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I am trying to disprove the following:

Let $X$ be a path-connected Hausdorff space which has a sequence of compact subsets

$$K_1\subseteq K_2\subseteq K_3\subseteq \cdots$$

such that $X=\bigcup_{n\geq 1} K_n$. Assume a subset $C$ is closed if and only if $C\cap K_n$ is compact for all $n$.

Claim: Every compact subset $C\subseteq X$ is contained in $K_n$ for some $n$.

Question: Is this false?

I believed $X=[0,1]$ with $K_n=[0,1-\frac{1}{n+1}]$ was a counterexample, but it's not. Any thoughts?

Note: this is not homework, at least not mine anyway. I found it while looking through old topology problem sheets.

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up vote 1 down vote accepted

The claim is true. To see this, let $C\subseteq X$. Suppose there is no $n\in\mathbb N$ such that $C\subseteq K_n$. Then we may choose for each $n\in\mathbb N$ an element $x_n\in C$ such that $x_n\notin K_n$. The sets $U_n = C\setminus\{x_n,x_{n+1},\ldots\}$ form an open cover for $C$, with no finite subcover, implying that $C$ is not compact. (Note that here we used the assumption that $A\subseteq X$ is closed if and only if $A\cap K_n$ is compact for each $n$. This implies that $\{x_n,x_{n+1},\ldots\}$ is closed in $X$ and therefore in $C$.) This proves the claim.

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Your example is not a counterexample, because the union of your sets $K_n$ is $[0,1)$, not $[0,1]$.

Added: Here’s a proof of the result.

Suppose that $X$ is Hausdorff, $X=\bigcup_{n\ge 1}K_n$, where each $K_n$ is compact, $K_n\subseteq K_{n+1}$ for each $n\in\Bbb Z^+$, and $C\subseteq X$ is closed iff $C\cap K_n$ is compact for all $n\in\Bbb Z^+$.

Let $C$ be a compact subset of $X$. Suppose that $C$ is not contained in any of the $K_n$; then $C\setminus K_n\ne\varnothing$ for each $n\in\Bbb Z^+$. Pick $x_1\in C\setminus K_1$; $x_1\in X$, so $x\in K_{n_1}$ for some $n_1\in\Bbb Z^+$. Pick $x_2\in C\setminus K_{n_1}$; then $x_2\in K_{n_2}$ for some $n_2\in\Bbb Z^+$. Continue in this fashion, recursively constructing a sequence $\langle n_k:k\in\Bbb Z^+\rangle$ of positive integers and a sequence $\langle x_k:k\in\Bbb Z^+\rangle$ of points of $C$ in such a way that $x_k\in K_{n_k}$ for $k\ge 1$ and $x_k\in C\setminus K_{n_{k-1}}$ for $k\ge 2$.

Let $A=\{x_k:k\in\Bbb Z^+\}$; then $A\cap K_m$ is finite for each $m\in\Bbb Z^+$. Indeed, for any $m\in\Bbb Z^+$ there is a $k\in\Bbb Z^+$ such that $n_k\ge m$, and and hence $x_i\notin K_{n_k}\supseteq K_m$ for $i>k$. Finite sets are compact, so $A\cap K_n$ is compact for each $n\in\Bbb Z^+$, and by hypothesis $A$ is closed in $X$. Let $S$ be any subset of $A$; then $S\cap K_n$ is finite and therefore compact for each $n\in\Bbb Z^*$, so $S$ is also closed in $X$. Thus, every subset of $A$ is closed. It follows immediately that every subset of $A$ is open in $A$ (as a subspace of $X$), i.e., that $A$ is an infinite closed discrete subset of $X$. But then $\big\{\{a\}:a\in A\big\}$ is a relatively open cover of $A$ with no finite subcover, contradicting the compactness of $A$. It follows that the compact set $C$ must have been contained in one of the sets $K_n$. $\dashv$

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I thought as much. I was unsure because I knew the sequence in the bracket converges to 1. – James Vincent Mar 8 '13 at 14:06
@James: It does, but $1$ isn’t actually in any of the sets $K_n$, so it doesn’t get picked up in the union either. – Brian M. Scott Mar 8 '13 at 14:07

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