# How does one definition of “countably compact” lead to another?

The fundamental definition of countably compact as given in my textbook ("Foundations of General Topology by Pervin) is: $X$ is countably compact if every infinite subset $E$ of $X$ has a limit point within $E$.

How does this definition lead to the modern definition of countably compact: every countable cover of a subset has a finite subcover?

Motivation: Let subset $E$ consist of an infinite countable number of points $\{x_{i}\}, i\in\Bbb{N}$. Let $\{x_{1},x_{2}\}$ be one open set, and each $\{x_{i}\},i\geq2$ be a distinct open set. Let every open set containing $x_{1}$ contain at least one of $x_{i},i\geq 2$. This shows that $x_{1}$ is a limit point of $E$, and also a part of $E$.

Clearly, if we take $\{x_{1},x_{2}\},\{x_{2}\},\{x_{3}\},\dots$ to cover $E$, we don't get a finite subcover of $E$.

EDIT: In another portion, the book says "a countably compact subset $E$ must have atleast one limit point within $E$".

• The thing you describe is not a topology. – Chris Eagle Jul 7 '13 at 13:28
• Yes, sorry the edits have been made. – fierydemon Jul 7 '13 at 13:31
• I think the first italicized statement is the definition of a limit point compact space. Countably compact does imply limit point compact, but the converse does not hold. – Cihan Jul 7 '13 at 13:45
• To confirm what Cihan said. The first statement is limit point or weakly countably compact. And that's not equivalent to countably compact in general. It is for a T1-space. – Julien Jul 7 '13 at 14:01

Pervin’s definition is of the Bolzano-Weierstrass property, and it can be stated more succinctly as follows:

$X$ has the Bolzano-Weierstrass property if it does not contain an infinite closed discrete subset.

Every countably compact space has the Bolzano-Weierstrass property. To see this, suppose that $X$ does not have the Bolzano-Weierstrass property. Then $X$ has an infinite closed discrete subset $D$. Let $D_0$ be a countably infinite subset of $D$; then $D_0$ is discrete and closed. For each $x\in D$ let $V_x$ be an open nbhd of $x$ such that $V_x\cap D_0=\{x\}$, and let $\mathscr{U}=\{X\setminus D_0\}\cup\{V_x:x\in D_0\}$; then $\mathscr{U}$ is a countable open cover of $X$ with no finite subcover, since for each $x\in D_0$, $V_x$ is the only member of $\mathscr{U}$ that contains $x$, and $X$ is not countably compact.

The converse is false in general. For each $n\in\Bbb N$ let $V_n=\{k\in\Bbb N:k<n\}$, and let $$\tau=\{\Bbb N\}\cup\{V_n:n\in\Bbb N\}\;;$$ then $\tau$ is a $T_0$ topology on $\Bbb N$. Suppose that $A\subseteq\Bbb N$ is a non-empty closed set; then $$A=\Bbb N\setminus V_n=\{k\in\Bbb N:k\ge n\}$$ for some $n\in\Bbb N$, and $A$ is not discrete: every open nbhd of $n+1$ contains $n$. Thus, the only closed discrete set in $\langle\Bbb N,\tau\rangle$ is the empty set, and therefore $\langle\Bbb N,\tau\rangle$ has the Bolzano-Weierstrass property. However, it’s easy to check that $\{V_n:n\in\Bbb N\}$ is an open cover of $X$ with no finite subcover, so $\langle\Bbb N,\tau\rangle$ is not countably compact.

The converse does hold in $T_1$-spaces, however: every $T_1$-space with the Bolzano-Weierstrass property is countably compact. To see this, suppose that $X$ is $T_1$ and that $\mathscr{U}=\{U_n:n\in\Bbb N\}$ is a countable open cover of $X$ with no finite subcover. For $n\in\Bbb N$ let $V_n=\bigcup_{k\le n}U_k$, and let $\mathscr{V}=\{V_n:n\in\Bbb N\}$. Then $\mathscr{V}$ is also a countable open cover of $X$ with no finite subcover, and $V_n\subseteq V_{n+1}$ for each $n\in\Bbb N$. Let $M=\{n\in\Bbb N:V_n\subsetneqq V_{n+1}\}$; if $M$ were finite, we’d have $V_{\max M}=X$, and $\mathscr{V}$ would have a finite subcover, so $M$ is infinite. For each $n\in M$ let $x_n\in V_{n+1}\setminus V_n$, and let $D=\{x_n:n\in M\}$; clearly $D$ is infinite, and I claim that it is closed and discrete as well.

To see this, let $x\in X$ be arbitrary; there is an $n\in M$ such that $x\in V_n$. By construction $V_n\cap D=\{x_k:k<n\}$; let $F_x=\{x\}\cup(V_n\cap D)$. $F_x$ is finite, and $X$ is $T_1$, so $F_x$ is closed, and therefore $V_n\setminus F_x$ is an open nbhd of $x$. Finally,

$$D\cap(V_n\setminus F_x)=\begin{cases} \varnothing,&\text{if }x\notin D\\ \{x\},&\text{if }x\in D\;, \end{cases}$$

so $D$ is indeed closed and discrete.

• +1, of course. I never really "used" the notion of countable compactness. What are the typical situations where it proves useful? – Julien Jul 7 '13 at 23:05
• @julien: In my experience it’s actually the B-W property that one uses more often. Proofs involving the (countable) compactness of a metrizable space often turn out to use just the B-W property. More generally, the supremum of sizes of closed, discrete subsets has proved useful enough to be given a name: it’s $e(X)$, the extent of $X$. (This isn’t actually quite enough to pin down the B-W property: $X$ has the B-W property if $e(X)=\omega$ and the extent is not attained.) – Brian M. Scott Jul 7 '13 at 23:11
• Ok, thank you. I did not know the concept of extent. – Julien Jul 7 '13 at 23:24
• Any chance you could look for me at math.stackexchange.com/questions/1576468/… ? – Tom Collinge Dec 15 '15 at 12:13

That equivalence (if you take a limit point of a set to be the point each neighbourhood of which contains infinitely many points from that set), as julien has stated, indeed holds in case of $T_1$ spaces and you can prove it in this case as follows.

So first assume a $T_1$ space $X$ to be countably compact in the sense of coverings, and assume that there's an infinite subset $E$ eithout limit points. You can take $E$ to be countable. Now since $X$ is $T_1$ and $E$ has no limit points, it follows that $E$ is closed, i.e. $X\backslash E$ is open (indeed, take any point $x\in X\backslash E$, by our assumption, it is not a limit point for $E$, hence $\exists U\subset X\backslash E$ - an open neighbourhood of $x$ which contains at most finitely many of points from $E$ - say, $x_1,...,x_n$. By $T_1$-property, you can choose an open neighbourhood $V$ of $x$ which does not contain any of $x_1,...x_n$. Then $U\cap V$ will be the neighbourhood of $x$ which does not intersect $E$. Hence, $X\backslash E$ is indeed open).

Also, for any $x_i\in E$ you can choose an open set $V_i$ with the property $V_i\cap E = \{x_i\}$ (again due to our assumption and $T_1$-property). Consider the family $\{X\backslash E, V_i|i\in \mathbb{N}\}$. It is a countable open cover (recall that we assumed $E$ to be countable) of $X$. Choose a finite subcover to obtain a contradiction ($E$ is infinite!).

The proof of the other part is as follows. You assume that there's a countable open cover ${\cal U} = \{U_1,...,U_n\}$ of $X$ that has no finite subcover. Hence you can assume without loss of generality that for every $k>1$ the set $U_{k+1}$ is not contained in the union of $U_1,...,U_k$ (since if it was contained, you would simply delete it from your cover, and considered the respective subcover. After doing all of the required deletions, you would obtain still an infinite subcover according to our assumption).

So now for any $k\in \mathbb{N}$ choose $x_k\in U_{k+1}\backslash \cup_{i=1}^k U_i$. Denote $E = \{x_1,...,x_n,...\}$. $E$ must have a limit point, say, $x_0\in X$. This point must be contained in $U_i$ for some $i\in \mathbb{N}$. But $U_i$ contains only finitely many points from $E$, namely, $x_1,...,x_{i-1}$, a contradiction with $x_0$ being a limit point of $E$.

I hope this helps...