Download 1. Theorem: If (X,d) is a metric space, then the following are

1. Theorem: If (X,d) is a metric space, then the following are equivalent:
(a) X is second countable
(b) X is Lindelöf.
(c) X is separable.
Proof:
(a) → (b) and (a) → (c) are true for general topological spaces by the tree from our notes. Therefore
they are true for metric spaces.
(c) → (a): Suppose X is Separable.
Let A ⊆ X be a countable, dense subset of X. Define: B = {Bd (a, 1/n)|a ∈ A, n ∈ Z + }. B is countable
since it is countable by countable. We now show that B is a basis.
Let x ∈ Ux ⊆ X, where Ux is open in X. Then we can find some > 0 s.t. x ∈ Bd (x, ) ⊆ Ux . Since
A is dense in X, we can find some a ∈ A s.t. a ∈ Bd (x, /2) ⊆ Bd (x, ). Let n∗ ≥ 2/. Then given
y ∈ Bd (a, 1/n∗ ), we have that d(x, y) ≤ d(x, a) + d(a, y) < /2 + 1/n∗ ≤ /2 + /2 = . Therefore
y ∈ Bd (x, ) ⊆ Ux . Hence x ∈ Bd (a, 1/n∗ ) ⊆ Bd (x, ) ⊆ Ux and by Lemma 13.2 B is a countable basis for
(X,d) and therefore X is second countable.
(b) → (a): Suppose X is Lindelöf.
Then for each n ∈ Z + define Un = {Bd (x, 1/n)|x ∈ X}. Now, clearly each Un covers X, so we can find a
countable subcollection of each that covers X. Also, since n is fixed for each Un , the countability of the
subcollection must come from a countable subset, Dn ⊆ X, over which we now index. Thus the countable
subcollections look like: Un0 = {Bd (x, 1/n)|x ∈ Dn }.
Now, define D =
S
n∈Z +
Dn . Then D is countable since it is a countable union of countable sets. We
simply need to show that ClX (D) = X.
Clearly ClX (D) ⊆ X. We just need to show that X ⊆ ClX (D).
Take x ∈ X. We want Ux
T
D 6= ∅ for all open Ux containing x. It will suffice to show that Bd (x, )
T
D 6= ∅
+
for all > 0. By the construction of D, we have that ∀n ∈ Z , ∃y ∈ Dn such that x ∈ Bd (y, 1/n).
Set n∗ ≥ 1/. Then x ∈ Bd (y, 1/n∗ ) for some y ∈ Dn∗ , thus y ∈ Bd (x, 1/n∗ ) ⊆ Bd (x, ). And thus
∅ =
6 Bd (x, )
T
Dn∗ ⊆ Bd (x, )
T
D. Therefore x ∈ ClX (D) and so X ⊆ ClX (D). Therefore D is a
countable dense subset of X, so X is separable, and by the above X is second countable.
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