The space Rℓ is not connected, as (−∞, 0) and [0, +∞) are disjoint non-empty
open sets in Rℓ whose union is all of Rℓ .
Exercise 23.8. First, we show that R𝜔 is not connected in the uniform topology. Let 𝐴 and 𝐵
denote the subsets of R𝜔 consisting of all bounded and all unbounded sequences respectively.
Then 𝐴 ∪ 𝐵 = R𝜔 and 𝐴 ∩ 𝐵 = ∅. If a ∈ R𝜔 , then the set
𝑈 (a, 1) = (𝑎 1 − 1, 𝑎 1 + 1) × · · · × (𝑎𝑛 − 1, 𝑎𝑛 + 1) × · · ·
contains the ball 𝐵 𝜌 (a, 𝜀) if 𝜀 < 1, and consists entirely of bounded sequences if a ∈ 𝐴, and of
unbounded sequences if a ∈ 𝐵. Thus 𝐴 and 𝐵 are open in R𝜔 . Since they are non-empty, it follows
that R𝜔 is not connected in the uniform topology.
Exercise 23.9. This is similar to the proof of Theorem 23.6. Take 𝑐 × 𝑑 ∈ (𝑋 \ 𝐴) × (𝑌 \ 𝐵). For
each 𝑥 ∈ 𝑋 \ 𝐴, the set
𝑈𝑥 = (𝑋 × {𝑑 }) ∪ ({𝑥 } × 𝑌 )
is connected since 𝑋 × {𝑑 } and {𝑥 } × 𝑌 are connected and have the common point 𝑥 × 𝑑. Then
Ð
𝑈 = 𝑥 ∈𝑋 \𝐴 𝑈𝑥 is connected because it is the union of the connected spaces 𝑈𝑥 which have the
point 𝑐 × 𝑑 in common. Similarly, for each 𝑦 ∈ 𝑌 \ 𝐵 the set
𝑉𝑦 = (𝑋 × {𝑦}) ∪ ({𝑐} × 𝑌 )
Ð
is connected, so 𝑉 = 𝑦 ∈𝑌 \𝐵 𝑉𝑦 is connected. Thus (𝑋 × 𝑌 ) \ (𝐴 × 𝐵) = 𝑈 ∪ 𝑉 is connected since
𝑐 × 𝑑 is a common point of 𝑈 and 𝑉 .
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Connected Subspaces of the Real Line
Exercise 23.10. (a) Let 𝐾 = {𝛼 1, . . . , 𝛼𝑛 } ⊂ 𝐽 . Then the function 𝑓 : 𝑋𝐾 → 𝑋𝛼 1 × · · · × 𝑋𝛼𝑛 given
by 𝑓 (x) = (𝑥𝛼 1 , . . . , 𝑥𝛼𝑛 ) is a homeomorphism. Since the latter is a finite product of connected
spaces, it is connected, and therefore so is 𝑋𝐾 .
(b) Since a is a point in common of the collection {𝑋𝐾 }𝐾 of connected spaces 𝑋𝐾 , for 𝐾 ⊂ 𝐽
finite, it follows that 𝑌 is connected.
Î
(c) Let x = (𝑥𝛼 )𝛼 ∈ 𝑋 \ 𝑌 and let 𝑈 be a (standard) basis element for 𝛼 𝑋𝛼 containing x, so
Î
that 𝑈 = 𝛼 𝑈𝛼 where 𝑈𝛼 is open in 𝑋𝛼 for all 𝛼 and 𝑈𝛼 = 𝑋𝛼 except for finitely many indices,
say 𝛼 1, . . . , 𝛼𝑛 . Let 𝐾 = {𝛼 1, . . . , 𝛼𝑛 }. Then 𝐾 is a finite subset of 𝐽 . As x ∉ 𝑌 , there exists some
Î
index 𝛽 ∈ 𝐽 \𝐾 such that 𝑥 𝛽 ≠ 𝑎 𝛽 . Let y = (𝑦𝛼 )𝛼 ∈ 𝛼 𝑋𝛼 be such that 𝑦𝛼𝑖 = 𝑥𝛼𝑖 for all 𝑖 = 1, . . . , 𝑛,
and 𝑦𝛼 = 𝑎𝛼 for all other indices. Then y ≠ x, and y ∈ 𝑈 ∩ 𝑋𝐾 ⊂ 𝑈 ∩ 𝑌 . Since 𝑈 was arbitrary,
we have x ∈ 𝑌 0 . It follows that 𝑋 = 𝑌 . We deduce that 𝑋 is connected from (b) and Theorem 23.4.
Exercise 23.11. Suppose that 𝑈 and 𝑉 constitute a separation of 𝑋 . If 𝑦 ∈ 𝑝 (𝑈 ), then 𝑦 = 𝑝 (𝑥)
for some 𝑥 ∈ 𝑈 , so that 𝑥 ∈ 𝑝 −1 ({𝑦}). Since 𝑝 −1 ({𝑦}) is connected and 𝑥 ∈ 𝑈 ∩ 𝑝 −1 ({𝑦}), we
have 𝑝 −1 ({𝑦}) ⊂ 𝑈 . Thus 𝑝 −1 ({𝑦}) ⊂ 𝑈 for all 𝑦 ∈ 𝑝 (𝑈 ), so that 𝑝 −1 (𝑝 (𝑈 )) ⊂ 𝑈 . The inclusion
𝑈 ⊂ 𝑝 −1 (𝑝 (𝑈 )) if true for any subset and function, so we have the equality 𝑈 = 𝑝 −1 (𝑝 (𝑈 )) and
therefore 𝑈 is saturated. Similarly, 𝑉 is saturated. Since 𝑝 is a quotient map, 𝑝 (𝑈 ) and 𝑝 (𝑉 ) are
disjoint non-empty open sets in 𝑌 . But 𝑝 (𝑈 ) ∪ 𝑝 (𝑉 ) = 𝑌 as 𝑝 is surjective, so 𝑝 (𝑈 ) and 𝑝 (𝑉 )
constitute a separation of 𝑌 , contradicting the fact that 𝑌 is connected. We conclude that 𝑋 is
connected.
Exercise 23.12. Suppose that 𝐶 and 𝐷 form a separation of 𝑌 ∪ 𝐴. Since 𝑌 ⊂ 𝐶 ∪ 𝐷 and 𝑌 is
connected, 𝑌 is entirely contained in either 𝐶 or 𝐷; suppose 𝑌 ⊂ 𝐶, so that 𝐷 ⊂ 𝐴. Since 𝐴 is open
and closed in 𝑋 \ 𝑌 , there exist 𝑈 open in 𝑋 and 𝐹 closed in 𝑋 such that 𝐴 = (𝑋 \ 𝑌 ) ∩ 𝑈 = 𝑈 \ 𝑌
and 𝐴 = (𝑋 \ 𝑌 ) ∩ 𝐹 = 𝐹 \ 𝑌 . Note that 𝐷 ⊂ 𝐴 ⊂ 𝑈 and 𝐷 ⊂ 𝐴 ⊂ 𝐹 . Since 𝐷 is open in 𝑌 ∪ 𝐴 and
𝑈 ⊂ (𝑈 − 𝑌 ) ∪ 𝑌 = 𝑌 ∪ 𝐴, it follows that 𝐷 is open in 𝑈 . Similarly, 𝐷 is closed in 𝐹 . Thus 𝐷 is
a non-empty subset open and closed in 𝑋, a contradiction. Hence 𝑌 ∪ 𝐴 is connected. Similarly,
𝑌 ∪ 𝐵 is connected.
24
Connected Subspaces of the Real Line
Exercise 24.1. (a) We follow the hint. If ℎ : [0, 1] → (0, 1] is a homeomorphism, then the restriction ℎ 0 : (0, 1) → (0, 1] \ {ℎ(0), ℎ(1)} of ℎ is a homeomorphism between a connected space and a
disconnected space, a contradiction. Similarly, (0, 1) is neither homeomorphic to [0, 1] nor (0, 1].
(b) The function 𝑓 : (0, 1) → (0, 1] given by 𝑓 (𝑥) = 𝑥/2 is continuous with image (0, 1/2),
and the restriction 𝑓 0 : (0, 1) → (0, 1/2) is a homeomorphism, so 𝑓 is an imbedding. Similarly,
𝑔 : (0, 1] → (0, 1) given by 𝑔(𝑥) = 𝑥/2 is an imbedding. But we know from (a) that (0, 1) and (0, 1]
are not homeomorphic.
(c) If 𝑓 : R𝑛 → R is a homeomorphism, the restriction 𝑓 0 : R𝑛 \ {0} → R \ {𝑓 (0)} is a homeomorphism, but R𝑛 \ {0} is connected (Example 4) while R \ {𝑓 (0)} can’t be connected. So there
is no homeomorphism between R𝑛 and R if 𝑛 > 1.
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Connected Subspaces of the Real Line
Exercise 24.2. Let 𝑓 : 𝑆 1 → R be continuous. Let 𝑥 ∈ 𝑆 1 . If 𝑓 (𝑥) = 𝑓 (−𝑥) we are done, so
assume 𝑓 (𝑥) ≠ 𝑓 (−𝑥). Define 𝑔 : 𝑆 1 → R by setting 𝑔(𝑥) = 𝑓 (𝑥) − 𝑓 (−𝑥). Then 𝑔 is continuous.
Suppose 𝑓 (𝑥) > 𝑓 (−𝑥), so that 𝑔(𝑥) > 0. Then −𝑥 ∈ 𝑆 1 and 𝑔(−𝑥) < 0. By the intermediate value
theorem, since 𝑆 1 is connected and 𝑔(−𝑥) < 0 < 𝑔(𝑥), there exists 𝑦 ∈ 𝑆 1 such that 𝑔(𝑦) = 0. i.e,
𝑓 (𝑦) = 𝑓 (−𝑦). Similarly, if 𝑓 (𝑥) < 𝑓 (−𝑥), then 𝑔(𝑥) < 0 < 𝑔(−𝑥) and again the intermediate
value theorem gives the result.
Exercise 24.3. If 𝑓 (0) = 0 or 𝑓 (1) = 1 we are done, so suppose 𝑓 (0) > 0 and 𝑓 (1) < 1. Let
𝑔 : [0, 1] → [0, 1] be given by 𝑔(𝑥) = 𝑓 (𝑥) − 𝑥 . Then 𝑔 is continuous, 𝑔(0) > 0 and 𝑔(1) < 0. Since
[0, 1] is connected and 𝑔(1) < 0 < 𝑔(0), by the intermediate value theorem there exists 𝑥 ∈ (0, 1)
such that 𝑔(𝑥) = 0, that is, such that 𝑓 (𝑥) = 𝑥 .
If 𝑋 equals [0, 1) or (0, 1) this is no longer true. For example 𝑓 : (0, 1) → (0, 1) given by 𝑓 (𝑥) =
𝑥/2 is continuous with no fixed points. Similarly 𝑔 : [0, 1) → [0, 1) given by 𝑔(𝑥) = 𝑥/2 + 1/2 is
continuous with no fixed points.
Exercise 24.4. First we prove that 𝑋 has the least upper bound property. Let 𝐴 be a non-empty
subset of 𝑋 that is bounded above. Suppose that 𝐴 does not have a supremum, so that the set 𝐵
of all upper bounds for 𝐴 does not have a smallest element. Consider the subsets
Ø
Ø
𝐶=
(−∞, 𝑎) and 𝐷 =
(𝑏, +∞)
𝑎 ∈𝐴
𝑏 ∈𝐵
of 𝑋 . Note 𝐶 and 𝐷 are non-empty since 𝐴 does not have a supremum, and they are union of open
rays, so they are open in 𝑋 . They are also disjoint, for suppose 𝑥 ∈ 𝐶 ∩ 𝐷. Then 𝑥 ∈ (𝑏, +∞) for
some 𝑏 ∈ 𝐵, so 𝑥 is an upper bound for 𝐴. Also, 𝑥 ∈ (−∞, 𝑎) for some 𝑎 ∈ 𝐴, so 𝑐 < 𝑎 ≤ 𝑐, absurd.
Thus 𝐶 ∩ 𝐷 = ∅. Finally, let 𝑥 ∈ 𝑋 . Then either 𝑥 ∈ 𝐵 or 𝑥 ∈ 𝑋 \ 𝐵. If 𝑥 ∈ 𝐵, since 𝐵 does not
have a smallest element, there exist 𝑏 ∈ 𝐵 such that 𝑏 < 𝑐, so 𝑐 ∈ (𝑏, +∞) ⊂ 𝐷. If 𝑥 ∈ 𝑋 \ 𝐵, there
exists 𝑎 ∈ 𝐴 such that 𝑥 < 𝑎, so 𝑥 ∈ (−∞, 𝑎) ⊂ 𝐶. It follows that 𝐶 and 𝐷 form a separation of 𝑋,
contradicting the fact that 𝑋 is connected. Hence, 𝐴 has a supremum. Since 𝐴 was arbitrary, 𝑋
has the least upper bound property.
Now, if 𝑥 < 𝑦 and there is no element 𝑧 such that 𝑥 < 𝑧 < 𝑦, then the sets (−∞, 𝑦) and (𝑥, +∞)
form a separation of 𝑋, a contradiction. We conclude that 𝑋 is a linear continuum.
Exercise 24.5. (a) We prove that 𝐴 ≔ Z+ × [0, 1) is a linear continuum. Let 𝑥 × 𝑦, 𝑧 × 𝑤 ∈ 𝐴
𝑦+𝑤
and suppose that 𝑥 × 𝑦 < 𝑧 × 𝑤 . If 𝑥 = 𝑧 and 𝑦 < 𝑤, then 𝑥 × 𝑦 < 𝑥 × 2 < 𝑧 × 𝑤 . If 𝑥 < 𝑧,
𝑥+𝑧
then 𝑥 × 𝑦 < 2 × 0 < 𝑧 × 𝑤 . Now, if 𝑆 ⊂ 𝐴 is non-empty and bounded above, let 𝜋 1 : 𝐴 →
Z+ and 𝜋 2 : 𝐴 → [0, 1) denote the projections onto the first and second factor respectively; let
𝑥 = max 𝜋1 (𝑆) and 𝑦 = sup 𝜋 2 (𝑆 ∩ ({𝑥 } × (0, 1])). If 𝑦 ≠ 1, then 𝑥 × 𝑦 = sup 𝑆. If 𝑦 = 1, then
(𝑥 + 1) × 0 = sup 𝑆. Since 𝑆 was arbitrary, 𝐴 have the least upper bound property. We conclude
that 𝐴 = Z+ × [0, 1) is a linear continuum.
(b) Since 0 × 1 < 0 × 2 but there is no element 𝑧 in [0, 1) × Z+ such that 0 × 1 < 𝑧 < 0 × 2, it
follows that [0, 1) × Z+ is not a linear continuum.
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(c) We show that 𝐶 ≔ [0, 1) × [0, 1] is a linear continuum. Similarly as in (a), for every 𝑥, 𝑦 ∈ 𝐵
such that 𝑥 < 𝑦, there exists 𝑧 ∈ 𝐶 such that 𝑥 < 𝑧 < 𝑦. Let 𝑆 ⊂ 𝐶 be non-empty and bounded
above, and let 𝜋1 : 𝐶 → [0, 1) and 𝜋 2 : 𝐶 → [0, 1] denote the projections. Let 𝑥 = sup 𝜋 1 (𝑆). If
𝑥 ∈ 𝜋1 (𝑆), let 𝑦 = sup 𝜋 2 (𝑆 ∩ ({𝑥 } × [0, 1])); then 𝑥 × 𝑦 = sup 𝑆. If 𝑥 ∉ 𝜋1 (𝑆), then 𝑥 × 0 = sup 𝑆.
Since 𝑆 was arbitrary, 𝐶 have the least upper bound property, so 𝐶 = [0, 1) × [0, 1] is a linear
continuum.
(d) [0, 1] × [0, 1) is not a linear continuum, since for any 𝑥 ∈ [0, 1), the subset 𝑆 = {𝑥 } × [0, 1)
is bounded above but does not have a least upper bound.
Exercise 24.6. Let 𝑋 be a well-ordered set. 𝑥 ×𝑦, 𝑧 ×𝑤 ∈ 𝑋 × [0, 1); suppose that 𝑥 ×𝑦 < 𝑧 ×𝑤 . If
𝑦+𝑤
𝑦+1
𝑥 = 𝑧 and 𝑦 < 𝑤, then 𝑥 ×𝑦 < 𝑥 × 2 < 𝑧 ×𝑤 . If 𝑥 < 𝑧, then 𝑥 ×𝑦 < 𝑥 × 2 < 𝑧 ×𝑤 . We prove that
𝑋 × [0, 1) have the greatest lower bound property, which is equivalent to having the least upper
bound property. Let 𝑆 ⊂ 𝑋 × [0, 1) be non-empty and bounded below; let 𝜋1 : 𝑋 × [0, 1) → 𝑋 and
𝜋2 : 𝑋 × [0, 1) → [0, 1) denote the projections onto the first and second factor respectively. Since
𝑋 is a well-ordered set, 𝜋 1 (𝑆) has a least element 𝑥 . Then 𝑥 × 𝑦, where 𝑦 = inf 𝜋2 (𝑆 ∩ ({𝑥 } × 𝑆)),
is a greatest lower bound for 𝑆. Since 𝑆 was arbitrary, 𝑋 × [0, 1) have the greatest lower bound
property. It follows that 𝑋 × [0, 1) is a linear continuum.
Exercise 24.7. (a) Since 𝑓 is order-preserving, it is injective, hence a bijection. Let (𝑎, +∞) be an
open ray in 𝑌 , 𝑥 = 𝑓 −1 (𝑎). Then 𝑓 −1 ((𝑎, +∞)) = (𝑥, +∞), so 𝑓 −1 ((𝑎, +∞)) is open in 𝑋 . Similarly,
𝑓 −1 ((−∞, 𝑏)) = (−∞, 𝑓 −1 (𝑏)) is open in 𝑋 for each 𝑏 ∈ 𝑌 . Since open rays constitute a subbasis
for the order topology on 𝑌 , and their preimages are open in 𝑋, it follows that 𝑓 is continuous.
Similarly, 𝑓 ((𝑐, +∞)) = (𝑓 (𝑐), +∞) and 𝑓 ((−∞, 𝑑)) = (−∞, 𝑓 (𝑑)) are open in 𝑌 , so the image of
every open ray in 𝑋 is open in 𝑌 and hence 𝑓 is an open map. Therefore, 𝑓 a homeomorphism.
(b) Given 𝑦 ∈ R+, the point 𝑦 1/𝑛 ∈ R+ satisfies 𝑓 (𝑦 1/𝑛 ) = 𝑦, so 𝑓 is surjective. If 𝑥, 𝑦 ∈ R+ and
𝑥 < 𝑦, then 𝑥 2 < 𝑥𝑦 < 𝑦 2 . Inductively, 𝑥 𝑘 < 𝑦𝑘 for all 𝑘 ∈ Z+, so 𝑓 is order-preserving. By (a), 𝑓
is a homeomorphism, so its inverse, the 𝑛th root function, is continuous.
(c) Let 𝑥 ∈ R. If 𝑥 ≥ 0, then 𝑓 (𝑥) = 𝑥 . If 𝑥 < 0, then 𝑓 (𝑥 − 1) = 𝑥 . Thus 𝑓 is surjective.
Furthermore, 𝑓 is clearly order-preserving. But 𝑓 can’t be a homeomorphism, since R is connected
while (−∞, −1) ∪ [0, +∞) is not. (Recall from Theorem 16.4 that if 𝑋 have the order topology
and 𝑌 ⊂ 𝑋 is convex, then the subspace topology on 𝑌 equals the order topology on 𝑌 . Since
(−∞, −1) ∪ [0, +∞) is not convex, we do not expect that this two topologies are equal. Indeed,
[0, +∞) is open in the subspace topology on (−∞, −1) ∪ [0, +∞) while is not open in the order
topology on (−∞, −1) ∪ [0, +∞).)
Exercise 24.8. (a) Let {𝑋𝛼 }𝛼 ∈𝐽 be a collection of path-connected spaces and consider the product
Î
Î
space 𝛼 𝑋𝛼 . Let x, y ∈ 𝛼 𝑋𝛼 . For each 𝛼, there exists a continuous function 𝑓𝛼 : [0, 1] → 𝑋𝛼
Î
such that 𝑓𝛼 (0) = 𝑥𝛼 and 𝑓𝛼 (1) = 𝑦𝛼 . Let 𝑓 : [0, 1] →
𝑋𝛼 be given by the equation 𝑓 (𝑡) =
(𝑓𝛼 (𝑡))𝛼 ∈𝐽 . Then 𝑓 is continuous by Theorem 19.6. Moreover, 𝑓 (0) = x and 𝑓 (1) = y, so x and
Î
Î
y can be joined by a path in 𝛼 𝑋𝛼 . Since x, y ∈ 𝛼 𝑋𝛼 were arbitrary, we conclude that the
Î
product space 𝛼 𝑋𝛼 is path-connected.
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Connected Subspaces of the Real Line
(b) It’s not true that 𝐴 path-connected implies 𝐴 path-connected. For instance, see Example
7: the subspace 𝐴 = {𝑥 × sin(1/𝑥) | 0 < 𝑥 ≤ 1} of R2 is path-connected, but its closure 𝑆 =
𝑆 ∪ ({0} × [−1, 1]), the topologist’s sine curve, is not.
(c) If 𝑓 (𝑥), 𝑓 (𝑦) ∈ 𝑓 (𝑋 ), there exists a continuous function 𝑔 : [0, 1] → 𝑋 such that 𝑔(0) = 𝑥
and 𝑔(1). Then 𝑓 ◦ 𝑔 : [0, 1] → 𝑓 (𝑋 ) is continuous and (𝑓 ◦ 𝑔) (0) = 𝑓 (𝑥), (𝑓 ◦ 𝑔) (0) = 𝑓 (𝑦). Thus
𝑓 (𝑋 ) is path-connected.
Ñ
Ð
(d) Let 𝑧 ∈ 𝛼 𝐴𝛼 . Let 𝑥, 𝑦 ∈ 𝛼 𝐴𝛼 . Then there exist indices 𝛽, 𝛾 such that 𝑥 ∈ 𝐴𝛽 and 𝑦 ∈ 𝐴𝛾 .
Since 𝐴𝛽 is path-connected, there exist a continuous function 𝑓 : [0, 1] → 𝐴𝛽 such that 𝑓 (0) = 𝑥
and 𝑓 (1) = 𝑧. Similarly, there is a continuous function 𝑔 : [1, 2] → 𝐴𝛾 such that 𝑔(1) = 𝑧 and
𝑔(2) = 𝑦. By Theorem 18.2(e), the functions 𝑓 0 : [0, 1] → 𝐴𝛽 ∪ 𝐴𝛾 and 𝑔 0 : [1, 2] → 𝐴𝛽 ∪ 𝐴𝛾
obtained by expanding the range of 𝑓 and 𝑔 respectively are continuous. By the pasting lemma
(Theorem 18.3), since [0, 1] and [1, 2] are closed in [0, 2] and 𝑓 0 (1) = 𝑔 0 (1), we can combine 𝑓 0
and 𝑔 0 to obtain a continuous function ℎ : [0, 2] → 𝐴𝛽 ∪ 𝐴𝛾 by setting ℎ(𝑥) = 𝑓 0 (𝑥) if 𝑥 ∈ [0, 1]
Ð
and ℎ(𝑥) = 𝑔 0 (𝑥) if 𝑥 ∈ [1, 2]. Applying Theorem 18.2(e) again, the function ℎ 0 : [0, 2] → 𝛼 𝐴𝛼
obtained by expanding the range of ℎ is continuous, and furthermore, ℎ 0 (0) = 𝑥 and ℎ 0 (2) = 𝑦.
Ð
Ð
Thus 𝑥 and 𝑦 can be joined by a path in 𝛼 𝐴𝛼 . It follows that 𝛼 𝐴𝛼 is path-connected.
Exercise 24.9. We use the hint. Let 𝑥, 𝑦 ∈ R2 \ 𝐴. There are uncountable many lines passing
through 𝑥, and 𝐴 is countable, so choose one of the lines that do not intersect 𝐴. Then choose a
line passing though 𝑦 that does not intersect 𝐴 but intersects the line passing though 𝑥 . These
two lines are path-connected and have a point in common, so their union is path-connected by
Exercise 24.8(d). Thus 𝑥 and 𝑦 can be joined by a path in R2 \𝐴. Since 𝑥, 𝑦 ∈ R2 \𝐴 were arbitrary,
it follows that that R2 \ 𝐴 is path-connected.
Exercise 24.10. We follow the hint. Given 𝑥 ∈ 𝑈 , let 𝐶𝑥 be the set of points that can be joined to
𝑥 by a path in 𝑈 . If 𝑦 ∈ 𝐶𝑥 , since 𝑈 is open, there exists a ball 𝐵𝑑 (𝑦, 𝑟 ) entirely contained within
𝑈 . Then every point in 𝐵𝑑 (𝑦, 𝑟 ) can be joined to 𝑦 by a path in 𝑈 , hence can be joined to 𝑥 . So
𝑦 ∈ 𝐵𝑑 (𝑦, 𝑟 ) ⊂ 𝐶𝑥 and 𝐶𝑥 is open in 𝑈 . Similarly, if 𝑧 ∈ 𝑈 \ 𝐶𝑥 , let 𝑟 0 > 0 be so that the ball
𝐵𝑑 (𝑧, 𝑟 0) is entirely contained within 𝑈 . Then none of the points in 𝐵𝑑 (𝑧, 𝑟 0) can be joined to 𝑥
by a path in 𝑈 , for then 𝑧 could be joined to 𝑥 . So 𝑧 ∈ 𝐵𝑑 (𝑧, 𝑟 0) ⊂ 𝑈 \ 𝐶𝑥 and 𝑈 \ 𝐶𝑥 is open in
𝑈 . Thus 𝐶𝑥 is non-empty and both open and closed in 𝑈 , which is connected, so 𝐶𝑥 = 𝑈 . Since 𝑥
was arbitrary, it follows that 𝑈 is path-connected.
Exercise 24.11. Let 𝐵 2 denote the closed unit ball in R2, and 𝐶 denote the closed ball with centre
2 × 0 and radius 1. Then 1 × 0 ∈ 𝐵 2 ∩ 𝐶, so 𝐴 = 𝐵 2 ∪ 𝐶 is connected. But Int 𝐴 is not connected,
since Int 𝐵 2 and Int 𝐶 form a separation of Int 𝐴. Furthermore, 𝐴 = (0, 1) is connected while
Bd 𝐴 = {0, 1} is not.
The converse does not hold either: Bd Q = R is connected while Q is not, and Int Q ∩ (0, 1) =
(0, 1) is connected while Q ∩ (0, 1) is not.
Exercise 24.12. (a) First assume that [𝑎, 𝑐) has the order type of [0, 1). There exists an orderpreserving bijection 𝑓 : [𝑎, 𝑐) → [0, 1), so its restriction to [𝑎, 𝑏) → [0, 𝑓 (𝑏)) is an order-preserving
bijection. Thus [𝑎, 𝑏 (𝑐)) has the order type of [0, 𝑓 (𝑏)), hence of [0, 1), and similarly for [𝑏, 𝑐).
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Components and Local Connectedness
Conversely, assume both [𝑎, 𝑏) and [𝑏, 𝑐) have the order type of [0, 1). Then we have orderpreserving bijections [𝑎, 𝑏) → [0, 1/2) and [𝑏, 𝑐) → [1/2, 1) which glue to an order-preserving
bijection [𝑎, 𝑐) → [0, 1).
(b) First suppose that [𝑥 0, 𝑏) has the order type of [0, 1). Then, for given 𝑖 we have 𝑥𝑖 <
𝑥𝑖+1 < 𝑏, so by part (a) we deduce that [𝑥𝑖 , 𝑥𝑖+1 ) has the order type of (0, 1]. Conversely, suppose
that each interval [𝑥𝑖 , 𝑥𝑖+1 ) has the order type of [0,1). Then for each 𝑖 ∈ N there is an order
1
, and they all glue to an order-preserving
preserving bijection 𝑓𝑖 : [𝑥𝑖 , 𝑥𝑖+1 ) → 1 − 21𝑖 , 1 − 2𝑖+1
bijection [𝑥 0, 𝑏) → [0, 1).
(c) We follows the hint. We prove the claim by transfinite induction. The base case is when
𝑎 is the successor of 𝑎 0, and it is clear. Now let 𝑎 have a immediate predecessor 𝑏 (greater than
𝑎 0 ) in 𝑆 Ω . Then [𝑎 0 × 0, 𝑏 × 0) has the order type of [0, 1) by induction hypothesis, and so does
[𝑏 × 0, 𝑎 × 0]. It then follows from (a) that [𝑎 0 × 0, 𝑎 × 0) has the order type of [0, 1). Now suppose
that 𝑎 is not a successor in 𝑆 Ω . Then there is an increasing sequence 𝑎𝑖 in 𝑆 Ω with 𝑎 = sup𝑖 {𝑎𝑖 }.
By induction, each interval [𝑎 0 × 0, 𝑎𝑖 × 0) has the order type of (0, 1]. Then so does [𝑎 0 × 0, 𝑎 × 0)
by part (b). We deduce by the principle of transfinite induction that the interval [𝑎 0 × 0, 𝑎 × 0] of
𝑆 Ω × [0, 1) has the order type of [0, 1) for all 𝑎 in 𝑆 Ω distinct from 𝑎 0 .
(d) Let 𝑎 1 denote the successor of 𝑎 0 in 𝑆 Ω . Consider the point 𝑎 1 × 0 ∈ 𝐿. As [𝑎 0 × 0, 𝑎 1 × 0)
has the order type of [0, 1) by (c), any point in this interval can be connected to 𝑎 1 × 0 by a path
in 𝐿. Given 𝑏 ≥ 𝑎 1, every point 𝑏 × 𝑡 for 𝑡 ∈ [0, 1) can be connected to the point 𝑏 × 0, and in turn
𝑏 × 0 can be connected to 𝑎 1 × 0 by parts (a) and (b). Thus every point of 𝐿 can be joined by a path
to 𝑎 1 × 0, so we deduce that 𝐿 is path-connected.
(e) This is clear for points of the form 𝑎 0 × 𝑡, 𝑡 ∈ [0, 1). Now let 𝑏 > 𝑎 0 and 𝑡 ∈ [0, 1). Then
(𝑎 0 × 0, 𝑏 × 0) ⊂ 𝐿 is homeomorphic to (0, 1) ⊂ R by part (c) and Exercise 24.7(a), and hence so is
(𝑎 0 × 0, 𝑏 × 𝑡).
(f) We follow the hint. Suppose that 𝐿 can be imbedded in R𝑛 for some 𝑛. In particular, 𝐿 is
second countable. Let 𝑌 = {(𝑎𝑖 , 𝑡𝑖 )}𝑖 ∈N be any countable subset of 𝑋 . Then there exists 𝑏 ∈ 𝑆 Ω
such that 𝑏 > 𝑎𝑖 for all 𝑖 ∈ N. Let 𝑏 0 = 𝑏 + 1 denote the successor of 𝑏. Then {𝑏 0 } × (0, 1) is
a neighbourhood of 𝑏 0 × 21 in 𝐿 which does not intersect 𝑌 . This is a contradiction by Theorem
30.3(b). It follows that 𝐿 cannot be imbedded in any R𝑛 .
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Components and Local Connectedness
Exercise 25.1. Let 𝐴 be a non-empty connected subspace of Rℓ and let 𝑎 ∈ 𝐴. Then [𝑎, +∞) ∩𝐴 is
non-empty and both open and closed in 𝐴, so [𝑎, +∞) ∩𝐴 = 𝐴. Thus, if 𝑏 ∈ 𝐴, then 𝑏 ∉ (−∞, 𝑎), so
𝑏 ≥ 𝑎. Similarly, 𝑎 ≥ 𝑏 for any 𝑏 ∈ 𝐴, so 𝐴 = {𝑎}. Since each path component lies in a connected
component, it follows that only the singletons {𝑎} are path-connected in Rℓ .
If 𝑓 : R → Rℓ is continuous, its image 𝑓 (R) must be connected in Rℓ , so must be a singleton.
Hence the continuous maps 𝑓 : R → Rℓ are the constant maps (cf. Exercise 18.7(b)).
Exercise 25.2. (a) The product space R𝜔 is connected by Example 7 of §13, so the whole space
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Components and Local Connectedness
is the only component. Since R is path-connected, the product space R𝜔 is path-connected by
Exercise 24.8(a), so the whole space is the only path component.
(b) We use the hint. Note that given x ∈ R𝜔 , the map 𝑓 : R𝜔 → R𝜔 given by 𝑓 (y) = x − y is a
homeomorphism. Therefore, x and y lie in the same component if and only if x−y lies in the same
component of 0. We prove that this occurs exactly when x − y is bounded. Recall from Exercise
23.8 that the set 𝐴 of all bounded sequences and the set 𝐵 of all unbounded sequences constitute
a separation of R𝜔 . Thus, if x − y lies in the same component of 0, it must belong to 𝐴, i.e., x − y
must be bounded. Conversely, assume that x − y is bounded; we show that x − y lies in the same
component of 0. Define 𝑓 : [0, 1] → R𝜔 by 𝑓 (𝑡) = 𝑡 (x − y). Let 𝑀 > 0 be such that |𝑥𝑛 − 𝑦𝑛 | < 𝑀
for all 𝑛 ∈ Z+ . Given 𝜀 > 0, let 𝛿 = 𝜀/𝑀. Thus, if |𝑡 − 𝑡 0 | < 𝛿, then 𝜌 (𝑓 (𝑡), 𝑓 (𝑡 0 )) < 𝛿𝑀 = 𝜀. Hence
𝑓 is a path in R𝜔 from 0 to 𝑓 (1) = x − y. So 0 and x − y lie in the same path component, and
in particular in the same component. We conclude that x − y is bounded if and only if it lies the
same component of 0, and this occurs if and only if x and y lie in the same component.
(c) Let R∞ denote the subspace of R𝜔 in the box topology consisting of all sequences which
are eventually zero. We have to prove that x and y lie in the same component of R𝜔 in the box
topology if and only if x − y ∈ R∞ .
First assume x − y ∈ R∞ . Define 𝑓 : [0, 1] → R𝜔 by 𝑓 (𝑡) = y + 𝑡 (x − y). Since x − y ∈ R∞, we
Î
have 𝑥𝑖 −𝑦𝑖 = 0 for all but finitely many indices 𝑖 1, . . . , 𝑖𝑛 . Let 𝑡 ∈ R and let 𝑈 = 𝑖 𝑈𝑖 be a standard
basis element for R𝜔 in the box topology containing 𝑓 (𝑡). For each 𝑗 = 1, . . . , 𝑛, let 𝜀 𝑗 > 0 be such
that the ball centred at the 𝑖 𝑗 -entry of 𝑓 (𝑡) and having radius 𝜀 𝑗 is entirely contained within 𝑈𝑖 𝑗 .
Let 𝜀 = min{𝜀 𝑗 }. Then 𝑉 = 𝑊 ∩ [0, 1], where 𝑊 = (𝑡 − 𝜀, 𝑡 + 𝜀) is a neighbourhood of 𝑡 in [0, 1]
and satisfies 𝑓 (𝑉 ) ⊂ 𝑈 . Therefore 𝑓 is continuous and hence a path in R𝜔 joining x and y.
Conversely, assume that x − y ∉ R∞ . Then there are infinitely many indices 𝑖 ∈ Z+ such that
𝑥𝑖 ≠ 𝑦𝑖 . Define ℎ(z) : R𝜔 → R𝜔 by the equation
(
𝑧𝑖 − 𝑥 𝑖 ,
if 𝑥𝑖 = 𝑦𝑖 ,
ℎ(z) = (𝑤 1, 𝑤 2, . . .), where 𝑤𝑖 =
𝑖
if 𝑥𝑖 ≠ 𝑦𝑖 .
|𝑥𝑖 −𝑦𝑖 | (𝑧𝑖 − 𝑥𝑖 ),
Then ℎ(x) = 0 is bounded. But ℎ(y) is not bounded since 𝑥𝑖 ≠ 𝑦𝑖 for infinitely many indices.
Furthermore, ℎ is a homeomorphism of R𝜔 onto itself by Exercise 19.8. Hence x and y do not lie
in the same component of R∞ .
Exercise 25.3. Let 𝐼 o2 denote the ordered square. If 𝑥 × 𝑦 ∈ 𝐼 o2 and 𝑈 is a neighbourhood of
𝑥 × 𝑦, then 𝑈 contains a basis interval. By Theorem 24.1, since 𝐼 o2 is a linear continuum, this basis
interval is a connected neighbourhood of 𝑥 × 𝑦 contained in 𝑈 . Hence 𝐼 o2 is locally connected.
Now we prove that 𝐼 o2 is not locally path-connected. Let 0 < 𝑝 < 1 and consider the point
𝑝 × 1 ∈ 𝐼 o2 . Let 𝑈 be a neighbourhood of 𝑝 × 0; suppose that 𝑈 contains a path-connected neighbourhood 𝑉 of 𝑝 × 1. Then 𝑉 contains a point 𝑞 × 0, where 𝑞 > 𝑝. Similarly as in Example 6
of §24, let 𝑓 : [0, 1] → 𝑉 be a path joining 𝑝 × 1 and 𝑞 × 0. By the intermediate value theorem.
𝑓 ([0, 1]) contains every point 𝑥 × 𝑦 between 𝑝 × 1 and 𝑞 × 0. Thus, for each 𝑝 < 𝑥 < 𝑞, the
set 𝑈𝑥 = 𝑓 −1 ({𝑥 } × (0, 1)) is non-empty and open in [0, 1], so we can choose a rational point
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𝑞𝑥 ∈ 𝑈𝑥 . Then the map 𝑥 ↦→ 𝑞𝑥 is an injective mapping from (𝑝, 𝑞) to Q, but the interval (𝑝, 𝑞) is
uncountable while Q is countable, a contradiction. Therefore, 𝐼 o2 is not locally path-connected at
the points 𝑝 × 1 with 0 < 𝑝 < 1. Similarly, 𝐼 o2 is not locally path-connected at the points 𝑞 × 0 with
0 < 𝑞 < 1. Hence the path components are the segments {𝑥 } × [0, 1] for 𝑥 ∈ [0, 1].
Exercise 25.4. Let 𝑈 be a open connected set in 𝑋 . By Theorem 25.4, each path component of 𝑈
is open in 𝑋, hence open in 𝑈 . Thus, each path component in 𝑈 is both open and closed in 𝑈 , so
must be empty or all of 𝑈 . It follows that 𝑈 is path-connected.
Exercise 25.5. (a) Each line segment joining 𝑝 to a point of 𝑋 is path-connected. Hence 𝑇 is
union of path-connected spaces with non-empty intersection, so 𝑇 is path-connected by Exercise
24.8(d).
The only connected open sets in 𝑋 are singletons, so they are all path-connected.
(b) In (a), 𝑇 is locally connected precisely at 𝑝: if 𝑞 ∈ 𝑇 \ {𝑝}, then any small neighbourhood
of 𝑞 intersects infinitely many lines, and has infinitely many components. We can extend the
example. Let 𝑌 = Q ∩ ( [0, 1] × {1}). Let 𝑅 be the union of all line segments joining the point 1 × 0
to points in 𝑌 . Similarly to (a), 𝑅 is path-connected. Then 𝑅 ∪𝑇 is path-connected as 1 × 0 ∈ 𝑅 ∩𝑇 .
But 𝑅 ∪ 𝑇 is locally connected at none of its points, as now every small neighbourhood of any
point, including 𝑝 and 1 × 0, has infinitely many components.
Exercise 25.6. We use the hint. Let 𝑈 be open in 𝑋 and 𝐶 be a component of 𝑈 . If 𝑥 ∈ 𝐶, there
is a connected subspace 𝐶𝑥 of 𝑋 contained in 𝑈 that contains a neighbourhood 𝑉𝑥 of 𝑥 . Since
𝐶𝑥 is connected, it must lie entirely within the component 𝐶 of 𝑈 . Then 𝑥 ∈ 𝑉𝑥 ⊂ 𝐶𝑥 ⊂ 𝐶, and
Ð
therefore 𝐶 = 𝑥 ∈𝐶 𝑉𝑥 is open in 𝑋 . Since 𝐶 was arbitrary, each component of 𝑈 is open in 𝑋 ;
since 𝑈 was arbitrary, it follows from Theorem 25.3 that 𝑋 .
Exercise 25.7. We use the hint. We describe the set 𝑋 . Without any loss of generality, we may
assume that 𝑝 is the origin of R2, 𝑝 = (0, 0). There is a sequence {(𝑎𝑖 , 0)} of points of R2 that lies
in the 𝑥-axis, so that 𝑎𝑖+1 < 𝑎𝑖 for all 𝑖 ∈ Z+, and such that (𝑎𝑖 , 0) → 𝑝 = (0, 0) when 𝑖 → ∞. The
line segment 𝐿 from 𝑝 to (𝑎 1, 0) is in 𝑋, so 𝑋 contains all points (𝑎𝑖 , 0). From each (𝑎𝑖 , 0) there are
emanating infinite lines 𝐿𝑖,𝑗 with endpoint (𝑎𝑖+1, 𝑟 𝑗 ) for some 𝑟 𝑗 > 0, and such that 𝑟 𝑗+1 < 𝑟 𝑗 for
all 𝑗 and 𝑟 𝑗 → 0 when 𝑗 → ∞. The space 𝑋 is the union of all this lines 𝐿𝑖,𝑗 and 𝐿.
First, we prove that 𝑋 is not locally connected at 𝑝. We show that any connected neighbourhood 𝑊 of 𝑝 must contain all the points (𝑎𝑖 , 0). Since 𝑊 is open, 𝑊 contains all but finitely many
(𝑎𝑖 , 0). Let 𝑁 ∈ Z+ be such that (𝑎𝑖 , 0) ∈ 𝑊 for all 𝑖 ≥ 𝑁 and (𝑎 𝑁 −1, 0) ∉ 𝑊 . Since 𝑊 is open
and contains (𝑎 𝑁 , 0), there is a ball 𝐵 ≔ 𝐵𝑑 ((𝑎 𝑁 , 0), 𝜀) in R2 such that (𝑎 𝑁 , 0) ∈ 𝐵 ∩ 𝑋 ⊂ 𝑊 .
Then 𝐵 ∩ 𝑋 intersects all but finitely many of the lines 𝐿𝑖,𝑗 . Each of these lines is connected and
intersects 𝑊 , so must lie entirely within 𝑊 . But then (𝑎 𝑁 −1, 0) ∈ 𝑊 , contradicting the choice of
𝑁 . Thus (𝑎𝑖 , 0) ∈ 𝑊 for all 𝑖 ∈ Z+ . Therefore, if 𝑈 is a neighbourhood of 𝑝 that does not contain all
(𝑎𝑖 , 0), then there is no connected neighbourhood of 𝑝 contained in 𝑈 . For example, 𝐵𝑑 (𝑝, 𝑟 ) ∩ 𝑋
where 𝑟 < 𝑑 (𝑝, (𝑎 1, 0)), is a neighbourhood of 𝑝 that contains no connected neighbourhood of 𝑝.
Hence 𝑋 is not locally connected at 𝑝.
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25
Components and Local Connectedness
Now we prove that 𝑋 is weakly locally connected at 𝑝. Let 𝑈 be a neighbourhood of 𝑝. Since
𝑈 is open in 𝑋, there exists an open set 𝑉 in R2 such that 𝑈 = 𝑋 ∩ 𝑉 . Then 𝑉 contains an open
ball 𝐵 ≔ 𝐵𝑑 (𝑝, 𝑟 ) for some 𝑟 > 0. Since 𝑎𝑖 → 0, 𝐵 must contain all but finitely many of the (𝑎𝑖 , 0).
Let 𝑁 ∈ Z+ be such that (𝑎𝑖 , 0) ∈ 𝐿𝑖,𝑗 ⊂ 𝐵 for all 𝑖 ≥ 𝑁 and all 𝑗 ∈ Z+, and such that (𝑎 𝑁 −1, 0) ∉ 𝐵.
Then
Ø
Ø 𝐿𝑖,𝑗 ∪ (𝐿 ∩ 𝑈 ) for each 𝑖,
𝐶=
𝐶𝑖 , where 𝐶𝑖 =
𝑖 ≥𝑁
𝑗 ∈Z+
is a contained in 𝐵, hence contained in 𝑈 . Furthermore, each 𝐶𝑖 is path-connected being a union
of path-connected spaces with the common point (𝑎𝑖 , 0), so 𝐶 is path-connected since 𝑝 ∈ 𝐶𝑖 for
all 𝑖. In particular, 𝐶 is connected. Moreover, 𝐶 contains the set 𝐵𝑑 (𝑝, 𝛿) ∩ 𝑋 for any 𝛿 < 𝑎 𝑁 , so 𝐶
contains a neighbourhood of 𝑝. Since 𝑈 was arbitrary, 𝑋 is weakly locally connected at 𝑝.
Exercise 25.8. We follow the hint. Let 𝑈 be open in 𝑌 and 𝐶 be a component of 𝑈 . We prove
that 𝐶 is open. If 𝐷 is a component of 𝑝 −1 (𝑈 ) that intersects 𝑝 −1 (𝐶), then 𝑝 (𝐷) is connected and
intersects 𝐶, so must lie entirely in 𝐶. Then 𝐷 ⊂ 𝑝 −1 (𝐶). This proves that 𝑝 −1 (𝐶) is union of
components of 𝑝 −1 (𝑈 ). Since 𝑋 is locally connected and 𝑝 −1 (𝑈 ) is open in 𝑋, each component of
𝑝 −1 (𝑈 ) is open in 𝑋 by Theorem 25.3, hence 𝑝 −1 (𝐶) is open in 𝑋 . Since 𝑝 is an quotient map, 𝐶 is
open in 𝑌 . Therefore, each component of 𝑈 is open in 𝑌 . It follows from Theorem 25.3 that 𝑌 is
locally connected.
Exercise 25.9. We use the hint. By Exercise 4 of Supplementary Exercises of Chapter 2, given
𝑥 ∈ 𝐺, the maps 𝑦 ↦→ 𝑥𝑦 and 𝑦 ↦→ 𝑦𝑥 are homeomorphisms of 𝐺 onto itself. Since 𝐶 is a
component, 𝑥𝐶 and 𝐶𝑥 are both components that contain 𝑥, so they are equal. Hence 𝑥𝐶 = 𝐶𝑥
for all 𝑥 ∈ 𝐺, so 𝐶 is a normal subgroup of 𝐺 .
Exercise 25.10. (a) The relation is clearly reflexive and symmetric. Now suppose 𝑥 ∼ 𝑦 and
𝑦 ∼ 𝑧. If 𝑥 𝑧, then there exists a separation 𝑋 = 𝐴 ∪ 𝐵 of 𝑋 into disjoints open sets such that
𝑥 ∈ 𝐴 and 𝑧 ∈ 𝐵. If 𝑦 ∈ 𝐴, then 𝑦 𝑧, in contradicting with the assumption. If 𝑦 ∈ 𝐵, then 𝑦 𝑥
and again we have a contradiction. Thus 𝑥 ∼ 𝑧 and the relation is transitive, hence an equivalence
relation.
(b) Let 𝐶 be a component of 𝑋 . Let 𝑥, 𝑦 ∈ 𝐶 and suppose 𝑥 𝑦. Then there is a separation
𝑋 = 𝐴 ∪ 𝐵 of 𝑋 into disjoint open sets such that 𝑥 ∈ 𝐴 and 𝑦 ∈ 𝐵. Then 𝐶 ∩ 𝐴 and 𝐶 ∩ 𝐵 form a
separation of 𝐶, contradicting the fact that 𝐶 is connected. Hence 𝑥 ∼ 𝑦 for all 𝑥, 𝑦 ∈ 𝐶 and thus
𝐶 lies within a quasicomponent of 𝑋 .
Now suppose that 𝑋 is locally connected. Let 𝐶 be a component of 𝑋 . If 𝑥 ∈ 𝐶, there exists
a connected neighbourhood 𝑈 of 𝑥 in 𝑋 such that 𝑥 ∈ 𝑈 . Since 𝑈 is connected and 𝐶 ∩ 𝑈 ≠ ∅,
then 𝑥 ∈ 𝑈 ⊂ 𝐶. This proves that 𝐶 is open in 𝑋 . Since 𝐶 is connected, we have 𝐶 = 𝐶 and hence
𝐶 is also closed in 𝑋 . Hence, the components of 𝑋 are both open and closed in 𝑋 . Now let 𝑄 be
a quasicomponent of 𝑋 ; we show that 𝑄 is connected. If 𝑥 ∈ 𝑄, there exists a component 𝐶 of 𝑋
such that 𝑥 ∈ 𝐶. This component must lie within a quasicomponent, so 𝐶 ⊂ 𝑄. If 𝑦 ∈ 𝑄, then 𝑦
lies in 𝐶 as well, for if 𝑦 ∈ 𝑋 \𝐶, then 𝐶 and 𝑋 \𝐶 would constitute a separation of 𝑋 into disjoint
open sets such that 𝑥 ∈ 𝐶 and 𝑦 ∈ 𝑋 \ 𝐶. Thus 𝑦 ∈ 𝐶 for all 𝑦 ∈ 𝑄, which proves 𝑄 = 𝐶. Hence
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Components and Local Connectedness
𝑄 is connected. This proves that the components and quasicomponents of 𝑋 are the same if 𝑋 is
locally connected.
(c) First consider 𝐴 = (𝐾 × [0, 1]) ∪ {0 × 0} ∪ {0 × 1}. Let 𝑇 = {0 × 0} ∪ {0 × 1}. For each
𝑛 ∈ Z+, let 𝑉𝑛 denote the vertical line 𝑉𝑛 = {1/𝑛} × [0, 1]. We show that the path components and
components of 𝐴 are {0 × 0}, {0 × 1} and each 𝑉𝑛 for 𝑛 ∈ Z+ . We show that the quasicomponents
1
of 𝐴 are each 𝑉𝑛 and 𝑇 . If 𝑥 ∈ 𝑉𝑛 , 𝑦 ∈ 𝑉𝑚 and 𝑛 < 𝑚, let 𝑡 ∈ [0, 1] be such that 𝑚1 < 𝑡 < 𝑚−1
. Then
𝐴 = 𝐷 ∪ 𝐸, where 𝐷 = ((−∞, 𝑡) × R) ∩𝐴 and 𝐸 = ((𝑡, +∞) × R) ∩𝐴 are disjoint open sets in 𝐴 such
1
that 𝑥 ∈ 𝐷 and 𝑦 ∈ 𝐸, so 𝑥 𝑦. Similarly, if 𝑥 ∈ 𝑇 and 𝑦 ∈ 𝑉𝑛 for some 𝑛, choose 𝑚1 < 𝑡 < 𝑚−1
;
then 𝐴 = 𝐷 ∪ 𝐸 where 𝐷 = ((−∞, 𝑡) × R) ∩ 𝐴 and 𝐸 = ((𝑡, +∞) × R) ∩ 𝐴 are disjoint open sets in 𝑋
such that 𝑥 ∈ 𝐷 and 𝑦 ∈ 𝐸, so 𝑥 𝑦. So, each quasicomponent of 𝐴 must lie within 𝑇 or some 𝑉𝑛 .
Since each 𝑉𝑛 is path-connected, is a path component, a component and a quasicomponent. Now
𝑇 splits into two more components (hence path components): {0 × 0} and {0 × 1}. We prove that
𝑇 is a quasicomponent. Suppose there is a separation 𝐴 = 𝐷 ∪ 𝐸 of 𝐴 into disjoint open sets in 𝐴
such that 0 × 0 ∈ 𝐷 and 0 × 1 ∈ 𝐸. Then 𝐷 and 𝐸 must intercept some common 𝑉𝑛 , so 𝐷 ∩ 𝑉𝑛 and
𝐸 ∩ 𝑉𝑛 is a separation of 𝑉𝑛 , contradicting the fact that 𝑉𝑛 is connected. Hence there is no such
separation, and 𝑇 is one single quasicomponent of 𝐴.
Now consider 𝐵 = 𝐴∪( [0, 1]×{0}). For each 𝑛 ∈ Z+, 𝑉𝑛 = {1/𝑛}×[0, 1] intersects ( [0, 1]×{0}).
Since these two subspaces are path-connected, their union 𝑇𝑛 path-connected. Since 𝐵 \ {0 × 1}
equals the union of the 𝑇𝑛 and 0 × 0 ∈ 𝑇𝑛 for all 𝑛, then 𝐵 \ {0 × 1} is path-connected, hence
connected. Since the closure of 𝐵 \ {0 × 1} is 𝐵, it follows that 𝐵 is connected, so it has only
one component and one quasicomponent, namely itself. We now prove that there are two path
components of 𝐵, namely 𝐵 \ {0 × 1} and {0 × 1}. Suppose that 0 × 1 and 0 × 0 can be joined
by a path in 𝐵, so that there is a continuous function 𝑓 : [0, 1] → 𝐵 such that 𝑓 (0) = 0 × 0 and
𝑓 (1) = 0 × 1. Let 𝑓1 = 𝜋 ◦ 𝑓 be the first coordinate function of 𝑓 . Then 𝑓1 is continuous and
𝑓1 (0) = 0, 𝑓1 (1) = 1, so there exists 𝑡 0 ∈ (0, 1) such that 𝑓 (𝑡 0 ) = 1/2 and 𝑓1 (𝑡) ≥ 1/2 for all 𝑡 ≥ 𝑡 0 .
Let 𝐶 = 𝐵 ∩ (R × [1/2, 1]) \ {0 × 1}. Then the restriction 𝑓 0 : [𝑡 0, 1] → 𝐶 of 𝑓 is continuous, so
0 × 1 can be joined by a path in 𝐶 to the point 𝑓 (𝑡 0 ) ∈ 𝐶 ⊂ 𝐴 \ {0 × 1}, which is impossible since
{0 × 1} is a path component of 𝐴. Therefore, there is no path between 0 × 0 and 0 × 1, and we
deduce that 𝐵 − {0 × 1} and {0 × 1} are the two path components of 𝐵.
Finally, consider 𝐶 = (𝐾 × [0, 1])∪(−𝐾 × [−1, 0])∪( [0, 1] ×−𝐾)∪( [−1, 0] ×𝐾). For each 𝑛 ∈ Z+,
let 𝑈𝑛 = [−1, 0] × {1/𝑛}, 𝑉𝑛 = {1/𝑛} × [0, 1], 𝑊𝑛 = [0, 1] × {−1/𝑛, }, and 𝑍𝑛 = {−1/𝑛} × [−1, 0].
Ð
Thus 𝐶 = 𝑛 (𝑈𝑛 ∪ 𝑉𝑛 ∪ 𝑊𝑛 ∪ 𝑍𝑛 ). The path components of 𝐶 are each of the sets 𝑈𝑛 , 𝑉𝑛 , 𝑊𝑛 , 𝑍𝑛 .
We prove that 𝐶 is connected. Suppose it is not, so that there is a separation 𝐶 = 𝐷 ∪ 𝐸 of 𝐶
into two non-empty disjoint open sets. Fix 𝑁 ∈ Z+ and consider the point 0 × (1/𝑁 ) ∈ 𝑈 𝑁 . We
can assume that 0 × (1/𝑁 ) ∈ 𝐷. Since 𝑈 𝑁 is (path) connected, 𝑈 𝑁 ⊂ 𝐷. Now, since 0 × (1/𝑁 ) is
Ð
a limit point of 𝑛 𝑉𝑛 and 𝐷 does not contain any limit point of 𝐸, all but finitely many 𝑉𝑛 are
contained in 𝐷. Let 𝑉 denote the union of those 𝑉𝑛 contained in 𝐷. Then each point 0 × (1/𝑛)
Ð
is a limit point of 𝑉 , so 0 × (1/𝑛) ∈ 𝐷 for all 𝑛, and hence 𝑛 𝑈𝑛 ⊂ 𝐷. Similarly, each point
Ð
Ð
(−1/𝑛) × 0 ∈ 𝑍𝑛 is a limit point of 𝑛 𝑈𝑛 , so 𝑛 𝑍𝑛 ⊂ 𝐷. Continuing this argument, it follows
Ð
that 𝐶 = 𝑛 (𝑈𝑛 ∪ 𝑉𝑛 ∪ 𝑊𝑛 ∪ 𝑍𝑛 ) ⊂ 𝐷, contradicting the fact that 𝐸 is non-empty. Thus 𝐶 is
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Solutions by positrón0802
26
Compact Spaces
connected, so it has only one component and one quasicomponent, namely itself.
26
Compact Spaces
Exercise 26.1. (a) If (𝑋, 𝒯 0) is compact, then (𝑋, 𝒯) is compact. The converse is not true in
general.
(b) Suppose that 𝒯 ⊂ 𝒯 0 . Let 𝐴 ∈ 𝒯 0 . Then 𝑋 \ 𝐴 is closed in (𝑋, 𝒯 0), so it is compact by
Theorem 26.2. Since 𝒯 ⊂ 𝒯 0, by (a) we have that 𝑋 \ 𝐴 is compact in (𝑋, 𝒯), hence closed in
(𝑋, 𝒯) by Theorem 26.3. Thus 𝐴 ∈ 𝒯, and hence 𝒯 = 𝒯 0 . This proves the claim.
Exercise 26.2. (a) Let 𝐵 be a subspace of R in the finite complement topology, and let 𝒜 be a
covering of 𝐵 by open sets in R. Fix 𝐴 ∈ 𝒜. Then R \ 𝐴 is finite, say R \ 𝐴 = {𝑐 1, . . . , 𝑐𝑛 }. For each
𝑖 = 1, . . . , 𝑛 such that 𝑐𝑖 ∈ 𝐵, choose an element of 𝒜 containing 𝑐𝑖 . Then these points 𝑐𝑖 and 𝐴
constitute a finite subcollection of 𝒜 that also covers 𝐵, so 𝐵 is compact by Lemma 26.1. Since 𝐵
was arbitrary, it follows that every subspace of R in the finite complement topology is compact.
(b) We show that [0, 1] is not compact in this topology. For each 𝑛 ∈ Z+, let 𝐵𝑛 = {1/𝑘 | 𝑘 ≥ 𝑛},
and let 𝐴𝑛 = [0, 1] \ 𝐵𝑛 . Note that, for each 𝑛, 𝐴𝑛 ⊂ 𝐴𝑛+1 . The collection 𝒜 = {𝐴𝑛 | 𝑛 ∈ Z+ } is an
open covering of [0, 1]; suppose that 𝒜 contains a finite subcollection that covers [0, 1]. Let 𝑁 be
the greatest positive number such that 𝐴𝑁 belongs to this subcollection. Then 1/𝑁 ∉ 𝐴𝑁 , so no
set of this subcollection contains 1/𝑁 , contradicting the fact that this subcollection covers [0, 1].
Therefore 𝒜 is an open covering of [0, 1] with no finite subcovering, so [0, 1] is not compact in
this topology.
Ð
Exercise 26.3. Let 𝑌1, . . . , 𝑌𝑛 be compact subspaces of 𝑋 . Let 𝒜 be a covering of 𝑛𝑖=1 𝑌𝑖 by sets
open in 𝑋 . For each 𝑖 = 1, . . . , 𝑛, 𝒜 is a covering of 𝑌𝑖 by sets open in 𝑋, so by Theorem 26.1, there
Ð
is a finite subcollection 𝒜𝑖 ≔ {𝐴𝑖1, . . . , 𝐴𝑛𝑖 𝑖 } of 𝒜 covering 𝑌𝑖 . Then 𝑖 𝒜𝑖 is a finite subcollection
Ð
Ð
of 𝒜 covering 𝑛𝑖=1 𝑌𝑖 , so 𝑛𝑖=1 𝑌𝑖 is compact by Theorem 26.1.
Exercise 26.4. Let 𝑌 be a compact subspace of a metric space with metric 𝑑. Since any metric
space is Hausdorff, 𝑌 is closed by Theorem 26.3. Fix 𝑦 ∈ 𝑌 . For each 𝑛 ∈ Z+, let 𝐵𝑛 ≔ 𝐵𝑑 (𝑦, 𝑛) be
the open ball centred at 𝑦 having radius 𝑛. Then the collection 𝒜 = {𝐵𝑛 }𝑛 ∈Z+ is an open covering
of 𝑌, so there is a finite subcollection of 𝒜 covering 𝑌 . If 𝑁 is the greatest positive number such
that 𝐵 𝑁 belongs to this subcollection, then 𝑌 ⊂ 𝐵 𝑁 , so 𝑌 is bounded.
Now, let 𝑋 be an infinite space with the discrete topology. Then 𝑋 is a metric space with the
discrete metric 𝑑. Now 𝑋 is closed in itself, and it is bounded since 𝑋 ⊂ 𝐵𝑑 (𝑥, 2) for any 𝑥 ∈ 𝑋 .
But 𝑋 is not compact, for the collection of singletons {𝑥 }, 𝑥 ∈ 𝑋, constitute an open covering
with no finite subcovering.
Exercise 26.5. For each 𝑥 ∈ 𝐴, by Lemma 26.4 there exist disjoint open sets 𝑈𝑥 and 𝑉𝑥 of 𝑋
containing 𝑥 and 𝐵, respectively. Then {𝑈𝑥 }𝑥 ∈𝐴 is an open covering of 𝐴, so there exist 𝑥 1, . . . , 𝑥𝑛 ∈
𝑛 also covers 𝐴. The corresponding 𝑉 are open in 𝑋 and contain 𝐵, so 𝑉 =
𝐴 such that {𝑈𝑥𝑖 }𝑖=1
𝑥𝑖
Ñ𝑛
Ð𝑛
𝑉
is
open
in
𝑋
and
contains
𝐵.
Let
𝑈
=
𝑈
.
Then
𝑈 and 𝑉 are open sets containing
𝑥
𝑥
𝑖
𝑖=1 𝑖
𝑖=1
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26
Compact Spaces
𝐴 and 𝐵 respectively. Furthermore, they are disjoint, for if 𝑐 ∈ 𝑈 ∩ 𝑉 , then 𝑐 ∈ 𝑈𝑥𝑖 ∩ 𝑉𝑥𝑖 = ∅ for
some 𝑖, a contradiction.
Exercise 26.6. Let 𝐶 be closed in 𝑋 . Then 𝐶 is compact by Theorem 26.2, so 𝑓 (𝐶) is compact by
Theorem 26.5. Thus 𝑓 (𝐶) is closed by Theorem 26.3, so 𝑓 is a closed map.
Exercise 26.7. Let 𝐶 be closed in 𝑋 × 𝑌 . Let 𝑥 0 ∈ 𝑋 \ 𝜋 1 (𝐶). Then the slice 𝑥 0 × 𝑌 of 𝑋 × 𝑌 is
contained in the open set (𝑋 ×𝑌 ) \𝐶 of 𝑋 ×𝑌 . Thus, by the tube lemma (Lemma 26.8), (𝑋 ×𝑌 ) \𝐶
contains some tube 𝑊 × 𝑌 about 𝑥 0 × 𝑌 , where 𝑊 is a neighbourhood of 𝑥 0 in 𝑋 . If 𝑥 ∈ 𝑊 , then
𝑥 ×𝑦 ∈ 𝑊 ×𝑌 for all 𝑦 ∈ 𝑌 , so 𝑥 ×𝑦 ∉ 𝐶 for all 𝑦 ∈ 𝑌 and hence 𝑥 ∉ 𝜋1 (𝐶). Then 𝑥 0 ∈ 𝑊 ⊂ 𝑋 \𝜋 1 (𝐶),
so 𝑋 \ 𝜋 1 (𝐶) is open in 𝑋 and 𝜋1 (𝐶) is closed in 𝑋 . Since 𝐶 was arbitrary, it follows that 𝜋 1 is a
closed map.
Exercise 26.8. First suppose that 𝑓 is continuous. We prove that (𝑋 × 𝑌 ) \ 𝐺 𝑓 is open in 𝑋 × 𝑌 .
Let 𝑥 0 × 𝑦0 ∈ (𝑋 × 𝑌 ) \ 𝐺 𝑓 , so that 𝑓 (𝑥 0 ) ≠ 𝑦0 . Since 𝑌 is Hausdorff, there exist disjoint open
sets 𝑈 and 𝑉 in 𝑌 such that 𝑓 (𝑥 0 ) ∈ 𝑈 and 𝑦0 ∈ 𝑉 . Then 𝑓 −1 (𝑈 ) × 𝑉 is a neighbourhood of
𝑥 0 × 𝑦0 . Furthermore, it does not intercept 𝐺 𝑓 , for if 𝑥 × 𝑦 ∈ (𝑓 −1 (𝑈 ) × 𝑉 ) ∩ 𝐺 𝑓 , then 𝑓 (𝑥) = 𝑦
and 𝑓 (𝑥) × 𝑦 ∈ 𝑈 ∩ 𝑉 = ∅, absurd. Thus 𝑥 0 × 𝑦0 ∈ 𝑓 −1 (𝑈 ) × 𝑉 ⊂ (𝑋 × 𝑌 ) \ 𝐺 𝑓 , so (𝑋 × 𝑌 ) \ 𝐺 𝑓 is
open in 𝑋 × 𝑌 and hence 𝐺 𝑓 is closed in 𝑋 × 𝑌 .
Conversely, assume that 𝐺 𝑓 is closed in 𝑋 × 𝑌 . We use the hint. Let 𝑥 0 ∈ 𝑋 and let 𝑉 be a
neighbourhood of 𝑓 (𝑥 0 ). Then 𝐶 ≔ 𝐺 𝑓 ∩ (𝑋 × (𝑌 \ 𝑉 )) is closed in 𝑋 × 𝑌 , so by Exercise 26.7,
since 𝑌 is compact, 𝜋 1 (𝐶) is closed in 𝑋 . We claim that 𝑈 ≔ 𝑋 \ 𝜋1 (𝐶) is a neighbourhood of 𝑥 0
such that 𝑓 (𝑈 ) ⊂ 𝑉 . Indeed, 𝑥 0 ∈ 𝑈 since 𝑓 (𝑥 0 ) ∉ 𝑌 \ 𝑉 . Let 𝑥 ∈ 𝑈 and suppose that 𝑓 (𝑥) ∉ 𝑉 .
Then 𝑥 × 𝑓 (𝑥) ∈ 𝐺 𝑓 ∩ (𝑋 × (𝑌 \ 𝑉 )) = 𝐶, so 𝜋1 (𝑥 × 𝑓 (𝑥)) = 𝑥 ∈ 𝜋1 (𝐶), contradicting the fact that
𝑥 ∈ 𝑈 . Therefore, 𝑈 is a neighbourhood of 𝑥 0 such that 𝑓 (𝑈 ) ⊂ 𝑉 , so 𝑓 is continuous.
Exercise 26.9. Let 𝑎 ∈ 𝐴 and consider the slice 𝑎 × 𝐵. Then 𝑎 × 𝐵 ⊂ 𝑁 . Cover 𝑎 × 𝐵 by standard
basis elements 𝑊 × 𝑍 for 𝑋 × 𝑌 lying in 𝑁 . Since 𝑎 × 𝐵 is compact, there are finitely many such
basis elements 𝑊1 × 𝑍 1, . . . ,𝑊𝑛 × 𝑍𝑛 that also cover 𝑎 × 𝐵, such that 𝑊𝑖 × 𝑍𝑖 actually intersects
𝑎 × 𝐵 for each 𝑖. Let
𝑛
𝑛
Ù
Ø
𝑈𝑎 =
𝑊𝑖 and 𝑉𝑎 =
𝑍𝑖 .
𝑖=1
𝑖=1
Then 𝑈𝑎 is a neighbourhood of 𝑎 and 𝑉𝑎 is open in 𝑌 and contains 𝐵. Furthermore, 𝑈𝑎 × 𝑉𝑎 ⊂ 𝑁 .
Indeed, if 𝑥 × 𝑦 ∈ 𝑈𝑎 × 𝑉𝑎 , then there exists 𝑖 ∈ {1, . . . , 𝑛} such that 𝑦 ∈ 𝑍𝑖 , and hence 𝑥 × 𝑦 ∈
𝑊𝑖 ×𝑍𝑖 ⊂ 𝑁 . Now, each 𝑈𝑎 ×𝑉𝑎 is open in 𝑋 ×𝑌 , so the collection {𝑈𝑎 ×𝑉𝑎 }𝑎 ∈𝐴 is an open covering
of 𝐴×𝐵. Since 𝐴×𝐵 is compact, a finite subcollection also covers 𝐴×𝐵, say 𝑈𝑎1 ×𝑉𝑎1 , . . . , 𝑈𝑎𝑘 ×𝑉𝑎𝑘 .
Let
𝑘
𝑘
Ø
Ù
𝑈 =
𝑈𝑎𝑘 and 𝑉 =
𝑉𝑎𝑘 ,
𝑖=1
𝑖=1
so that 𝑈 is open in 𝑋 and 𝑉 is open in 𝑌 . We claim that 𝐴×𝐵 ⊂ 𝑈 ×𝑉 ⊂ 𝑁 . Indeed, if 𝑎 ×𝑏 ∈ 𝐴×𝐵,
then 𝑎 × 𝑏 ∈ 𝑈𝑎𝑖 × 𝑉𝑎𝑖 for some 𝑖 = 1, . . . , 𝑘, so 𝑎 ∈ 𝑈 . Since each 𝑉𝑎 contains 𝐵, we have 𝐵 ⊂ 𝑉 ,
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so 𝑏 ∈ 𝑉 . Thus 𝑎 × 𝑏 ∈ 𝑈 × 𝑉 . Finally, if 𝑥 × 𝑦 ∈ 𝑈 × 𝑉 , then 𝑦 ∈ 𝑉𝑎𝑖 for some 𝑖 = 1, . . . , 𝑘, so
𝑥 × 𝑦 ∈ 𝑈𝑎𝑖 × 𝑉𝑎𝑖 ⊂ 𝑁 .
Exercise 26.10. (a) Note that 𝑓𝑛 (𝑥) ≤ 𝑓 (𝑥) for all 𝑛 ∈ Z+ and all 𝑥 ∈ 𝑋 . Let 𝜀 > 0. For each 𝑥 ∈ 𝑋,
there exists 𝑁𝑥 ∈ Z+ such that 𝑓 (𝑥) − 𝑓𝑛 (𝑥) < 𝜀 for all 𝑛 ≥ 𝑁𝑥 . Since 𝑓 and 𝑓𝑛 are continuous,
there exists a neighbourhood 𝑈𝑥 of 𝑥 such that 𝑓 (𝑧) − 𝑓𝑛 (𝑧) < 𝜀 for all 𝑧 ∈ 𝑈𝑥 . Now {𝑈𝑥 }𝑥 ∈𝑋 is an
𝑛 also
open covering of 𝑋, so there exist 𝑥 1, . . . , 𝑥𝑛 ∈ 𝑋 such that the finite subcollection {𝑈𝑥𝑖 }𝑖=1
covers 𝑋 . Let 𝑁 = max{𝑁𝑥 1 , . . . , 𝑁𝑥𝑛 }. Given 𝑥 ∈ 𝑋, there exists 𝑖 ∈ {1, . . . , 𝑛} such that 𝑥 ∈ 𝑈𝑥𝑖 ;
since 𝑁 ≥ 𝑁𝑥𝑖 , we have 𝑓 (𝑥) − 𝑓𝑛 (𝑥) < 𝜀 for all 𝑛 ≥ 𝑁 . Thus 𝑓 (𝑥) − 𝑓𝑛 (𝑥) < 𝜀 for all 𝑛 ≥ 𝑁 and
all 𝑥 ∈ 𝑋, so 𝑓𝑛 converges uniformly to 𝑓 .
(b) In Exercise 21.9 we proved that the sequence of functions 𝑓𝑛 : R → R given by
𝑓𝑛 (𝑥) =
𝑛 3 [𝑥
1
− (1/𝑛)] 2 + 1
converges to the zero function, but not uniformly as the sequence 𝑥𝑛 = 1/𝑛 converges to 𝑥 = 0
but 𝑓𝑛 (𝑥𝑛 ) = 1 for all 𝑛. We can restrict its domain to any compact interval [0, 1], and the convergence is still not uniform. So the theorem fails if we delete the requirement that the sequence be
monotone.
Now consider 𝑓𝑛 : R → R given by 𝑓𝑛 (𝑥) = arctan(𝑥 + 𝑛). The sequence (𝑓𝑛 )𝑛 converges
point-wise to the constant (hence continuous) function 𝑓 ≡ 𝜋/2, and the sequence is monotone.
Since 𝑓𝑛 (−𝑛) = 0 for all 𝑛, the convergence is not uniform. So the theorem fails if we delete the
requirement that 𝑋 be compact.
Exercise 26.11. We use the hint. Since each 𝐴 ∈ 𝒜 is closed, 𝑌 is closed. Suppose that 𝐶 and 𝐷
form a separation of 𝑌 . Then 𝐶 and 𝐷 are closed in 𝑌 , hence closed in 𝑋 . Since 𝑋 is compact, 𝐶
and 𝐷 are compact by Theorem 26.2. Since 𝑋 is Hausdorff, by Exercise 26.5, there exist 𝑈 and 𝑉
open in 𝑋 and disjoint containing 𝐶 and 𝐷, respectively. We show that
Ù
(𝐴 \ (𝑈 ∪ 𝑉 ))
𝐴∈𝒜
is not empty. Let {𝐴1, . . . , 𝐴𝑛 } be a finite subcollection of elements of 𝒜. We may assume that
𝐴𝑖 ( 𝐴𝑖+1 for all 𝑖 = 1, . . . , 𝑛 − 1. Then
𝑛
Ù
(𝐴𝑖 \ (𝑈 ∪ 𝑉 )) = 𝐴1 \ (𝑈 ∪ 𝑉 ).
𝑖=1
Suppose that 𝐴1 \ (𝑈 ∪ 𝑉 ) = ∅. Then 𝐴1 ⊂ 𝑈 ∪ 𝑉 . Since 𝐴1 is connected and 𝑈 ∩ 𝑉 = ∅, 𝐴1 lies
within either 𝑈 or 𝑉 , say 𝐴1 ⊂ 𝑈 . Then 𝑌 ⊂ 𝐴1 ⊂ 𝑈 , so that 𝐶 = 𝑌 ∩𝐶 ⊂ 𝑌 ∩𝑉 = ∅, contradicting
Ñ
the fact that 𝐶 and 𝐷 form a separation of 𝑌 . Hence, 𝑛𝑖=1 (𝐴𝑖 \ (𝑈 ∪ 𝑉 )) is non-empty. Therefore,
the collection {𝐴 \ (𝑈 ∪ 𝑉 ) | 𝐴 ∈ 𝒜} has the finite intersection property, so
Ù Ù
(𝐴 \ (𝑈 ∪ 𝑉 )) =
𝐴 \ (𝑈 ∪ 𝑉 ) = 𝑌 \ (𝑈 ∪ 𝑉 )
𝐴∈𝒜
𝐴∈𝒜
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is non-empty by Theorem 26.9. So there exists 𝑦 ∈ 𝑌 such that 𝑦 ∉ 𝑈 ∪ 𝑉 ⊂ 𝐶 ∪ 𝐷, contradicting
the fact that 𝐶 and 𝐷 form a separation of 𝑌 . We conclude that there is no such separation, so that
𝑌 is connected.
Exercise 26.12. We use the hint. We first show that if 𝑈 is an open set containing 𝑝 −1 ({𝑦}), then
there is a neighbourhood 𝑊 of 𝑦 such that 𝑝 −1 (𝑊 ) is contained in 𝑈 . Since 𝑋 − 𝑈 is closed in 𝑋,
𝑝 (𝑋 − 𝑈 ) is closed in 𝑌 and does not contain 𝑦, so 𝑊 = 𝑌 \ 𝑝 (𝑋 \ 𝑈 ) is a neighbourhood of 𝑦.
Moreover, since 𝑋 \ 𝑈 ⊂ 𝑝 −1 (𝑝 (𝑋 \ 𝑈 )) (by elementary set theory), we have
𝑝 −1 (𝑊 ) = 𝑝 −1 (𝑌 \ 𝑝 (𝑋 \ 𝑈 )) = 𝑝 −1 (𝑌 ) \ 𝑝 −1 (𝑝 (𝑋 \ 𝑈 )) ⊂ 𝑋 \ (𝑋 \ 𝑈 ) = 𝑈 .
Now let 𝒜 be an open covering of 𝑋 . For each 𝑦 ∈ 𝑌 , let 𝒜𝑦 be a subcollection of 𝒜 such that
Ø
𝑝 −1 ({𝑦}) ⊂
𝐴.
𝐴∈𝒜𝑦
Since 𝑝 −1 ({𝑦}) is compact, there exists a finite subcollection of 𝒜𝑦 that also covers 𝑝 −1 ({𝑦}), say
Ð𝑛 𝑦 𝑖
𝑛
𝐴𝑦 is open and contains 𝑝 −1 ({𝑦}), so there exists a neighbourhood 𝑊𝑦
{𝐴1𝑦 , . . . , 𝐴𝑦𝑦 }. Thus 𝑖=1
Ð𝑛 𝑦 𝑖
−1
of 𝑦 such that 𝑝 (𝑊𝑦 ) is contained in 𝑖=1
𝐴𝑦 . Then {𝑊𝑦 }𝑦 ∈𝑌 is an open covering of 𝑌 , so there
𝑘
exist 𝑦1, . . . , 𝑦𝑘 ∈ 𝑌 such that {𝑊𝑦 𝑗 } 𝑗=1 also covers 𝑌 . Then
−1
𝑋 = 𝑝 (𝑌 ) ⊂ 𝑝
−1
𝑘
Ø
𝑊𝑦 𝑗 =
𝑗=1
so
𝑘
Ø
−1
𝑝 (𝑊𝑦 𝑗 ) ⊂
𝑗=1
n
o
𝐴𝑖𝑦 𝑗
𝑛𝑦 𝑗
𝑘 Ø
Ø
𝑗=1
𝐴𝑖𝑦 𝑗 ,
𝑖=1
𝑗=1,...,𝑘.
𝑖=1,...,𝑛 𝑦 𝑗 .
is a finite subcollection of 𝒜 that also covers 𝑋 . Therefore, 𝑋 is compact.
Exercise 26.13. (a) We follow the hint. Let 𝑐 ∈ 𝐺 not in 𝐴 · 𝐵. Let 𝑏 ∈ 𝐵. Then 𝑐 · 𝑏 −1 ∉ 𝐴. By
Supplementary Exercise 7(c) of Chapter 2, there exists a neighbourhood 𝑈 of 𝑐 · 𝑏 −1 disjoint from
𝐴. Then 𝑊 = 𝑈 · 𝐵 is a neighbourhood of 𝑐 disjoint from 𝐴 · 𝐵. It follows that 𝐴 · 𝐵 is closed in 𝐺 .
(b) Let 𝐴 be closed in 𝐺 . Let 𝑥 ∈ 𝐺 be such that 𝑥𝐻 ∉ 𝑝 (𝐴). Then 𝑥 ∉ 𝐴 · 𝐻, which in closed in
𝐺 by compactness of 𝐻 and part (a), so there exists a neighbourhood 𝑈 of 𝑥 disjoint from 𝐴 · 𝐻 .
By Supplementary Exercise 5(c) of Chapter 2, 𝑝 is an open map, so 𝑝 (𝑈 ) is a neighbourhood of
𝑥𝐻 disjoint from 𝑝 (𝐴). It follows that 𝑝 is a closed map.
(c) The projection 𝑝 : 𝐺 → 𝐺/𝐻 is continuous and surjective, and it is closed by part (b). For
every 𝑥 ∈ 𝐺, the subspace 𝑝 −1 (𝑥𝐻 ) = 𝑥𝐻 is compact, being homeomorphic to 𝐻 (Supplementary
Exercise 4 of Chapter 2). We conclude by Exercise 26.12 that 𝐺 is compact.
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Exercise 27.1. Let 𝐴 be a subset of 𝑋 bounded above. Then 𝐴 is closed, hence compact by
assumption. Applying Theorem 27.4 to the inclusion 𝜄 : 𝐴 → 𝑋, there exists 𝑏 ∈ 𝐴 such that 𝑎 ≤ 𝑏
for all 𝑎 ∈ 𝐴. We prove that 𝑏 is the least upper bound of 𝐴. Indeed, since 𝐴 ⊂ 𝐴, we have 𝑎 ≤ 𝑏
for all 𝑎 ∈ 𝐴. Suppose that there exists 𝑐 ∈ 𝑋 such that 𝑐 < 𝑏 and 𝑎 ≤ 𝑐 for all 𝑎 ∈ 𝐴. Then 𝑐 ∈ 𝐴,
for if (𝑐 − 𝜀, 𝑐 + 𝜀) is a neighbourhood of 𝑐 that does not intercept 𝐴, then the neighbourhood
(𝑏 − 𝑐 − 𝜀, 𝑏 + 𝜀) of 𝑏 does not intercept 𝐴 either, contradicting the fact that 𝑏 ∈ 𝐴. Thus 𝑐 ∈ 𝐴, so
𝑐 ≤ 𝑏, contrary to the assumption. Hence 𝑏 is the least upper bound of 𝐴. It follows that 𝑋 has
the least upper bound property.
Exercise 27.2. (a) If 𝑑 (𝑥, 𝐴) = 𝛿 > 0, then 𝐵𝑑 (𝑥, 𝛿) ∩ 𝐴 = ∅, so 𝑥 ∉ 𝐴. Conversely, if 𝑥 ∉ 𝐴, there
exists 𝛿 > 0 such that 𝐵𝑑 (𝑥, 𝛿) ∩ 𝐴 = ∅, and hence 𝑑 (𝑥, 𝐴) > 𝛿/2 > 0.
(b) The distance function 𝑑 : 𝐴 × 𝐴 → R is continuous by Exercise 20.3(a). Thus, given 𝑥 ∈ 𝑋,
the function 𝑑𝑥 : 𝐴 → R given by 𝑑𝑥 (𝑎) = 𝑑 (𝑥, 𝑎) is continuous by Exercise 19.11. Since 𝐴 is
compact, by the extreme value theorem (Theorem 27.4), there exists 𝑎 ∈ 𝐴 such that 𝑑𝑥 (𝑎) =
𝑑 (𝑥, 𝑎) ≤ 𝑑 (𝑥, 𝑎 0) for all 𝑎 0 ∈ 𝐴. Then 𝑑 (𝑥, 𝐴) = 𝑑 (𝑥, 𝑎).
(c) Let 𝑦 ∈ 𝑈 (𝐴, 𝜀). Then 𝑑 (𝑦, 𝐴) = inf {𝑑 (𝑦, 𝑎) | 𝑎 ∈ 𝐴} < 𝜀, so there exists 𝑎 ∈ 𝐴 such
Ð
that 𝑑 (𝑦, 𝑎) < 𝜀, so that 𝑦 ∈ 𝐵𝑑 (𝑎, 𝜀). Thus 𝑈 (𝐴, 𝜀) ⊂ 𝑎 ∈𝐴 𝐵𝑑 (𝑎, 𝜀). Conversely, let 𝑎 ∈ 𝐴 and
Ð
𝑥 ∈ 𝐵𝑑 (𝑎, 𝜀). Then 𝑑 (𝑥, 𝐴) ≤ 𝑑 (𝑥, 𝑎) < 𝜀, so 𝑥 ∈ 𝑈 (𝐴, 𝜀). It follows that 𝑈 (𝐴, 𝜀) = 𝑎 ∈𝐴 𝐵𝑑 (𝑎, 𝜀).
(d) For each 𝑥 ∈ 𝐴, there exists 𝜀𝑥 > 0 such that 𝐵𝑑 (𝑥, 𝜀𝑥 ) ⊂ 𝑈 . Then {𝐵𝑑 (𝑥, 𝜀𝑥 )}𝑥 ∈𝐴 is an open
𝑛 also covers 𝐴. Let 𝛿 > 0 be
covering of 𝐴, so there exist 𝑥 1, . . . , 𝑥𝑛 ∈ 𝐴 such that {𝐵𝑑 (𝑥𝑖 , 𝜀𝑥𝑖 )}𝑖=1
the Lebesgue number for this finite covering, and let 𝜀 = 𝛿/2. Then for any 𝑎 ∈ 𝐴 the ball 𝐵𝑑 (𝑎, 𝜀)
has diameter less than 𝛿, so there exists 𝑖 ∈ {1, . . . , 𝑛} such that 𝐵𝑑 (𝑎, 𝜀) ⊂ 𝐵𝑑 (𝑥𝑖 , 𝜀𝑥𝑖 ). Therefore
𝑈 (𝐴, 𝜀) =
Ø
𝐵𝑑 (𝑎, 𝜀) ⊂
𝑎 ∈𝐴
𝑛
Ø
𝐵𝑑 (𝑥𝑖 , 𝜀𝑥𝑖 ) ⊂ 𝑈 .
𝑖=1
(e) Let 𝐶 = {𝑥 × tan 𝑥 | − 𝜋2 < 𝑥 < 𝜋2 }. Then 𝐶 is closed in 𝑋 = R2, but is not compact (as not
bounded). Let 𝑈 = (− 𝜋2 , 𝜋2 ) × R. Then 𝑈 is open in R2, but for any 𝜀 > 0, there exists 𝑐 ∈ 𝐶 such
that 𝐵𝑑 (𝑐, 𝜀) ∩ (R2 − 𝑈 ) ≠ ∅. So no 𝜀-neighbourhood of 𝐶 is contained in 𝑈 .
Exercise 27.3. (a) The 𝐾-topology and the standard topology on R are both Hausdorff. Furthermore, [0, 1] is compact in the standard topology. If [0, 1] were compact in R𝐾 , these topologies
on [0, 1] would be equal by Exercise 26.1(b), but the 𝐾-topology is strictly finer than the standard
topology. Thus [0, 1] is not compact as a subspace of R𝐾 .
(b) We use the hint. Clearly (−∞, 0) inherits its usual topology as subspace of R𝐾 . Now, if
𝑉 is a standard basis element for (0, +∞) as a subspace of R𝐾 , then 𝑉 = 𝑈 ∩ (0, +∞) for 𝑈 a
standard basis element for R𝐾 . If 𝑈 = (𝑎, 𝑏), then clearly 𝑉 is open in (0, +∞) with the standard
topology. Let 𝐾 denote the closure of 𝐾 in the standard topology. If 𝑈 = (𝑎, 𝑏) − 𝐾, then 𝑉 =
(𝑎, 𝑏) ∩ (0, +∞) − 𝐾 = (𝑎, 𝑏) ∩ (0, +∞) − 𝐾, so 𝑉 is open in (0, +∞) with the standard topology
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as well. Thus (0, +∞) inherits its usual topology as a subspace of R𝐾 . Hence (−∞, 0) and (0, +∞)
are connected as subspaces of R𝐾 , so their closures (−∞, 0] and [0, +∞) are connected and have
the common point 0. It follows from Theorem 23.3 that (−∞, 0] ∪ [0, +∞) = R𝐾 is connected.
(c) Let R denote the real numbers with the standard topology. Suppose that there is a path
from 0 to 1 in R𝐾 , that is, a continuous function 𝑓 : [0, 1] → R𝐾 such that 𝑓 (0) = 0 and 𝑓 (1) = 1.
Since R𝐾 is finer than R, 𝑓 is continuous when considered as a function to R. By the intermediate
value theorem (Theorem 24.3), the interval [0, 1] is contained in 𝑓 ( [0, 1]). Since [0, 1] is compact
in R and 𝑓 is continuous, it follows that 𝑓 ( [0, 1]) is compact in R𝐾 . Thus [0, 1] is closed in R𝐾 and
contained in the compact space 𝑓 ( [0, 1]), so [0, 1] is compact in R𝐾 , contradicting (a). We deduce
that there is no path from 0 to 1 in R𝐾 and hence R𝐾 is not path-connected.
Exercise 27.4. The distance function 𝑑 : 𝑋 × 𝑋 → R is continuous by Exercise 20.3(a), so given
𝑥 ∈ 𝑋, the function 𝑑𝑥 : 𝑋 → R given by 𝑑𝑥 (𝑦) = 𝑑 (𝑥, 𝑦) is continuous by Exercise 19.11. Since
𝑋 is connected, the image 𝑑𝑥 (𝑋 ) is a connected subspace of R, and contains 0 since 𝑑𝑥 (𝑥) = 0.
Thus, if 𝑦 ∈ 𝑋 and 𝑦 ≠ 𝑥, then 𝑑𝑥 (𝑋 ) contains the set [0, 𝛿], where 𝛿 = 𝑑𝑥 (𝑦) > 0. Therefore 𝑋
must be uncountable.
Exercise 27.5. We follow the hint. Let 𝑈 be any non-empty open set in 𝑋 . We prove that 𝑈 has
Ð
a point that is not in 𝑛 𝐴𝑛 . Since 𝐴1 has empty interior, we have 𝑈 ⊄ 𝐴1 . We show that there is
a non-empty open set 𝑉1 in 𝑋 such that 𝑉1 ⊂ 𝑈 \ 𝐴1 . Indeed, let 𝑥 ∈ 𝑈 \ 𝐴1 . Then 𝐴1 ∪ (𝑋 \ 𝑈 )
is closed and does not contain 𝑥 . Since 𝑋 is compact, 𝐴1 ∪ (𝑋 \ 𝑈 ) is compact by Theorem 26.2.
Thus, by Lemma 26.4, there exist disjoint open sets 𝑉1 and 𝑊 of 𝑋 containing 𝑥 and 𝐴1 ∪ (𝑋 \ 𝑈 )
respectively. Hence 𝑥 ∈ 𝑉1 ⊂ 𝑉1 ⊂ 𝑋 \ (𝐴1 ∪ (𝑋 \ 𝑈 )) = 𝑈 \ 𝐴1 . Similarly, 𝑉1 is a non-empty
open set in 𝑋 not contained in 𝐴2, so there exists a non-empty open set 𝑉2 such that 𝑉2 ⊂ 𝑉1 \ 𝐴2 .
Inductively we obtain a nested sequence
𝑉1 ⊂ 𝑉1 ⊂ · · ·
Ñ∞
of non-empty closed sets of 𝑋 . Since 𝑋 is compact, 𝑖=1
𝑉𝑖 is non-empty by Theorem 26.9. So
Ñ∞
there exists 𝑥 ∈ 𝑖=1 𝑉𝑖 ⊂ 𝑉1 ⊂ 𝑈 , and 𝑥 ∉ 𝐴𝑖 for all 𝑖 ∈ Z+ . Thus 𝑈 has a point that is not in
Ð
Ð
𝑛 𝐴𝑛 . Since 𝑈 was arbitrary, we deduce that 𝑛 𝐴𝑛 has empty interior.
Exercise 27.6. (a) Let 𝐵 be a connected subspace of 𝐶. Suppose that 𝐵 contains two different
points 𝑥 ≠ 𝑦. By the Archimedean property, there exists 𝑛 ∈ Z+ such that 1/3𝑛 < |𝑥 − 𝑦|. Thus
𝑥 and 𝑦 belong to different closed intervals in 𝐴𝑛 . Let 𝐼 ⊂ 𝐴𝑛 be the interval containing 𝑥 . Then
𝐵 ∩ 𝐼 and 𝐵 ∩ (𝐶𝐼 ) form a separation of 𝐵, contradicting the fact that 𝐵 is connected. It follows
that 𝐶 is totally disconnected.
(b) Each 𝐴𝑛 is closed in [0, 1], so 𝐶 is an intersection of closed sets, hence closed. Thus 𝐶 is a
closed subspace of the compact space [0, 1], so 𝐶 is compact by Theorem 26.2
(c) We use induction. Clearly 𝐴1 = [0, 31 ] ∪ [ 32 , 3] satisfies this property. At each step, each
closed interval of length 1/3𝑛 of 𝐴𝑛 is divided in three closed intervals of length 1/3𝑛+1, and the
interior of the middle interval is removed, leaving in 𝐴𝑛+1 two closed intervals of length 1/3𝑛+1 .
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28
Limit Point Compactness
At each step only the interior of closed intervals are removed, so the endpoints of the intervals
are never removed and they lie in 𝐶.
(d) Let 𝑥 ∈ 𝐶. Then 𝑥 ∈ 𝐴𝑛 for each 𝑛, so there exist intervals 𝐼𝑛 ⊂ 𝐴𝑛 containing 𝑥 . For each
interval 𝐼𝑛 , we can choose an endpoint 𝑥𝑛 of the interval not equal to 𝑥 . These endpoints 𝑥𝑛 are
in 𝐶 by (c), and 𝑥𝑛 → 𝑥 . So 𝑥 is not an isolated point of 𝑋 . Since 𝑥 was arbitrary, we deduce that
𝐶 has no isolated points.
(e) 𝐶 is non-empty compact Hausdorff with no isolated points, hence uncountable by Theorem
27.7.
28
Limit Point Compactness
Exercise 28.1. Consider the subset 𝑌 = {0, 1}𝜔 ⊂ [0, 1] 𝜔 of sequences whose entries are 0’s and
1’s. 𝑌 is clearly infinite. We show that 𝑌 has no limit point in [0, 1] 𝜔 . Let x ∈ [0, 1] 𝜔 . If x ∉ 𝑌 , then
there exists 𝑖 ∈ Z+ such that 𝑥𝑖 ≠ 0, 1. Let 𝛿 = min{1 − 𝑥𝑖 , 𝑥𝑖 }. Then 𝐵 𝜌 (x, 𝛿) is a neighbourhood
of x for which 𝐵 𝜌 (x, 𝛿) ∩ 𝑌 = ∅, so x ∉ 𝑌 0 . Now, if x ∈ 𝑌 and y ∈ 𝑌 with x ≠ y, then there
exists 𝑖 ∈ Z+ such that |𝑥𝑖 − 𝑦𝑖 | = 1. Thus, if x ∈ 𝑌 and 𝜀 < 1, we have 𝐵 𝜌 (x, 𝛿) ∩ 𝑌 = {x} and
hence x ∉ 𝑌 0 . It follows that 𝑌 = {0, 1}𝜔 is an infinite subset of [0, 1] 𝜔 that has no limit point, as
claimed.
Exercise 28.2. Let 𝐴 = {1 − 1/𝑛 | 𝑛 ∈ Z+ }. Then 𝐴 is infinite. Clearly any point 𝑥 ∈ [0, 1) is not
a limit point of 𝐴: if 1 − 1/𝑛 ≤ 𝑥 < 1 − 1/(𝑛 + 1), then [𝑥, 1 − 1/(𝑛 + 1)) is a neighbourhood of 𝑥
with no point in 𝐴 \ {𝑥 }. Furthermore, if 𝑥 = 1, then {1} = [1, 2) ∩ [0, 1] is a neighbourhood of 𝑥
with no point in 𝐴 \ {𝑥 }. Hence 𝐴 is an infinite subset of [0, 1] with no limit point, so [0, 1] is not
limit point compact as a subspace of Rℓ .
Exercise 28.3. (a) No. Consider 𝑋 = Z+ ∪𝑌 of Example 1: 𝑌 is a set of two points in the indiscrete
topology. Then 𝑋 is limit point compact and 𝜋 1 : 𝑋 → Z+ is continuous, but 𝜋1 (𝑋 ) = Z+ is not
limit point compact.
(b) Yes. If 𝐴 is closed, then 𝐴 contains all its limit points. If 𝐵 ⊂ 𝐴 is infinite, then 𝐵 has a limit
point 𝑥 in 𝑋 . Then 𝑥 is also a limit point of 𝐴, so 𝑥 ∈ 𝐴 0 ⊂ 𝐴. Therefore, every infinite subset of
𝐴 has a limit point in 𝐴, so 𝐴 is limit point compact.
(c) No. Consider 𝑆 Ω in the order topology. It is Hausdorff by Exercise 17.10. By Example 2,
𝑆 Ω is a limit point compact subspace of 𝑆 Ω , but is is not closed since it does not contain its limit
point Ω.
Exercise 28.4. First let 𝑋 be a countable compact space. Note that if 𝑌 is a closed subset of
𝑋, then 𝑌 is countable compact as well, for if {𝑈𝑛 }𝑛 ∈Z+ is a countable open covering of 𝑌 , then
{𝑈𝑛 }𝑛 ∈Z+ ∪ (𝑋 \ 𝑌 ) is a countable open covering of 𝑋 ; there is a finite subcovering of 𝑋, hence
a finite subcovering of 𝑌 . Now let 𝐴 be an infinite subset. We show that 𝐴 has a limit point. Let
𝐵 be a countable infinite subset of 𝐴. Suppose that 𝐵 has no limit point, so that 𝐵 is closed in 𝑋 .
Then 𝐵 is countable compact. Since 𝐵 has no limit point, for each 𝑏 ∈ 𝐵 there is a neighbourhood
𝑈𝑏 of 𝑏 that intersects 𝐵 in the point 𝑏 alone. Then {𝑈𝑏 }𝑏 ∈𝐵 is an open covering of 𝐵 with no finite
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28
Limit Point Compactness
subcovering, contradicting the fact that 𝐵 is countable compact. Hence 𝐵 has a limit point, so that
𝐴 has a limit point as well. Since 𝐴 was arbitrary, we deduce that 𝑋 is limit point compact. (Note
that the 𝑇1 property is not necessary in this direction.)
Now assume that 𝑋 is a limit point compact 𝑇1 space. We show that 𝑋 is countable compact.
Suppose, on the contrary, that {𝑈𝑛 }𝑛 ∈Z+ is a countable open covering of 𝑋 with no finite subcovering. For each 𝑛, take a point 𝑥𝑛 in 𝑋 not in 𝑈 1 ∪ · · · ∪ 𝑈𝑛 . By assumption, the infinite set
𝐴 = {𝑥𝑛 | 𝑛 ∈ Z+ } has a limit point 𝑦 ∈ 𝑋 . Since {𝑈𝑛 }𝑛 ∈Z+ covers 𝑋, there exists 𝑁 ∈ Z+ such that
𝑦 ∈ 𝑈 1 ∪ · · · ∪ 𝑈 𝑁 . Now 𝑋 is 𝑇1, so for each 𝑖 = 1, . . . , 𝑁 there exists a neighbourhood 𝑉𝑖 of 𝑦 that
does not contain 𝑥𝑖 (see Exercise 17.15). Then
𝑉 = (𝑉1 ∩ · · · ∩ 𝑉𝑁 ) ∩ (𝑈 1 ∪ · · · ∪ 𝑈 𝑁 )
is a neighbourhood of 𝑦 that does not contain any of the points 𝑥𝑖 , contradicting the fact that 𝑦 is a
limit point of 𝐴. It follows that every countable open covering of 𝑋 must have a finite subcovering,
so 𝑋 is countable compact.
Exercise 28.5. We could imitate the proof of Theorem 26.9, but we prove directly each direction.
First let 𝑋 be countable compact and let 𝐶 1 ⊂ 𝐶 2 ⊂ · · · be a nested sequence of closed nonempty sets of 𝑋 . For each 𝑛 ∈ Z+, 𝑈𝑛 = 𝑋 \ 𝐶𝑛 is open in 𝑋 . Then {𝑈𝑛 }𝑛 ∈Z+ is a countable
collection of open sets with no finite subcollection covering 𝑋, for if 𝑈𝑖 1 ∪ · · · ∪𝑈 1𝑛 covers 𝑋, then
𝐶𝑖 1 ∩ · · · ∩ 𝐶𝑖𝑛 is empty, contrary to the assumption. Hence {𝑈𝑛 }𝑛 ∈Z+ does not cover 𝑋, so there
Ð
Ñ
Ñ
exist 𝑥 ∈ 𝑋 \ 𝑛 ∈Z+ 𝑈𝑛 = 𝑛 ∈𝑍 + (𝑋 \ 𝑈𝑛 ) = 𝑛 ∈𝑍 + 𝐶𝑛 .
Conversely, assume that every nested sequence 𝐶 1 ⊂ 𝐶 2 ⊂ · · · of closed non-empty sets of 𝑋
has a non-empty intersection and let {𝑈𝑛 }𝑛 ∈Z+ be a countable open covering of 𝑋 . For each 𝑛,
let 𝑉𝑛 = 𝑈 1 ∪ · · · ∪ 𝑈𝑛 and 𝐶𝑛 = 𝑋 \ 𝑉𝑛 . Suppose that no finite subcollection of {𝑈𝑛 }𝑛 ∈Z+ covers
𝑋 . Then each 𝐶𝑛 is non-empty, so 𝐶 1 ⊂ 𝐶 2 ⊂ · · · is a nested sequence of non-empty closed sets
Ñ
Ñ
and 𝑛 ∈Z+ 𝐶𝑛 is non-empty by assumption. Then there exists 𝑥 ∈ 𝑛 ∈Z+ 𝐶𝑛 , so that 𝑥 ∉ 𝑉𝑛 for
all 𝑛, contradicting the fact that {𝑈𝑛 }𝑛 ∈Z+ covers 𝑋 . It follows that there exists 𝑁 ∈ Z+ such that
𝐶 𝑁 = ∅, so that 𝑋 = 𝑉𝑁 and hence some finite subcollection of {𝑈𝑛 }𝑛 ∈Z+ covers 𝑋 . We deduce
that 𝑋 is countable compact.
Exercise 28.6. We follow the hint. Note that 𝑓 is an imbedding by Exercise 21.2. It remains to
prove that 𝑓 is surjective. Suppose it is not, and let 𝑎 ∈ 𝑓 (𝑋 ). Since 𝑋 is compact, 𝑓 (𝑋 ) is compact
and hence closed (every metric space is Hausdorff). Thus, there exists 𝜀 > 0 such that the 𝜀neighbourhood of 𝑎 is contained in 𝑋 \ 𝑓 (𝑋 ). Set 𝑥 1 = 𝑎, and inductively 𝑥𝑛+1 = 𝑓 (𝑥𝑛 ) for 𝑛 ∈ Z+ .
We show that 𝑑 (𝑥𝑛 , 𝑥𝑚 ) ≥ 𝜀 for 𝑛 ≠ 𝑚. Indeed, we may assume 𝑛 < 𝑚. If 𝑛 ≥ 1, then 𝑑 (𝑥𝑛 , 𝑥𝑚 ) =
𝑑 (𝑓 −1 (𝑥𝑛 ), 𝑓 −1 (𝑥𝑚 )) = 𝑑 (𝑥𝑛−1, 𝑥𝑚−1 ). By induction it follows that 𝑑 (𝑥𝑛 , 𝑥𝑚 ) = 𝑑 (𝑥𝑛−𝑖 , 𝑥𝑚−𝑖 ) for
all 𝑖 ≥ 1, and hence 𝑑 (𝑥𝑛 , 𝑥𝑚 ) = 𝑑 (𝑎, 𝑥𝑚−𝑛 ) = 𝑑 (𝑎, 𝑓 (𝑥𝑚−𝑛−1 )). Since 𝑓 (𝑥𝑚−𝑛−1 ) ∈ 𝑓 (𝑋 ) and
𝐵(𝑎, 𝜀) ∩ 𝑓 (𝑋 ) = ∅, we have 𝑑 (𝑥𝑛 , 𝑥𝑚 ) ≥ 𝜀, as claimed. Thus {𝑥𝑛 }𝑛 ∈Z+ is a sequence with no
convergent subsequence, so 𝑋 is not sequentially compact. By Theorem 28.2, this contradicts the
fact that 𝑋 is compact. Therefore 𝑓 is surjective and hence a homeomorphism.
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28
Limit Point Compactness
Exercise 28.7. First we prove (b), which implies (a).
(b) We use the hint. For each 𝑛 ∈ Z+, let 𝐴𝑛 = 𝑓 𝑛 (𝑋 ). Since 𝑋 is compact, each 𝐴𝑛 is compact,
Ñ
hence closed. Thus 𝐴 = 𝑛 ∈Z+ 𝐴𝑛 is closed. Moreover, note that 𝐴𝑛+1 ⊂ 𝐴𝑛 for all 𝑛, so 𝐴 is
non-empty by Theorem 26.9. Now we prove that 𝑓 (𝐴) = 𝐴. Let 𝑦 ∈ 𝑓 (𝐴). There exists 𝑥 ∈ 𝐴
such that 𝑦 = 𝑓 (𝑥). Let 𝑛 ∈ Z+ be such that 𝑥 ∈ 𝐴𝑛 . If 𝑛 = 1, then 𝑦 ∈ 𝑓 (𝑋 ) = 𝐴1 because 𝑥 ∈ 𝑋 .
If 𝑛 > 1, then 𝑦 ∈ 𝑓 𝑛 (𝑋 ) = 𝐴𝑛 because 𝑥 ∈ 𝐴𝑛−1 . Thus 𝑦 ∈ 𝐴𝑛 for all 𝑛, so that 𝑦 ∈ 𝐴 and hence
𝑓 (𝐴) ⊂ 𝐴. Now let 𝑥 ∈ 𝐴. Then 𝑥 ∈ 𝑓 𝑛 (𝑋 ) for all 𝑛, so there exists a sequence {𝑥𝑛 }𝑛 ∈Z+ such
that 𝑓 𝑛+1 (𝑥𝑛 ) = 𝑥 for each 𝑛. Let 𝑦𝑛 = 𝑓 𝑛 (𝑥𝑛 ), 𝑛 ∈ Z+ . Since 𝑋 is a compact metric space, it is
sequentially compact by Theorem 28.2, so {𝑦𝑛 }𝑛 ∈Z+ has a convergent subsequence, say {𝑧𝑛 }𝑛 ∈Z+
with limit 𝑎 ∈ 𝑋 . Then 𝑧𝑛 → 𝑎 and 𝑧𝑛 ∈ 𝐴 for all 𝑛, so 𝑎 ∈ 𝐴 by the sequence lemma (Lemma
21.2). Since 𝐴 is closed, we have 𝑎 ∈ 𝐴; since 𝑓 is continuous and 𝑓 (𝑧𝑛 ) = 𝑥 for all 𝑛, we have
𝑓 (𝑎) = 𝑥 . Thus 𝑥 ∈ 𝑓 (𝐴) and hence 𝐴 ⊂ 𝑓 (𝐴). Therefore 𝐴 = 𝑓 (𝐴). Moreover, 𝐴 has only one
point. Indeed, suppose that 𝐴 has more than one point. Since the distance function 𝑑 : 𝐴 × 𝐴 → R
is continuous (Exercise 20.3(a)), it follows from the extreme value theorem (Theorem 27.4) that
there exists 𝑥 ≠ 𝑦 in 𝐴 such that 𝑑 (𝑥 0, 𝑦 0) ≤ 𝑑 (𝑥, 𝑦) for all 𝑥 0, 𝑦 0 ∈ 𝐴. Let 𝑥 = 𝑓 (𝑎) and 𝑦 = 𝑓 (𝑏)
with 𝑎, 𝑏 ∈ 𝐴. Then 𝑑 (𝑥, 𝑦) = 𝑑 (𝑓 (𝑎), 𝑓 (𝑏)) < 𝑑 (𝑎, 𝑏) ≤ 𝑑 (𝑥, 𝑦), a contradiction. Hence 𝐴 contains
only one point, and is a fixed point of 𝑓 . Now 𝑓 has only one fixed point, for if 𝑥 and 𝑦 are two
different fixed points of 𝑓 , then 𝑑 (𝑥, 𝑦) = 𝑑 (𝑓 (𝑥), 𝑓 (𝑦)) < 𝑑 (𝑥, 𝑦), absurd. Thus the point in 𝐴 is
the only fixed point of 𝑓 .
(c) Note that
|𝑓 (𝑥) − 𝑓 (𝑦)| = |𝑥 − 𝑦| 1 −
𝑥 +𝑦
2
𝑥+𝑦
for all 𝑥, 𝑦 ∈ 𝑋 . Since |1 − 2 | < 1 for all 𝑥, 𝑦 ∈ 𝑋 with 𝑥 ≠ 𝑦, 𝑓 is a shrinking map. We show
that 𝑓 is not a contraction. Suppose there exists 𝛼 < 1 such that
|𝑓 (𝑥) − 𝑓 (𝑦)| ≤ 𝛼 |𝑥 − 𝑦|
for all 𝑥, 𝑦 ∈ 𝑋 . Note that
𝑥
|𝑓 (𝑥) − 𝑓 (0)| = (𝑥 − 0) 1 −
2
𝑥
for all 𝑥 ∈ 𝑋 . Thus, taking 𝑥 ≠ 0 such that 𝛼 < 1 − 2 , we see that the condition for being a
contraction is not satisfied.
(d) Note that
(𝑥 2 + 1) 1/2 − (𝑦 2 + 1) 1/2
𝑥 +𝑦
= 2
1/2
𝑥 −𝑦
(𝑥 + 1) + (𝑦 2 + 1) 1/2
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29
Local Compactness
for all 𝑥, 𝑦 ∈ R, 𝑥 ≠ 𝑦. Thus
|𝑥 − 𝑦|
(𝑥 2 + 1) 1/2 − (𝑦 2 + 1) 1/2
1+
2
𝑥 −𝑦
|𝑥 − 𝑦|
𝑥 +𝑦
=
1+ 2
1/2
2
(𝑥 + 1) + (𝑦 2 + 1) 1/2
|𝑥 − 𝑦| |𝑥 − 𝑦| 1 1
<
+
+ = |𝑥 − 𝑦|,
2
2
2 2
|𝑓 (𝑥) − 𝑓 (𝑦)| =
for all 𝑥, 𝑦 ∈ R, 𝑥 ≠ 𝑦, so 𝑓 is a shrinking map. Now suppose that there exists 𝛼 < 1 such that
𝑓 (𝑥) −
1
= |𝑓 (𝑥) − 𝑓 (0)| ≤ 𝛼 |𝑥 |
2
for all 𝑥 ∈ R. Since 𝑓 is strictly increasing, we have 𝑓 (𝑥) > 1/2 for all 𝑥 > 0. Thus
!
|𝑥 |
𝑥
𝑥
1
1
1+ 2
=
1+
𝑓 (𝑥) − = |𝑓 (𝑥) − 𝑓 (0)| =
2
2
2
(𝑥 + 1) 1/2
(1 + 𝑥12 ) 1/2
for 𝑥 > 0. So, if 𝑥 > 0 satisfies
!
1
1
1+
> 𝛼,
2
(1 + 𝑥12 ) 1/2
then the condition for being a contraction is not satisfied, and we can find such 𝑥 since the above
expression tends to 1 as 𝑥 → ∞. So 𝑓 is not a contraction. Furthermore, it has no fixed point,
since
𝑥 + (𝑥 2 + 1) 1/2 𝑥 + |𝑥 |
𝑓 (𝑥) =
>
>𝑥
2
2
for all 𝑥 ∈ R.
29
Local Compactness
Exercise 29.1. First, we prove that each set Q ∩ [𝑎, 𝑏], where 𝑎, 𝑏 are irrational numbers, is not
compact. Indeed, since Q∩ [𝑎, 𝑏] is countable, we can write Q∩ [𝑎, 𝑏] = {𝑞 1, 𝑞 2, . . .}. Then {𝑈𝑖 }𝑖 ∈Z+ ,
where 𝑈𝑖 = Q ∩ [𝑎, 𝑞𝑖 ) for each 𝑖, is an open covering of Q ∩ [𝑎, 𝑏] with no finite subcovering. Now
let 𝑥 ∈ Q and suppose that Q is locally compact at 𝑥 . Then there exists a compact set 𝐶 containing
a neighbourhood 𝑈 of 𝑥 . Then 𝑈 contains a set Q ∩ [𝑎, 𝑏] where 𝑎, 𝑏 are irrational numbers. Since
this set is closed and contained in the compact 𝐶, it follows Q ∩ [𝑎, 𝑏] is compact, a contradiction.
Therefore, Q is not locally compact.
Î
Exercise 29.2. (a) Recall that the projections 𝜋 𝛽 : 𝛼 𝑋𝛼 → 𝑋 𝛽 are continuous open maps.
Î
Î
Assume that 𝛼 𝑋𝛼 is locally compact. Let x ∈ 𝛼 𝑋𝛼 . Then there exist a compact subspace
Î
𝐶 of 𝛼 𝑋𝛼 containing a neighbourhood of x. This neighbourhood contains a standard basis
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Î
Î
element for 𝛼 𝑋𝛼 containing x, say 𝑈 = 𝛼 𝑈𝛼 where 𝑈𝛼 = 𝑋𝛼 for all but finitely many indices,
say 𝛼 1, . . . , 𝛼𝑛 . Let 𝛽 ≠ 𝛼𝑖 for all 𝑖 = 1, . . . , 𝑛. Then 𝜋 𝛽 (𝐶) is compact and contains 𝜋 𝛽 (𝑈 ) = 𝑋 𝛽 ,
so 𝜋 𝛽 (𝐶) = 𝑋 𝛽 and 𝑋 𝛽 is compact. Hence 𝑋𝛼 is compact for all but finitely many values of 𝛼 .
Î
It remains to prove that 𝑋𝛼𝑖 is locally compact for 𝑖 = 1, . . . , 𝑛. Let 𝑥 ∈ 𝑋𝛼𝑖 . Let x ∈ 𝛼 𝑋𝛼 be
such that 𝑥𝛼𝑖 = 𝑥 . There is a compact 𝐶 containing a basis neighbourhood 𝑈 of x. Then 𝜋𝛼𝑖 (𝐶)
is a compact subspace of 𝑋𝛼𝑖 containing the neighbourhood 𝜋𝛼𝑖 (𝑈 ) of 𝑥 . Thus the spaces 𝑋𝛼𝑖 , for
𝑖 = 1, . . . , 𝑛, are indeed locally compact.
(b) Suppose that each 𝑋𝛼 is locally compact and 𝑋𝛼 is compact for all but finitely many 𝛼 . Let
Î
x ∈ 𝛼 𝑋𝛼 . For each 𝛼, there exists a compact subspace 𝐶𝛼 of 𝑋𝛼 containing a neighbourhood 𝑈𝛼
of 𝑥𝛼 . By assumption, for all but finitely many indices we may assume that 𝐶𝛼 = 𝑈𝛼 = 𝑋𝛼 . By the
Î
Î
Tychonoff theorem, 𝛼 𝐶𝛼 is compact, and it contains the neighbourhood 𝛼 𝑈𝛼 of x. It follows
Î
that 𝛼 𝑋𝛼 is locally compact.
Exercise 29.3. If 𝑓 is continuous but not open, 𝑓 (𝑋 ) is not necessarily locally compact. Indeed,
consider Q𝑑 , the rational numbers having the discrete topology, and let Q denote the rational
numbers in the usual topology. Then Q𝑑 is locally compact, for if 𝑥 ∈ Q𝑑 , then {𝑥 } is open and
compact. Let 𝑖 : Q𝑑 → Q be the identity map. Then 𝑖 is continuous, but 𝑖 (Q𝑑 ) = Q is not locally
compact by Exercise 29.1.
Now, if 𝑓 is continuous and open, then 𝑓 (𝑋 ) is locally compact. Indeed, let 𝑦 ∈ 𝑓 (𝑋 ), so that
𝑦 = 𝑓 (𝑥) for some 𝑥 ∈ 𝑋 . Since 𝑋 is locally compact, there exists a compact subspace 𝐶 of 𝑋
containing a neighbourhood 𝑈 of 𝑥 . Since 𝑓 is continuous, 𝑓 (𝐶) is compact and since 𝑓 is open,
𝑓 (𝑈 ) is open. Thus 𝑓 (𝐶) is compact and contains the neighbourhood 𝑓 (𝑈 ) of 𝑦, so 𝑓 (𝑋 ) is locally
compact.
Exercise 29.4. Consider 0 ∈ [0, 1] 𝜔 and suppose that [0, 1] 𝜔 is locally compact at 0. Then there
exists a compact 𝐶 containing an open ball 𝐵 = 𝐵 𝜌 (0, 𝜀) ⊂ [0, 1] 𝜔 . Note that 𝐵 = [0, 𝜀] 𝜔 . Then
[0, 𝜀] 𝜔 is closed and contained in the compact 𝐶, so it is compact. But [0, 𝜀] 𝜔 is homeomorphic to
[0, 1] 𝜔 , which is not compact by Exercise 28.1. This contradiction proves that [0, 1] 𝜔 is not locally
compact in the uniform topology.
Exercise 29.5. Let 𝑌1 = 𝑋 1 ∪ {∞1 } and 𝑌2 = 𝑋 2 ∪ {∞2 } be the respective one-point compactifications. Define 𝑔 : 𝑌1 → 𝑌2 by 𝑔|𝑋 1 = 𝑓 and 𝑔(∞1 ) = ∞2 . Then 𝑔 is bijective. By symmetry of the
problem, it suffices to prove that 𝑔 is open. Let 𝑈 be an open set of 𝑌1 . If 𝑈 is open in 𝑋 1, then
𝑔(𝑈 ) = 𝑓 (𝑈 ) is open in 𝑋 1, hence in 𝑌1 . If 𝑈 = 𝑋 1 \ 𝐶, where 𝐶 is a compact subspace of 𝑋, then
𝑔(𝑈 ) = 𝑔(𝑌1 ) \ 𝑔(𝐶) = 𝑌2 \ 𝑔(𝐶). Since 𝑓 is continuous and 𝐶 ⊂ 𝑋, it follows 𝑔(𝐶) = 𝑓 (𝐶) is
compact, so 𝑌2 \ 𝑔(𝐶) is open in 𝑌2 . Thus 𝑔 is an open map, hence a homeomorphism.
Exercise 29.6. The function 𝑓 : R → (0, 1) given by
𝑓 (𝑥) =
1
1 + 2−𝑥
is a homeomorphism, and the map 𝑔 : (0, 1) → 𝑆 1 \ {𝑝}, where 𝑝 = 1 × 0, defined by
𝑔(𝑥) = cos(2𝜋𝑥) × sin(2𝜋𝑥)
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is also a homeomorphism (as it is a bijective local homeomorphism). Thus R is homeomorphic
to 𝑆 1 \ {𝑝} via ℎ ≔ 𝑔 ◦ 𝑓 . Furthermore, since 𝑆 1 is compact Hausdorff and the one-point compactification is uniquely determined up to homeomorphism, the one-point compactification of
𝑆 1 \ {𝑝} is precisely 𝑆 1 . By Exercise 29.5, ℎ extends to a homeomorphism between the one-point
compactification of R and 𝑆 1 .
Exercise 29.7. Since the one-point compactification is uniquely determined up to homeomorphism, it suffices to prove that 𝑆 Ω is compact Hausdorff. It is Hausdorff since it has the order topology
(see Exercise 17.10). Now let 𝒜 be an open covering of 𝑆 Ω . Let 𝐴 ∈ 𝒜 contain Ω. Then 𝐴 contains
an interval (𝑎, +∞). Let 𝑎 0 be the least element of 𝑆 Ω . Since 𝑆 Ω is well-ordered, [𝑎 0, 𝑎] is compact
by Theorem 27.1. Thus finitely many elements of 𝒜 cover [𝑎 0, 𝑎], and these elements together
with 𝐴 cover 𝑆 Ω , so 𝑆 Ω is compact. We conclude that the one-point compactification of 𝑆 Ω is
homeomorphic with 𝑆 Ω .
Exercise 29.8. Let 𝑋 = {1/𝑛 | 𝑛 ∈ Z+ }. As subspaces of R, the map 𝑓 : Z+ → 𝑋 given by
𝑓 (𝑛) = 1/𝑛 is a homeomorphism. Since 𝑋 ∪{0} is closed and bounded in R, it is compact (Theorem
27.3). It is also Hausdorff, so (up to homeomorphism) it is the one-point compactification of 𝑋 .
By Exercise 29.5, 𝑓 extends to a homeomorphism between the one-point compactification of Z+
and 𝑋 ∪ {0}.
Exercise 29.9. Let 𝑝 : 𝐺 → 𝐺/𝐻 be the quotient map. Then 𝑝 is continuous, and it is open by
Exercise 5(c) of Supplementary Exercises of Chapter 2. Hence 𝑝 (𝐺) = 𝐺/𝐻 is locally compact by
Exercise 29.3.
Exercise 29.10. Let 𝑈 be a neighbourhood of 𝑥 . Since 𝑋 is locally compact at 𝑥, there exists a
compact subspace 𝐶 of 𝑋 containing a neighbourhood 𝑊 of 𝑥 . Then 𝑈 ∩ 𝑊 is open in 𝑋, hence
in 𝐶. Thus, 𝐶 \ (𝑈 ∩ 𝑊 ) is closed in 𝐶, hence compact by Theorem 26.2. Since 𝑋 is Hausdorff, by
Lemma 26.4 there exist disjoint open sets 𝑉1 and 𝑉2 of 𝑋 containing 𝑥 and 𝐶 \ (𝑈 ∩𝑊 ) respectively.
Let 𝑉 = 𝑉1 ∩ 𝑈 ∩ 𝑊 . Since 𝑉 is closed in 𝐶, it is compact. Furthermore, 𝑉 is disjoint from
𝐶 \ (𝑈 ∩ 𝑊 ) ⊂ 𝐶 \ 𝑈 , so 𝑉 ⊂ 𝑈 .
Exercise 29.11. (a) Since 𝑝 and 𝑖𝑍 are continuous and surjective, it follows that 𝜋 is continuous
(Exercise 18.10) and surjective. It remains to prove that 𝐴 is open in 𝑌 × 𝑍 if 𝜋 −1 (𝐴) is open in
𝑋 × 𝑍 . Let 𝑦 × 𝑧 ∈ 𝐴 with 𝑦 × 𝑧 = 𝜋 (𝑥 × 𝑧). Then 𝑥 × 𝑧 ∈ 𝜋 −1 (𝐴), so we can take 𝑈 1 × 𝑊 a basis
neighbourhood of 𝑥 × 𝑧 contained in 𝜋 −1 (𝐴). Since 𝑊 is a neighbourhood of 𝑦, by Exercise 29.10
there is a neighbourhood 𝑉 of 𝑦 such that 𝑉 is compact and 𝑉 ⊂ 𝑊 . Hence 𝑥 × 𝑧 ∈ 𝑈 1 × 𝑉 ⊂
𝜋 −1 (𝐴). Note that 𝜋 (𝑈 1 × 𝑉 ) = 𝑝 (𝑈 1 ) × 𝑉 ⊂ 𝜋 (𝜋 −1 (𝐴)) ⊂ 𝐴, so 𝑝 −1 (𝑝 (𝑈 1 )) × 𝑉 ⊂ 𝜋 −1 (𝐴).
Now, for each 𝑢 ∈ 𝑝 −1 (𝑝 (𝑈 1 )) use the tube lemma (Lemma 28.6) to find a neighbourhood 𝑊𝑢 of
Ð
𝑢 such that 𝑝 −1 (𝑝 (𝑈 1 )) × 𝑉 contains the tube 𝑊𝑢 × 𝑉 . Then 𝑈 2 ≔ 𝑢 𝑊𝑢 contains 𝑝 −1 (𝑝 (𝑈 1 ))
and 𝑈 2 × 𝑉 ⊂ 𝜋 −1 (𝐴). For each 𝑖 ≥ 3 we repeat this process to find an open set 𝑈𝑖+1 containing
Ð
𝑝 −1 (𝑝 (𝑈𝑖 )) such that 𝑈𝑖+1 × 𝑉 ⊂ 𝜋 −1 (𝐴). Let 𝑈 = 𝑖 𝑈𝑖 . We show that 𝑈 is saturated. We have
𝑈 ⊂ 𝑝 −1 (𝑝 (𝑈 )) by elementary set theory. Now, if 𝑥 ∈ 𝑝 −1 (𝑝 (𝑈 )) then 𝑝 (𝑥) = 𝑝 (𝑤) for some
some 𝑤 ∈ 𝑈 . Then 𝑤 ∈ 𝑈𝑖 for some 𝑖, so 𝑥 ∈ 𝑝 −1 (𝑝 (𝑈𝑖 )) ⊂ 𝑈𝑖+1 ⊂ 𝑈 . Thus 𝑝 −1 (𝑝 (𝑈 )) ⊂ 𝑈 .
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Therefore 𝑈 = 𝜋 −1 (𝑝 (𝑈 )), so 𝑈 is saturated. Since 𝑝 is a quotient map, 𝑝 (𝑈 ) is open in 𝑌 . Now
𝑥 × 𝑧 ∈ 𝑈 × 𝑉 ⊂ 𝜋 −1 (𝐴), so 𝜋 (𝑈 × 𝑉 ) = 𝑝 (𝑈 ) × 𝑉 is an open set containing 𝑦 × 𝑧 and contained
in 𝜋 (𝜋 −1 (𝐴)) ⊂ 𝐴, so 𝐴 is open in 𝑌 × 𝑍 . We conclude that 𝜋 is a quotient map.
(b) Since 𝑝 : 𝐴 → 𝐵 is a quotient map and 𝐶 is locally compact Hausdorff, it follows from (a)
that 𝑝 × 𝑖𝐶 : 𝐴 × 𝐶 → 𝐵 × 𝐶 is a quotient map. Similarly, 𝑖 𝐵 × 𝑞 : 𝐵 × 𝐶 → 𝐵 × 𝐷 is a quotient map.
Therefore, 𝑝 × 𝑞 = (𝑖 𝐵 × 𝑞) ◦ (𝑝 × 𝑖𝐶 ) is a quotient map being the composite of quotient maps.
Supplementary Exercises: Nets
Supplementary Exercise 1. (a) If 𝑎 and 𝑏 are elements of a simply ordered set, then either 𝑎 ≤ 𝑏
or 𝑏 ≤ 𝑎, so that 𝑎 ≤ max{𝑎, 𝑏} and 𝑏 ≤ max{𝑎, 𝑏}.
(b) If 𝐴 ⊂ 𝑆 and 𝐵 ⊂ 𝑆, then 𝐴 𝐴 ∪ 𝐵 and 𝐵 𝐴 ∪ 𝐵.
(c) If 𝐴, 𝐵 ∈ 𝒜, then 𝐴 𝐴 ∩ 𝐵 and 𝐵 𝐴 ∩ 𝐵.
(d) Similarly as in (b), as finite unions of closed sets is closed.
Supplementary Exercise 2. We have that 𝐾 inherits the partial order of 𝐽 . Let 𝛽, 𝛾 ∈ 𝐾 . Then
there exists 𝛼 ∈ 𝐽 such that 𝛽 𝛼 and 𝛾 𝛼, and in turn 𝛿 ∈ 𝐾 such that 𝛼 𝛿. Thus 𝐾 is
directed.
Supplementary Exercise 3. The set Z+ is simply ordered, hence directed by Supplementary
Exercise 1(a). The given definitions are precisely the usual ones in the case of sequences.
Supplementary Exercise 4. Let 𝑈 be a neighbourhood of 𝑥 and 𝑉 a neighbourhood of 𝑦, so that
𝑈 × 𝑉 is a (standard) basis neighbourhood of 𝑥 × 𝑦 in 𝑋 × 𝑌 . By assumption, there exist 𝛼, 𝛽 ∈ 𝐽
such that 𝑥𝛾 ∈ 𝑈 for all 𝛾 𝛼 and 𝑦𝛾 ∈ 𝑈 for all 𝛾 𝛽. As 𝐽 is directed, there exists 𝛿 ∈ 𝐽 such
that 𝛿 𝛼 and 𝛿 𝛽. Then 𝑥𝛾 × 𝑦𝛾 ∈ 𝑈 × 𝑉 for all 𝛾 𝛿. It follows that (𝑥𝛼 × 𝑦𝛼 )𝛼 ∈𝐽 → 𝑥 × 𝑦 in
𝑋 ×𝑌.
Supplementary Exercise 5. Let (𝑥𝛼 )𝛼 ∈𝐽 be a net in 𝑋 and suppose that 𝑥𝛼 → 𝑥 and 𝑥𝛼 →
𝑦, where 𝑥 ≠ 𝑦. Since 𝑋 is Hausdorff, there exist disjoint neighbourhoods 𝑈 of 𝑥 and 𝑉 of 𝑦
respectively. Then there exist 𝛼, 𝛽 ∈ 𝐽 such that 𝑥𝛾 ∈ 𝑈 for all 𝛾 𝛼 and 𝑥𝛾 ∈ 𝑉 for all 𝛾 𝛽.
Since 𝐽 is directed, there is 𝛿 ∈ 𝐽 such that 𝛿 𝛼 and 𝛿 𝛽. But then 𝑥𝛾 ∈ 𝑈 ∩ 𝑉 = ∅ for all
𝛾 𝛿, a contradiction.
Supplementary Exercise 6. First suppose (𝑥𝛼 )𝛼 ∈𝐽 is a net in 𝑋 of points of 𝐴 converging to 𝑥 .
If 𝑈 is a neighbourhood of 𝑥, then it contains some 𝑥𝛼 ∈ 𝐴. Thus 𝑥 ∈ 𝐴. Conversely, assume
that 𝑥 ∈ 𝐴. We follow the hint. Let 𝒜 denote the collection of all neighbourhoods of 𝑥, partially
ordered by reverse inclusion. Given 𝑈 ∈ 𝒜, there exists 𝑥𝑈 ∈ 𝑈 ∩𝐴 by assumption. Then (𝑥𝑈 )𝑈 ∈𝒜
is a net in 𝑋 of points of 𝐴. Moreover, if 𝑉 is a neighbourhood of 𝑥, then 𝑥𝑈 ∈ 𝑉 for all 𝑈 ⊂ 𝑉 ,
that is, for all 𝑈 𝑉 . Thus 𝑥𝑈 → 𝑥 .
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Supplementary Exercise 7. First assume 𝑓 is continuous and let (𝑥𝛼 )𝛼 ∈𝐽 be a net in 𝑋 converging to 𝑥 . Let 𝑈 be a neighbourhood of 𝑓 (𝑥). Then 𝑓 −1 (𝑈 ) is a neighbourhood of 𝑥, so there exists
𝛼 ∈ 𝐽 such that 𝑥𝛾 ∈ 𝑓 −1 (𝑈 ) for all 𝛾 𝛼, so that 𝑓 (𝑥𝛾 ) ∈ 𝑈 for all such 𝛾 . Thus 𝑓 (𝑥𝛼 ) → 𝑓 (𝑥).
Conversely, suppose that 𝑓 is not continuous. Then there exists 𝑈 open in 𝑌 such that 𝑓 −1 (𝑈 )
is not open in 𝑋 . Then there exists 𝑥 ∈ 𝑓 −1 (𝑈 ) such that no neighbourhood of 𝑥 is contained
in 𝑓 −1 (𝑈 ), so that 𝑥 ∈ 𝑋 \ 𝑓 −1 (𝑈 ). It follows from Supplementary Exercise 6 that there is a net
(𝑥𝛼 )𝛼 ∈𝐽 in 𝑋 of points of 𝑋 \ 𝑓 −1 (𝑈 ) converging to 𝑥 . As no point 𝑓 (𝑥𝛼 ), 𝛼 ∈ 𝐽, belongs to 𝑈 , it
follows that the net (𝑓 (𝑥𝛼 ))𝛼 ∈𝐽 in 𝑌 does not converge to 𝑓 (𝑥) ∈ 𝑈 .
Supplementary Exercise 8. Let 𝑓 : 𝐽 → 𝑋 be a be a net in 𝑋, 𝑓 (𝛼) = 𝑥𝛼 for 𝛼 ∈ 𝐽, converging
to 𝑥 and let 𝑔 : 𝐾 → 𝐽 be a function form 𝐾 a directed set such that 𝑓 ◦ 𝑔 : 𝐾 → 𝑋 is a subnet of 𝑓 .
Let 𝑈 be a neighbourhood of 𝑥 in 𝑋 . Then there exists 𝛼 ∈ 𝐽 such that 𝑥𝛾 ∈ 𝑈 for all 𝛾 𝛼 . Since
𝑔(𝐾) is cofinal in 𝐽, there exists 𝛽 ∈ 𝐾 such that 𝑔(𝛽) 𝛼 . Thus, for all 𝛾 ∈ 𝐾 such that 𝛾 𝛽, we
have 𝛼 𝑔(𝛽) 𝑔(𝛾), so that 𝑥𝑔 (𝛾 ) ∈ 𝑈 for all such 𝛾 . It follows that 𝑓 ◦ 𝑔 converges to 𝑥 .
Supplementary Exercise 9. Let (𝑥𝛼 )𝛼 ∈𝐽 be a net in 𝑋 . first assume that the net has the point 𝑥 as
an accumulation point. For each neighbourhood 𝑈 of 𝑥, let 𝐽𝑈 denote the set of those 𝛼 for which
𝑥𝛼 ∈ 𝑈 , which is cofinal in 𝐽 by assumption. We follow the hint. Let 𝐾 be the set of all pairs (𝛼, 𝑈 )
where 𝛼 ∈ 𝐽 and 𝑈 is a neighbourhood of 𝑥 containing 𝑥𝛼 . Define (𝛼, 𝑈 ) (𝛽, 𝑉 ) if 𝛼 𝛽 and
𝑉 ⊂ 𝑈 . Then 𝐾 is partially ordered. Moreover, let (𝛼, 𝑈 ) and (𝛽, 𝑉 ) belong to 𝐾 . As 𝐽 is directed,
there exists 𝛾 ∈ 𝐽 such that 𝛾 𝛼 and 𝛾 𝛽. Since 𝐽𝑈 ∩𝑉 is cofinal in 𝐽, there exists 𝛿 ∈ 𝐽𝑈 ∩𝑉 such
that 𝛿 𝛾 . Then (𝛼, 𝑈 ) (𝛿, 𝑈 ∩ 𝑉 ) and (𝛽, 𝑉 ) (𝛿, 𝑈 ∩ 𝑉 ) and therefore 𝐾 is directed. Now
define 𝑔 : 𝐾 → 𝐽 by 𝑔(𝛼, 𝑈 ) = 𝛼 . Then clearly 𝑔(𝛼, 𝑈 ) 𝑔(𝛽, 𝑉 ) for (𝛼, 𝑈 ) (𝛽, 𝑉 ). Moreover,
given 𝛼 ∈ 𝐽 and 𝑈 any neighbourhood of 𝑥, then we may find 𝛽 ∈ 𝐽𝑈 such that 𝛼 𝛽 = 𝑔(𝛽, 𝑈 ).
Thus 𝑔(𝐾) = 𝐽 is cofinal in 𝐽 . It remains to prove that the subnet (𝑥𝑔 (𝛼,𝑈 ) )(𝛼,𝑈 ) ∈𝐾 of (𝑥𝛼 )𝛼 ∈𝐽
converges to 𝑥 . So let 𝑈 be a neighbourhood of 𝑥 . Then there exists 𝛼 ∈ 𝐽𝑈 , so that (𝛼, 𝑈 ) ∈ 𝐾 . If
(𝛽, 𝑉 ) (𝛼, 𝑈 ), then 𝑥𝑔 (𝛽,𝑉 ) = 𝑥 𝛽 ∈ 𝑉 ⊂ 𝑈 . It follows that 𝑥𝑔 (𝛼,𝑈 ) → 𝑥 .
Conversely, assume that some subnet (𝑥𝑔 (𝛽) )𝛽 ∈𝐾 (for some 𝐾 directed and 𝑔 : 𝐾 → 𝐽 ) of
(𝑥𝛼 )𝛼 ∈𝐽 converges to 𝑥 . Let 𝑈 be a neighbourhood of 𝑥 and let 𝐽𝑈 denote the set of those 𝛼 ∈ 𝐽
for which 𝑥𝛼 ∈ 𝑈 . By assumption, there exists 𝛽 ∈ 𝐾 such that 𝑥𝑔 (𝛾 ) ∈ 𝑈 for all 𝛾 𝛽. Let 𝛼 ∈ 𝐽 .
By cofinality of 𝑔(𝐾) in 𝐽, there exists 𝛿 ∈ 𝐾 such that 𝑔(𝛿) 𝛼 . Then, taking 𝛾 ∈ 𝐾 such that
𝛾 𝛽 and 𝛾 𝛿, we have 𝑔(𝛾) ∈ 𝐽𝑈 and 𝑔(𝛾) 𝑔(𝛿) 𝛼 . It follows that 𝐽𝑈 is cofinal in 𝐽 . We
deduce that 𝑥 is an accumulation point of the net (𝑥𝛼 )𝛼 ∈𝐽 .
Supplementary Exercise 10. We follow the hint. First assume that 𝑋 is compact and let (𝑥𝛼 )𝛼 ∈𝐽
be a net in 𝑋 . Given 𝛼 ∈ 𝐽, let 𝐵𝛼 = {𝑥 𝛽 | 𝛼 𝛽}. Given indices 𝛼 1, . . . , 𝛼𝑛 ∈ 𝐽, there exists 𝛾 such
that 𝛾 𝛼𝑖 for all 𝑖 = 1, . . . , 𝑛, so that 𝑥𝛾 ∈ ∩𝑛𝑖=1 𝐵𝛼𝑖 . Thus, the collection {𝐵𝛼 }𝛼 ∈𝐽 has the finite
intersection property. Since 𝑋 is compact, it follows from Theorem 26.9 that there exists a point
𝑥 ∈ ∩𝛼 𝐵𝛼 . Let 𝑈 be a neighbourhood of 𝑥 . Then, given 𝛼 ∈ 𝐽 we have 𝑥 ∈ 𝐵𝛼 , so there exists
𝛽 𝛼 such that 𝑥 𝛽 ∈ 𝑈 . Thus 𝑥 is an accumulation point of the net (𝑥𝛼 )𝛼 ∈𝐽 by definition, and it
follows from Supplementary Exercise 9 that the net (𝑥𝛼 )𝛼 ∈𝐽 has a convergent subnet.
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Local Compactness
Conversely, suppose every net in 𝑋 has a convergent subnet. Let 𝒜 be a collection of closed
sets having the finite intersection property, and let ℬ denote the collection of all finite intersections of elements of 𝒜, partially ordered by reverse inclusion. Then, for each 𝐵 ∈ ℬ there exists a
point 𝑥 𝐵 ∈ 𝐵. Then (𝑥 𝐵 )𝐵 ∈ℬ is a net in 𝑋 . By assumption and Supplementary Exercise 9, (𝑥 𝐵 )𝐵 ∈ℬ
has an accumulation point 𝑥 . Let 𝐴 ∈ 𝒜 ⊂ ℬ and let 𝑈 be a neighbourhood of 𝑥 . Then there
exists 𝐵 ∈ ℬ such that 𝑥 𝐵 ∈ 𝑈 and 𝐴 𝐵, i.e. 𝐵 ⊂ 𝐴. Thus 𝑥 𝐵 ∈ 𝑈 ∩ 𝐴. Since 𝐴 and 𝑈 were
arbitrary, it follows that 𝑥 ∈ 𝐴 = 𝐴 for all 𝐴 ∈ 𝒜, so that 𝑥 ∈ ∩𝐴∈𝒜 𝐴. We conclude that 𝑋 is
compact by Theorem 26.9.
Supplementary Exercise 11. (Note that this was proved in Exercise 26.13(a).) Let 𝑥 ∈ 𝐴 · 𝐵. By
Supplementary Exercise 6, there is a net (𝑥𝛼 )𝛼 ∈𝐽 of points of 𝐴 · 𝐵 converging to 𝑥 . For each 𝛼 ∈ 𝐽,
we have 𝑥𝛼 = 𝑦𝛼 · 𝑧𝛼 for some 𝑦𝛼 ∈ 𝐴 and 𝑧𝛼 ∈ 𝐵. Now (𝑧𝛼 )𝛼 ∈𝐽 is a net in 𝐵, so by Supplementary
Exercise 10 it has a convergent subnet (𝑧𝑔 (𝛽) )𝛽 ∈𝐾 (for some 𝐾 directed and 𝑔 : 𝐾 → 𝐽 ), say 𝑧𝑔 (𝛽) →
𝑏 ∈ 𝐵. Recall from Supplementary Exercise 1 of Chapter 2 that 𝐺 × 𝐺 → 𝐺, (𝑔 × ℎ) ↦→ 𝑔 · ℎ −1,
is continuous. It follows from Supplementary Exercise 7 that (𝑦𝑔 (𝛽) )𝛽 ∈𝐾 = (𝑥𝑔 (𝛽) · (𝑧𝑔 (𝛽) ) −1 )𝛽 ∈𝐾
converges to 𝑥 · 𝑏 −1 . Since 𝐴 is closed, we have 𝑎 B 𝑥 · 𝑏 −1 ∈ 𝐴 by Supplementary Exercise 6, so
that 𝑥 = 𝑎 · 𝑏 ∈ 𝐴 · 𝐵. We conclude that 𝐴 · 𝐵 is closed in 𝐺 .
Supplementary Exercise 12. None of the solutions given above for these exercises made use
of condition (2).
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