if A is a compact subset of X,the inter section of all open sets which include A is a compact subset of X

Question

I want to show that if A is a compact subset of X,the inter section of all open sets which include A is a compact subset of X,but intersection of all closed sets which include A is not necessarily compact.

I tried to prove if closure of every compact space is compact... But...

Answer 1

Let $B$ be the intersection of all open subsets containing $A$. Assume we have an open cover $B\subseteq \bigcup_{i\in I}U_i$. As this also covers $A$, there is a finite subcover, $A\subseteq \bigcup_{k=1}^n U_{i_k}$. On the right hand we have an open set containing $A$, hence this set contains $B$, i.e., we have found a finite subcover for $B$. This shows that $B$ is compact.

Answer 2

Hagen has answered your first question. For the second, recall that compact subsets of Hausdorff spaces are closed, so the desired example will have to be a non-closed compact subset $A$ of a non-Hausdorff space $X$. One possibility that suggests itself is to make $A$ a compact set that is dense in a non-compact space $X$: then the only closed set containing $A$ will be $X$ itself. One simple way to do this is to put the particular point topology on some infinite set. For instance, we can let $X=\Bbb N$ with the topology

$$\tau=\{\varnothing\}\cup\{U\subseteq X:0\in U\}\;,$$

and let $A=\{0\}$. $A$ is finite, so it’s certainly compact. However, every non-empty open set in $X$ contains $0$, so $\operatorname{cl}A=X$, and $X$ is therefore the only closed set containing $A$. It only remains to check that $X$ is not compact. To see this, let

$$\mathscr{U}=\big\{\{0,n\}:n\in\Bbb Z^+\big\}\;.$$

Each member of $\mathscr{U}$ contains $0$, so each member of $\mathscr{U}$ is open. Certainly $\bigcup\mathscr{U}=X$, so $\mathscr{U}$ is an open cover of $X$. But no proper subset of $\mathscr{U}$ covers $X$, so $\mathscr{U}$ certainly has no finite subcover, and $X$ is therefore not compact.

That space is $T_0$ but not $T_1$; it’s just a little more difficult to construct a $T_1$ example. Let $X=\Bbb R$ with the topology

$$\tau=\{\varnothing\}\cup\{U\subseteq\Bbb R:\Bbb Z\setminus U\text{ is finite}\}\;;$$

that is, the non-empty open sets of $X$ are the sets that contain all but at most finitely many of the integers. As subsets of $X$, the set of non-integers bears the discrete topology, and the set of integers bears the cofinite topology. It’s easy to check that this space is $T_1$ and that $\Bbb Z$ is a (hereditarily!) compact subset. However, $\operatorname{cl}\Bbb Z=X$, since every non-empty open set in $X$ intersects $\Bbb Z$. Thus, the intersection of all closed sets containing $\Bbb Z$ is $X$, and I’ll leave it to you to verify that $X$ is not compact; you can use an argument very similar to the one that I used in the first example.