Math 3210-3
HW 10
Solutions
NOTE: You are only required to turn in problems 1-5, and 8-9.
Topology of the Reals
1. Find the interior of each set.
(a) n1 : n ∈ N
1
Let n ∈ N. Since the irrationals are dense in R, there exists an i ∈ R r Q such that n+1
<i<
1
1
Thus for all neighborhoods N of n , N * { n : n ∈ N}. Therefore, the interior of this set is ∅.
(b) [0, 3] ∪ (3, 5)
The interior is (0, 5).
√
(c) ♣ {r ∈ Q : 0 < r < 2}
The interior is ∅ since the irrationals are dense in R.
√
(d) {r ∈ Q : r ≥ 2}
Once again, the interior is ∅.
(e) [0, 2] ∩ [2, 4]
Since [0, 2] ∩ [2, 4] = {2}, the interior is ∅.
2. ♣ Find the boundary of each set in exercise 1.
(a) { n1 : n ∈ N} ∪ {0}
(b) {0, 5}
√
(c) [0, 2]
√
(d) [ 2, ∞)
(e) {2}
3. Classify each of the following sets as open, closed, neither, or both.
(a) n1 : n ∈ N neither
(b) N closed
(c) Q neither
1
(d) ♣ ∩∞
n=1 0, n both. This is the empty set in disguise.
(e) x : |x − 5| ≤ 21 closed
(f) {x : x2 > 0} open. This is R r {0}.
4. ♣ Find the closure of each set in exercise 3.
(a) { n1 : n ∈ N} ∪ {0}
(b) N
(c) R
(d) ∅
(e) {x : |x − 5| ≤ 21 }
(f) R
1
n.
5. If A is open and B is closed, prove that A r B is open and B r A is closed.
Proof: By Theorem 5, we know A r B = A ∩ (R r B), and the intersection of two open sets is open
by Theorem 42. For the second equality, we have B r A = B ∩ (R r A), and the intersection of two
closed sets is closed by Corollary 1.
˜
6. Let S and T be subsets of R. Prove the following:
(a) cl(cl S) = cl S
Proof: By Theorem 43(b) and (c) we have cl S is closed, so cl S = cl(cl S).
˜
(b) cl(S ∪ T ) = (cl S) ∪ (cl T )
Proof: First we will show that (S ∪ T )′ = S ′ ∪ T ′ . So x ∈ (S ∪ T )′ iff for all deleted neighborhoods
N of x, N ∩ (S ∪ T ) 6= ∅ iff (N ∩ S) ∪ (N ∩ T ) 6= ∅, by Theorem 6(e), iff N ∩ S 6= ∅ or N ∩ T 6= ∅
iff x ∈ S ′ or x ∈ T ′ .
Now to prove our statement. We know cl(S ∪ T ) = (S ∪ T ) ∪ (S ∪ T )′ = S ∪ T ∪ S ′ ∪ T ′ =
(S ∪ S ′ ) ∪ (T ∪ T ′ ) = cl S ∪ cl T .
˜
(c) cl(S ∩ T ) ⊆ (cl S) ∩ (cl T )
Proof: First we will show that (S ∩ T )′ ⊆ S ′ ∩ T ′ . Let x ∈ (S ∩ T )′ . Then for every deleted
neighborhood N of x, N ∩ (S ∩ T ) 6= ∅. Thus (N ∩ S) ∩ (N ∩ T ) 6= ∅, and x ∈ S ′ ∩ T ′ .
To prove the statement, we have
cl(S ∩ T ) =
⊆
=
=
⊆
=
(S ∩ T ) ∪ (S ∩ T )′
′
by definition of closure
′
(S ∩ T ) ∪ (S ∩ T ) by what we just proved
[(S ∩ T ) ∪ S ′ ] ∩ [(S ∩ T ) ∪ T ′ ] by Theorem 6, part 5
[(S ∪ S ′ ) ∩ (T ∪ S ′ )] ∩ [(S ∪ T ) ∩ (T ∪ T ′ )] by Theorem 6, part 4
(S ∪ S ′ ) ∩ (T ∪ T ′ ) by a property of intersection
(cl S) ∩ (cl T )
by definition of closure
˜
(d) Find an example to show that equality need not hold in part (c).
Let S = (0, 1) and let T = (1, 2). Then S ∩ T = ∅, cl(S ∩ T ) = ∅, but (cl S) ∩ (cl T ) = {1}.
7. For any set S ⊆ R, let S denote the intersection of all the closed sets containing S.
(a) Prove that S is a closed set.
Proof: Let A = {Aα : Aα ⊇ S and Aα is closed}. Then S = ∩A which is closed by Corollary 1.
˜
(b) Prove that S is the smallest closed set containing S. That is, show that S ⊆ S, and if C is any
closed set containing S, then S ⊆ C.
Proof: Let x ∈ S. Then x ∈ Aα for every α, so x ∈ ∩A = S. Thus S ⊆ S. Now suppose there is
some closed set C with S ⊆ C ⊆ S. But C ⊇ S, which implies that C ∈ A, so S ⊆ C =⇒ S = C.
Therefore, S is the smallest closed set containing S.
˜
(c) Prove that S = cl s.
Proof: We know that cl S is closed, and cl S = S ∪ S ′ , which implies that S ⊆ cl S =⇒ S ⊆ cl S.
Conversely, let x ∈ cl S. If x ∈ S, then x ∈ S. If x ∈ S ′ r S, then every neighborhood N of x
intersects S. Thus x ∈ Aα for every α. Therefore x ∈ S. And we can finally say that cl S = S.
˜
(d) If S is bounded, prove that S is bounded.
Proof: Let m, n ∈ R such that m ≤ s, n ≥ s for all s ∈ S. Then [m − 1, n + 1] ⊇ S =⇒ S ⊆
[m − 1, n + 1] =⇒ m − 1 is a lower bound for S, and n + 1 is an upper bound for S. Therefore S
is bounded.
˜
Compact Sets
8. Show that each subset of R is not compact by describing an open cover for it that has no finite subcover.
(a) [1, 3)
Let F = {An } where An = (0, 3 − n1 ), for n ∈ N. Then F is an open cover for [1, 3), which has no
finite subcover.
(b) N
Let G = {Bn } where Bn = (−n, n) for n ∈ N. Then G is an open cover of N which has no finite
subcover.
(c) ♣ { n1 : n ∈ N}
Let H = {Cn } where Cn = ( n1 , 2) for n ∈ N. Then H is an open cover of our set which has no
finite subcover.
(d) {x ∈ Q : 0 ≤ x ≤ 2}
Because the rationals
are dense in R we can construct an infinite collection
of rational points that
√
√
get close to 2, but never reach it. Let x1 = 21 . Since x1 and 2 are real numbers, there is
a rational number
√ number
√ between them, call it x2 . Now since x2 ∈ R, there is some rational
between x2 and 2, call it x3 . Continue to get x1 < x2 < x3 < · · · < xn < · · · < 2. Now let
An = (1, xn ). Then J = {An } is an open cover of our set, but it has no finite subcover.
9. ♣ Prove that the intersection of any collection of compact sets is compact.
Proof: Let A = {Ai } be a collection of compact sets. Then by the Heine-Borel Theorem, each Ai is
closed and bounded. By Corollary 1, the intersection of a collection of closed sets is closed, so ∩A is
closed. Now we must just show ∩A is bounded. Let m, n ∈ R with m a lower bound for A1 and n
an upper bound for A1 . Since ∩A ⊆ A1 m must be a lower bound for ∩A, and n must be an upper
bound for ∩A. Hence ∩A is closed and bounded, so by the Heine-Borel Theorem, ∩A is compact.
˜