# Every compact subset must be closed?

This is an exercise from a topological book.

In $T_1$ space, every compact subset must be closed? For any two compact subset, their intersection must be compact?

Thanks for any help:)

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What does $T_1$ mean? With the usual definition, this is false. –  Chris Eagle Jul 29 '12 at 12:57
...and what have you tried, what are your thoughts on the matter? –  t.b. Jul 29 '12 at 12:59
I know. Only in the Hausdorff space, every compact subset is closed. However, i failed to find out the counterexample. –  Paul Jul 29 '12 at 13:04
That is also false: there are non-Hausdorff spaces in which every compact subset is closed. Anyway, any thoughts you have on the question should be in the question, not waiting to be drawn out in the comments. –  Chris Eagle Jul 29 '12 at 13:09
@Chris thank you for your reminding. I will from now on. Because i just a new user in this sites, there are many things I don't know. Thanks again:) –  Paul Jul 29 '12 at 13:14

A space that gives negative answers to both questions is the segment $[-1,1]$ with doubled origin.

More precisely, take two copies $I_0 = [-1,1]\times \{0\}$ and $I_1 = [-1,1] \times \{1\}$ of that interval and identify $(t,0)$ with $(t,1)$ whenever $t \neq 0$, that is, we consider the quotient space $Q$ of $[-1,1] \times \{0,1\}$ modulo the equivalence relation generated by $(t,0) \sim (t,1)$ whenever $t \neq 0$.

The space $Q$ is $T_1$: a set in $Q$ is closed if and only if its pre-image under the quotient map is closed, and a pre-image of a point consists of one or two points. Since continuous images of compact sets are compact, the images $C_i$ of $I_i = [-1,1] \times \{i\}$ under the quotient map are both compact, but they are open and non-closed (look at the pre-images under the quotient map again). The intersection $C_0 \cap C_1$ is homeomorphic to $[-1,1] \smallsetminus \{0\}$, in particular it is non-compact.

Note: In order to obtain an example you need to intersect two non-closed compact sets, otherwise the claims in @blindman's answer show that the intersection is compact.

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Thanks t.b. This is another counterexample. The example is new for me and very interesting:) –  Paul Jul 31 '12 at 6:47
It's a variant of a standard example: the line with two origins, see also here. It is a bit (but only very vaguely) related to your question on the Alexandrov double segment space, but much simpler (that's one reason why I chose that example). –  t.b. Jul 31 '12 at 6:53
Thanks for the link. I will spend some time to read it carefully:) –  Paul Jul 31 '12 at 7:06

Consider the finite complement topology on an infinite set. Is it $T_1$? Which subsets are compact? Which are closed?

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The topological space of course is $T_1$. All the finite subsets are closed. I don't know which subsets are compact. –  Paul Jul 29 '12 at 13:16
@Paul: Every subset is compact, wich you can easily prove by applying te definition in terms of open covers. –  Michael Greinecker Jul 29 '12 at 14:43
@Michael Yes: For any open cover of the space and for any subset, choose any point from this set, then there exists an open set of the cover whose complement is finite. Then it is obvious... Thanks:) –  Paul Jul 31 '12 at 6:39

I would like to advertise the following result, and some consequences, just in case it is not so well reported:

Theorem: Let $B,C$ be compact subsets of $X,Y$ respectively and let $\mathcal W$ be a cover of $B \times C$ by sets open in $X \times Y$. Then $B,C$ have open neighbourhoods $U,V$ respectively such that $U \times V$ is covered by a finite number of sets of $\mathcal W$.

This has a number of consequences from "the product of two compact spaces is compact" to "a compact subset of a Hausdorff space is closed". (Two others are listed in my book "Topology and groupoids", p. 86, but I forget where I first learned of the theorem, back in the 1960s!)

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Thanks Ronnie Brown. This theorem is very surprising for me. –  Paul Jul 31 '12 at 6:53
The other two corollaries are: "Let $B,C$ be compact subsets of $X,Y$ and let $W$ be an open subset of $X \times Y$ containing $B \times C$. Then $B,C$ have open neighbourhoods $U,V$ such that $U \times V \subseteq W$." "If $B,C$ are disjoint compact subsets of the Hausdorff space $X$, then $B,C$ have disjoint open neighbourhoods in $X$." Don't other books on general topology give the theorem, and corollaries? –  Ronnie Brown Jul 31 '12 at 11:16
These corollaries are obviously right and given by the other books on general topology, for example, the Engelking's book. Thanks again for the answer:) –  Paul Aug 1 '12 at 6:47

Claim 1. If $X$ is a compact subset of a Hausdorff space $(Y, \tau)$ then $X$ is closed.

Proof. It sufficies to show that $Y\setminus X$ is open. To show that $Y\setminus X$ is open, let $y \in Y \setminus X$. For each $x \in X$ fix disjoint $U_x, V_x \in \tau$ so that $x \in U_x$ and $y \in V_x$. From the open cover $\{U_x : x \in X\}$ of $X$ extract a finite subcover, say $\{U_{x_1},U_{x_2},\ldots,U_{x_n}\}$. Then $V_{x_1}\cap V_{x_1},\cap\ldots,\cap V_{x_n}$ is a neighborhood of $y$ in $Y$ that does not intersect $X$.

Claim 2. If $X$ is compact subset of a topological space $(Y, \tau)$ and $K\subset X$ is a closed subset then $K$ is compact.

Proof. Suppose that $\{V_\alpha: \alpha\in I\}\subset \tau$ is an arbitrary open cover for $K$. Since $K$ is closed, $Y\setminus K$ is open. Then $X\subset Y$ is covered by $V_\alpha(\alpha\in I)$ and $Y\setminus K$. Since $X$ is compact, it can be covered by a finite number of open subsets $V_{\alpha_1},\ldots, V_{\alpha_n}$ and $Y\setminus K$, and so is $K\subset X$. Hence $K$ is compact.

Claim 3. If $X_1, X_2$ are closed subsets of a topological space $Y$ then $X_1\cap X_2$ is also closed.

Claim 4. If $X_1, X_2$ are compact subsets of a Hausdorff space $Y$ then $X_1\cap X_2$ is also compact.

Counterexample. If $(Y,\tau)$ is not a Hausdorff space then Claim 1. is not valid. Indeed, let $Y=\{a,b\}$ and $\tau=\{\emptyset, X\}$. Note that $Y$ is not a Hausdorff space. Moreover, $\{a\}$ is a compact subset of $Y$ but $\{a\}$ is not closed.

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I've edited your post to use the built-in Markdown formatting for bold and italics. Here is an explanation of how to use Markdown formatting. –  Zev Chonoles Jul 29 '12 at 14:38
@Zev Chonoles: Thank you for your helping and revising my answer. –  blindman Jul 29 '12 at 14:41
I do not know why I got -1 in my answer. –  blindman Jul 29 '12 at 14:44
Someone has apparently downvoted you. Perhaps you should make clearer what you are claiming is the answer to the question, or add further reasoning. If your answer is incorrect, that will also usually bring downvotes. –  Zev Chonoles Jul 29 '12 at 14:51