4

I'm studying the Topology-Munkres book and there is a Theorem that states that Every compact subspace of a Hausdorff space is closed and I was wondering if there is any example where the "Hausdorff" condition is not needed, I mean could you give an example where the statement is fulfill only by Every compact subspace of a space $X$ is closed

Is just that I'm really bad finding examples and I want to know this, thank you.

user89935
  • 61
  • If you have "Every compact subspace of a space X is closed"... every point is closed (since is compact). So, you find that X is Hausdorff. – user99126 Oct 08 '13 at 11:18
  • 3
    @user99126: You are almost right, except that closed points means T1, not Hausdorff. – Moishe Kohan Oct 08 '13 at 11:58

2 Answers2

4

Let $X$ be an uncountable set with the cocountable topology, i.e., say that $A\subset X$ is closed iff $A=X$ or $A$ is countable. Certainly $X$ is not Hausdorff.

Which subsets $A\subset X$ are compact? Certainly $A$ is compact if it is finite. If $A$ is infinite, let $\{a_1,a_2,\dots\}$ be a countably infinite subset. Then the open sets $X\backslash\{a_i,a_{i+1},\dots\}$ cover $A$ but have no finite subcover, so $A$ is not compact. Since every finite set is closed, every compact subspace of $X$ is closed.

Sean Eberhard
  • 11,182
1

If every subspace is closed (like in the discrete topology) then the statement is trivially fulfilled.

Kaa1el
  • 2,058