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first countable implies compactly
2013-03-22 3:22:20
Proposition 1. Any first countable topological space is compactly generated.
Proof. Suppose X is first countable, and A ⊆ X has the property that, if C
is any compact set in X, the set A ∩ C is closed in C. We want to show tht
A is closed in X. Since X is first countable, this is equivalent to showing that
any sequence (xi ) in A converging to x implies that x ∈ A. Let C = {xi | i =
1, 2, . . .} ∪ {x}.
Lemma 1. C is compact.
Proof. Let {Uj | j ∈ J} be a collection of open sets covering C. So x ∈ Uj for
some j. Since Uj is open, there is a positive integer k such that xi ∈ Uj for all
i ≥ k. Now, each xi ∈ Ud(i) for i = 1, . . . , k. So C is covered by Ud(1) , . . . , Ud(k) ,
and Uj , showing that C is compact.
In addition, as a subspace of X, C is also first countable. By assumption,
A ∩ C is closed in C. Since xi ∈ A ∩ C for all i ≥ 1, we see that x ∈ A ∩ C as
well, since C is first countable. Hence x ∈ A, and A is closed in X.
∗ hFirstCountableImpliesCompactlyGeneratedi
created: h2013-03-2i by: hCWooi version:
h42065i Privacy setting: h1i hExamplei h54E99i
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