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In the Euclidean spaces compact sets are always closed.

This is not true for general topological spaces. Can we characterise when it is possible?

Is it true for metric spaces?

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It would be true when the space is Hausdorff. – Arctic Char Mar 7 '14 at 3:53
up vote 12 down vote accepted

As mentioned by John in the comments, in Hausdorff (T2) spaces, compact subsets are always closed.

Suppose $X$ is Hausdorff, and $K \subseteq X$ is compact. To show that $K$ is closed, it suffices to show that $X \setminus K$ is open; that is, show that each $x \in X \setminus K$ has an open neighbourhood $U$ with $U \subseteq X \setminus K$. So given $x \in X \setminus K$, by Hausdorffness for each $y \in K$ there are disjoint open neighbbourhoods $U_y , V_y$ of $x,y$, respectively. Clearly $\{ V_y : y \in K \}$ is an open cover of $K$, and so there are finitely many $y_1 , \ldots , y_n \in K$ such that $K \subseteq V_{y_1} \cup \cdots \cup V{y_n}$. It is not too difficult to show that $U = U_{y_1} \cap \cdots \cap U_{y_n}$ has the desired property.

As metric spaces are Hausdorff, it follows that compact subsets of metric spaces are closed.

A space in which every compact subset is closed is called a KC space. It is easy to show that every KC space is T1 (since finite subsets are always compact, in a KC space all finite subsets are closed which is equivalent to being T1). So we have implications $$\text{T}_2 \Longrightarrow \text{KC} \Longrightarrow \text{T}_1.$$ However neither of these arrows reverse.

Example 1. Let $X$ be an uncountable set with the co-countable topology (i.e., the open sets are $\varnothing$ and all $A \subseteq X$ such that $X \setminus A$ is countable). Since every two nonempty open sets have nonempty intersection, it follows that $X$ is not T2. Note that $X$ is clearly T1. Suppose that $A \subseteq X$ is infinite. Taking a countably infinite subset $A_0 = \{ a_i : i \in \mathbb{N} \} \subseteq A$, note that for each $n$ the set $U_n = X \setminus \{ a_i : i \neq n \}$ is open. Clearly $A \subseteq \bigcup_{n=1}^\infty U_n$, however as $U_n \cap A_0 = \{ a_n \}$ for each $n$ it follows that no finite subcollection (in fact, no proper subcollection) of the $U_n$ covers $A$. Thus the only compact subsets are the finite subsets, which are closed. Therefore, $X$ is a KC space.

Example 2. Let $X$ be an infinite set with the co-finite topology (i.e., the open sets are $\varnothing$ and all $A \subseteq X$ such that $X \setminus A$ is finite). Since every finite subset is closed, it follows that $X$ is T1. However every subset of $X$ is compact: given any nonempty family $\mathcal{U}$ of nonempty open subsets of $X$, picking $U \in \mathcal{U}$ as $X \setminus U$ is finite we need only take finitely many more $V_1 , \ldots , V_n \in \mathcal{U}$ so that $U \cup V_1 \cup \cdots \cup V_n = \bigcup \mathcal{U}$. As any proper infinite subset of $X$ is not closed, $X$ is not a KC space.

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So $KC$ space can be called $T_{1.5}$ ^^. – Arctic Char Mar 9 '14 at 3:16
@John: This would only make sense if this was the only (or most) natural property between T$_1$ and Hausdorff. Another property is that all sequences have at most one limit, and such a space is called a US space; this actually lies between KC and T$_1$. And there are yet more such properties between T$_1$ and Hausdorffness. – arjafi Mar 9 '14 at 4:53

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