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While looking up information on compact operators I came across these two conflicting posts.

  1. If a set is compact then it is closed
  2. Topology: Example of a compact set but its closure not compact

So the first link says that if a set $U$ is compact then it is closed. $U$ closed means $U = \overline{U}$ and hence $\overline{U}$ is compact. This seems to be in direct contradiction with the second post?

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csss
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    A compact subspace of a Hausdorff space is closed. – Watson Aug 25 '16 at 10:08
  • @Watson That should be an answer, not just a comment :) – 5xum Aug 25 '16 at 10:13
  • @csss: see also http://math.stackexchange.com/questions/83355/how-to-prove-that-a-compact-set-in-a-hausdorff-topological-space-is-closed?rq=1, http://math.stackexchange.com/questions/176458 – Watson Aug 25 '16 at 12:13

2 Answers2

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This is not a contradiction, because the main property is:

A compact subspace $K$ of a Hausdorff space $X$ is closed.

Indeed, we show that for every $x \in X \setminus K$, there is an open set $U$ such that $x \in U \subset X \setminus K$. Fix such an $x$.

As $X$ is Hausdorff ($T2$), for every $y \in K$, there are disjoint open sets $U_y,V_y$ such that $x \in U_y$ and $y \in V_y$. Now you can use the compactness of $K \subset \bigcup_{y \in K} V_y$, so that $$K \subset \bigcup_{y \in E} V_y$$ for some finite subset $E \subset K$.

Define $U = \bigcap_{y \in E} U_y$, which is an open set because $E$ is finite. You can finally show that $U$ satisfies the desired conditions. I leave it to!

Watson
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  • And I take it a non-Hausdorff space is a very uncommon space in applications? That is, when dealing with spaces such as $\mathbb{R}^n$ and Hilbert spaces such as $L^2(\Omega)$ and $W^{1, 2}(\Omega)$ then the closure of a compact set will be compact? – csss Aug 25 '16 at 10:45
  • @csss: Yes, in general, e.g. in (functional) analysis, we are dealing with metric spaces, which are Hausdorff, so that the above property always holds. – Watson Aug 25 '16 at 10:46
  • If $X$ is a topological space such that every compact subspace is closed, then it is clearly $T1$. This doesn't imply $T2$, see http://math.stackexchange.com/questions/518787/. – Watson Aug 25 '16 at 12:10
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    @csss: In some branches of mathematics, non-Hausdorff spaces are more common. For example, in dynamics --- the study of group actions on topological spaces --- the orbit space or quotient of a group action is often non-Hausdorff. – Lee Mosher Aug 25 '16 at 15:10
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Actually, the co-finite topology on R, the set of real numbers. Every subset of R is compact. However, it is not every subset of R is closed. There is a special spaces that assumed that every compact set is closed. Certainly, in Hausdorff spaces, every compact set is closed. For your the second question, Take X=R, the set of real numbers. Define topology on X by U is a subset of X is open if and only if U contains 1 or U is the empty set.Then K={1,2} is compact, but cl(K)=R which is not compact with this topology. Actually, every point x in not K, and any open subset U contain x then 1 is also in U. So, the intersection of U and K is not empty, then x is in the closure of K. Therefore, cl(K)=R. The collection U(x)={1,x} forms an open cover of X, obvious there is no finite sub-collection of it cover X.

Murtadha Sarray
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