Midterm Solutions
Problem 1(a)
Proposition. Let f : X → Y be a continuous surjection, and suppose that A is dense in X.
Then f (A) is dense in Y .
Proof. By part (2) of Theorem 18.1, we know that f (A) ⊃ f (Ā). But Ā = X since A is
dense, and f (X) = Y since f is surjective, so f (A) ⊃ Y .
Problem 1(b)
Proposition. If A and B are dense open subsets of X, then A ∩ B is dense in X.
Proof. Let x ∈ X, and let U be a neighborhood of x. Since A is dense, x lies in the closure
of A, so U intersects A at a point y. Note that U ∩ A is a neighborhood of y, being the
intersection of two open sets. Since B is dense, y lies in the closure of B, so U ∩A intersects B
at a point z. Then z ∈ U ∩ A ∩ B, which proves that U intersects A ∩ B, and therefore x
lies in the closure of A ∩ B.
Problem 1(c)
Proposition. Let X be a metric space, and suppose that X has a countable dense subset.
Then X has a countable basis for its topology.
Proof. Let A be a countable dense subset of X, and let
B = {B(a, 1/n) | a ∈ A and n ∈ Z+ }.
Clearly B is countable. We claim that B is a basis for the topology on X.
Let x ∈ X, and let U be a neighborhood of x. Then there exists a δ > 0 so that
B(x, δ) ⊂ U . Let n ∈ Z+ so that 1/n < δ/2. Since A is dense, x lies in the closure of A, so
B(x, 1/n) intersects A at some point a. Then d(x, a) < 1/n, so x ∈ B(a, 1/n). Moreover, if
y is any point in B(a, 1/n), then
d(y, x) ≤ d(y, a) + d(a, x) <
1
1
δ δ
+
< + = δ,
n n
2 2
and thus B(a, 1/n) ⊂ B(x, δ) ⊂ U . By Lemma 13.2, we conclude that B is a basis for the
topology on X.
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Problem 1(d)
Proposition. Let X be the space Rω under the product topology. Then X has a countable
dense subset.
Proof. Let A be the set
{(x1 , x2 , x3 , . . .) ∈ Qω | xn = 0 for all but finitely many n}.
We claim that A is countable and dense in X.
To prove that A is countable, observe that A =
S∞
n=1
An , where
An = {(x1 , x2 , x3 , . . .) ∈ A | xk = 0 for all k > n}.
But each An is countable, since (x1 , . . . , xn ) 7→ (x1 , . . . , xn , 0, 0, . . .) is a bijection from Qn
to An . It follows that A is countable, being a countable union of countable sets.
To prove that A is dense, let x be a point in X, and let U be a basic open set containing x.
Then U has the form
U = S(1, U1 ) ∩ S(2, U2 ) ∩ · · · ∩ S(n, Un )
for some n ∈ Z+ and some open sets U1 , . . . , Un in R. Since Q is dense in R, each xi lies in the
closure of Q, and hence each open set Ui contains a point qi ∈ Q. Then (q1 , . . . , qn , 0, 0, . . .)
is an element of A ∩ U , which proves that x lies in the closure of A.
Problem 2(a)
Proposition. Let X be a topological space, and let A ⊂ X. Then A is discrete in X if and
only if A is closed in X and the topology that A inherits as a subspace of X is the discrete
topology.
Proof. Suppose first that A is discrete in X. Since A has no limit points, no point of X − A
is a limit point of A, and therefore A is closed. Moreover, if a ∈ A, then a cannot be a limit
point of A, so there must exist a neighborhood U of a that contains no other points of A;
then U ∩ A = {a} is open in A, which proves that the subspace topology on A is the discrete
topology.
For the converse, suppose that A is closed and the subspace topology on A is discrete.
Since A is closed, no point of X − A can be a limit point of A. Furthermore, if a ∈ A, then
{a} is open in A, so {a} = A ∩ U for some open set U in X, and hence a is not a limit point
of A. Then A has no limit points, so A is discrete in X.
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Problem 2(b)
Proposition. The set A = {1/n | n ∈ Z+ } is not discrete in R, but the subspace topology
on A is the discrete topology.
Proof. Note that 0 is a limit point of A, since each ball B(0, δ) = (−δ, δ) contains a point
of A. Thus A is not discrete in R. However, {1} = A ∩ (1/2, ∞) is open in A and
1
1
{1/n} = A ∩
,
n+1 n−1
is open in A for all n > 1, so the subspace topology on A is the discrete topology.
Problem 2(c)
Proposition. Let X be a metric space and let A ⊂ X. Suppose there exists an > 0 so that
d(a1 , a2 ) ≥ for every pair of distinct points a1 , a2 ∈ A. Then A is discrete in X.
Proof. Suppose to the contrary that A has at least one limit point x, and consider the
neighborhood U = B(x, /2) of x. Since X is a metric space, it is Hausdorff, so it follows
from Theorem 17.9 that U contains infinitely many points of A. Let a1 and a2 be two such
points. Then
d(a1 , a2 ) ≤ d(a1 , x) + d(x, a2 ) <
+ = ,
2 2
a contradiction.
Problem 2(d)
Proposition. Let X be the space Rω under the uniform topology. Then X has an uncountable discrete subset.
Proof. Let A be the set {0, 1}ω . Then A is uncountable by Theorem 7.7. Moreover, it is
easy to see that ρ̄(x, y) = 1 for every pair of distinct points x, y ∈ A, so A is discrete in X
by part (c).
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