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first countable implies compactly generated∗ CWoo† 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 † This text is available under the Creative Commons Attribution/Share-Alike License 3.0. You can reuse this document or portions thereof only if you do so under terms that are compatible with the CC-BY-SA license. 1