Math 3240
Topology 1, final exam study questions
Q
Question: Let {Xi }i∈N be separable spaces. Show that X = i∈N Xi is also separable.
Solution: Let Di be dense Q
in Xi for all
Q
Q i. We need to produce a countable dense subset in
i∈N Xi , and we cannot use
i∈N Di ⊂
i∈N Xi because a countable product of countable sets is
not countable.
So for each i, choose pi ∈ Xi . For n ∈ N, define Bn according to
Y
Y
Bn =
Di ×
{pi }
i≤n
i>n
Q
which consists of sequences (xi ) ∈ i∈N Xi for which xi ∈ Di if i ≤ n and xi = pi otherwise. Each
Bn is a product
of finitely many countable sets and infinitely many singletons, and so it is countable.
S
Let B = n∈N Bn , which is also countable since it is a countable union of countable sets.
Next we show that B
Q is dense in X. It suffices to show that any basic open set intersects B
nontrivially, so let U = i∈N Ui be a basic open set in the product topology (therefore Ui = Xi for
all but finitely many i). Choose N ∈ N such that for all i > N we have Ui = Xi .
Since Di is dense in Xi , there exists di ∈ Ui ∩ Di for all i ≤ N . Set a = (ai ) where ai = di if
i ≤ N and ai = pi if i > N . Then a ∈ BN ⊂ B and by construction, a ∈ U . Therefore B ∩ U 6= ∅.
Box
Q
Question: Show that if the family {Xi }i∈I is Hausdorff, then so is i∈I Xi .
Box
Q
Solution: Let (zi ), (wi ) ∈ i∈I Xi be given, assume they are distinct points. Since Xi is Hausdorff
for all i, whenever wi 6= zi choose open sets Ui and Vi such that zi ∈ Ui , wi ∈ Vi and Ui ∩ Vi = ∅ for
all i. If wi = zi then set Ui = Vi = Xi . Then set
Y
Y
U=
Ui , V =
Vi
i∈I
i∈I
Since (zi ) 6= (wi ) there exists i ∈ I such that zi 6= wi and so Ui ∩ Vi = ∅. Therefore U ∩ V = ∅, and
Box
Q
moreover (zi ) ∈ U and (wi ) ∈ V , and therefore i∈I Xi is Hausdorff.
Question:If X is an n-manifold and A ⊆ X is an open subspace of X, show that A is also an
n-manifold.
Solution: Show that A is Hausdorff: Choose any x, y ∈ A : x 6= y. Since x, y ∈ X and X is
Hausdorff, ∃ Ux , Uy ⊆ X such that x ∈ Ux , y ∈ Uy , and Ux ∩ Uy = ∅. Now let Vx = Ux ∩ A and
Vy = Uy ∩ A. By definition of the subspace topology, Vx and Vy are open in A. Also, x ∈ Vx , y ∈ Vy ,
and Vx ∩ Vy = ∅. So A is Hausdorff.
Show that A is separable: Since X is separable, it has a countable dense set D. Choose any
open V ⊆ A. By definition of the subspace topology, ∃ open U ⊆ X such that V = U ∩ A. Since
A is open in X and finite intersections of open sets are open, V is open in X. Since D is dense in
X, V ∩ D 6= ∅. Note that since V ⊆ A, we have V ∩ D ∩ A = V ∩ D 6= ∅. So D ∩ A is dense in
A. Also, since D ∩ A ⊆ D and D is countable, D ∩ A is also countable. Since D ∩ A is a countable
dense subset of A, A is separable.
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Finally, choose any x ∈ A. Since x ∈ X, it has an open neighbourhood U in X which is
homeomorphic to the open unit n-ball B n (0, 1), say with the homeomorphism h : U → B n (0, 1).
Now let V = U ∩ A. Since A is open in X, V must be open in U by definition of the subspace
topology. Since h is a homeomorphism, h(V ) is open in B n (0, 1). So there is some open n-ball
B n (a, r) such that h(x) ∈ B n (a, r) ⊆ h(V ). Let S = h−1 (B n (a, r)). Since h(x) ∈ B n (a, r) ⊆ h(V ),
we have x ∈ S ⊆ V ⊆ A. Note also that S is open in U , since B n (a, r) is open in B n (0, 1) and
h is continuous. Clearly S ' B n (a, r) under the homeomorphism h, and B n (a, r) ' B n (0, 1) since
they are both open n-balls. So S is an open neighbourhood of x in A which is homeomorphic to
B n (0, 1).
Question:Let X and Y be topological spaces, and ∼ be an equivalence relation on X with the
quotient map q : X → X/ ∼. Show that f : X/ ∼ → Y is continuous if and only if f ◦ q : X → Y
is continuous.
Solution:(⇒) Let f be continuous. By definition of the quotient topology, q is continuous (for any
open U ⊆ X/ ∼, q −1 (U ) is open in X). So the composition f ◦ q is continuous.
(⇐) Let f ◦ q be continuous. Choose any open U ⊆ Y . So (f ◦ q)−1 (U ) is open in X. Since
(f ◦ q)−1 = q −1 ◦ f −1 , q −1 (f −1 (U )) is open in X. By definition of the quotient topology, f −1 (U ) must
be open in X/ ∼. So f is continuous.
Question: Let A and B be closed subsets of a space X, with A ∩ B and A ∪ B connected subsets
of X. Show that both A and B are connected.
Solution: WLOG for contradiction we can suppose that one of the two sets has a separation, say
A is not connected and has a separation {S, T }. Then
A ∩ B = (S ∪ T ) ∩ B = (S ∩ B) ∪ (T ∩ B)
Since A ∩ B is connected, we know that at least one of S ∩ B or T ∩ B is empty. Suppose it is
T ∩ B that is empty. Then {S ∪ B, T } is a pair of disjoint nonempty sets whose union is A ∪ B, for
contradiction we need only show that they are either both closed or both open.
It suffices to observe that S and T are closed in A since they are sets in a separation, and since
A is closed they are therefore both closed in X. Thus S ∪ B and T are closed, a contradiction.
Question and solution: http://bit.ly/1mbTgMy ←clickable
www.proofwiki.org/wiki/Connected Open Subset of Euclidean Space is Path-Connected
Question: Prove the Extreme Value Theorem, which states: If X is compact, and f : X → R is
continuous, then f attains a maximum and minimum value on X.
Solution: Let f : X → R be continuous on X. Since X is compact, f (X) is a compact set in R. By
the Heine-Borel Theorem, f (X) is closed and bounded. Denote α = sup f (X) and β = inf f (X).
Then since f (X) is closed, we have α, β ∈ f (X), and so there are x1 , x2 such that f (x1 ) = α and
f (x2 ) = β. That is, f attains its maximum and minimum on X.
Question: Determine whether each of the following spaces are: compact, countably compact, or
Lindelof.
(a) The space W = N ∪ {x1 } ∪ {x2 } with the discrete topology on N, and the only open sets
containing x1 or x2 are N ∪ {x1 } and N ∪ {x2 }, respectively.
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(b) The space X = R, where open sets are any sets whose complement is countable or includes 0.
(c) The space Y = R with the compact complement topology- that is, sets are open if their
complement is compact in the usual Euclidean topology.
(d) The space Z = Z with the evenly spaced integer topology- that is the topology generated by
sets of the form a + kZ = {a + λk : λ ∈ Z} for a ∈ Z, k ∈ Z, k 6= 0.
Solution:
(a) Any open cover of X must include open sets covering x1 and x2 , so contains N ∪ {x1 } and
N ∪ {x2 }. Hence X is covered by the (finite) subcover that is these two sets, so is compact,
and therefore countably compact and Lindelof.
(b) W is Lindelof, since if C is an open cover, and U ∈ C, then W − U is countable and hence can
be covered by at most a countable subcover from C. However, it is not countably compact,
since the covering C = {R − {Z − {0}}} ∪ {z} : z ∈ Z is a countable cover of W with no finite
subcover. Hence W is not compact.
(c) If C is an open cover for Y , and U ∈ C, then by definition R − U is compact in the regular
Euclidean topology. Then a finite number of elements from C must cover R − U , so we have
a finite subcover of R from C. Thus Y is compact and therefore countably compact and
Lindelof.
(d) If P is the infinite set of all primes congruent to 3 (mod 4), then P is an infinite set with no
accumulation point, so Z is not countably compact, nor compact. However, it is Lindelof,
since if C is an open cover for Z, then C contains an open set covering each z ∈ Z, so we can
choose a countable subcover.
Question: Let D be a dense subset of X, let Y be Hausdorff, and let f, g : X → Y be continuous
maps. Show that if f |D = g|D , then f = g.
Solution: Suppose not, so that there is an x ∈ X such that f (x) 6= g(x). As y is Hausdorff,
there exists open neighbourhoods U , V of f (x) and g(x) respectively such that U ∩ V = ∅. Then
by continuity we get open neighbourhoods f −1 (U ) and g −1 (U ) of X, so that f −1 (U ) ∩ g −1 (U ) is
nonempty. Thus f −1 (U ) ∩ g −1 (U ) contains a point d of the dense set D.
But then f (d) ∈ U and g(d) ∈ V , and f (d) = g(d) since f |D = g|D . So f (d) = g(d) ∈ U ∩ V , a
contradiction.
Question: Show that f : X → Y is continuous if and only if f (A) = f (A) for all A ⊂ X.
Solution: (⇒) First suppose that f is continuous and x ∈ A. We want to show that for every
open neighbourhood U of f (x), we have U ∩ f (A) 6= ∅. Let U be an open neighbourhood of f (x).
Then f −1 (U ) is an open neighbourhood of x ∈ A. Therefore
A ∩ f −1 (U ) 6= ∅ ⇒ f (A) ∩ U 6= ∅ ⇒ x ∈ f (A) ⇒ f (A) ⊂ f (A).
(⇐) Suppose that for all A ⊂ X, we have f (A) ⊂ f (A). We will show that for all closed sets
U ⊂ Y , f −1 (U ) is closed in X, which is equivalent to saying that the preimage of open sets is open
(and thus the map will be continuous).
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As A ⊂ A, we also have U = f (A) ⊂ f (A). Thus, U = f (A) and so A ⊂ A, so A = f −1 (U ) is
closed. Thus f is continuous.