Proof help of backwards implication: A subset K of a metric space $X_d$ is compact iff every infinite subset of K has a limit point in K.

Question

I'm currently reading Rudin's Principles of Mathematical Analysis. At page 38, Theorem 2.37 says that Every infinite subset of a compact set K has a limit point in K. I wonder whether the converse still hold(I guess it holds too), but I don't know how to prove it.

Answer 1

This is indeed the case and is the subject of Exercise $26$ of that chapter, on p. $45$. The exercise reads:

$26.$ Let $X$ be a metric space in which every infinite subset has a limit point. Prove that $X$ is compact. Hint: By Exercises $23$ and $24$, $X$ has a countable base. It follows that every open cover of $X$ has a countable subcover $\{G_n\}$, $n=1,2,3,\ldots$ . If no finite subcollection of $\{G_n\}$ covers $X$, then the complement $F_n$ of $G_1\cup\cdots\cup G_n$ is nonempty for each $n$, but $\bigcap F_n$ is empty. If $E$ is a set which contains a point from each $F_n$, consider a limit point of $E$, and obtain a contradiction.

Exercises $23$ and $24$ read:

$23.$ A collection $\{V_\alpha\}$ of open subsets of $X$ is said to be a base for $X$ if the following is true: For every $x\in X$ and every open set $G\subset X$ such that $x\in G$, we have $x\in V_\alpha\subset G$ for some $\alpha$. In other words, every open set in $X$ is the union of a subcollection of $\{V_\alpha\}$. Prove that every separable metric space has a countable base. Hint: Take all neighborhoods with rational radius and center in some countable dense subset of $X$.

$24.$ Let $X$ be a metric space in which every infinite subset has a limit point. Prove that $X$ is separable. Hint: Fix $\delta\gt0$, and pick $x_1\in X$. Having chosen $x_1,\ldots,x_j\in X$, choose $x_{j+1}\in X$, if possible, so that $d(x_i,x_{j+1})\ge\delta$ for $i=1,\ldots,j$. Show that this process must stop after a finite number of steps, and that $X$ can therefore be covered by finitely many neighborhoods of radius $\delta$. Take $\delta=1/n$ ($n=1,2,3,\ldots$), and consider the centers of the corresponding neighborhoods.

Since these are exercises in the book you’re reading, I’ll leave it to you to follow these hints, but feel free to ask again if you get stuck.

Note that the premise that every infinite subset has a limit point is used twice, once in Exercise $24$ and once in Exercise $26$. What makes the proof somewhat difficult to find is that it’s not obvious how to combine these two applications into one. If we argue only that $X$ can be covered by a finite set of open balls of constant radius, they might not be contained in elements of the open cover; and if we argue only that we can find an infinite set outside of any countable subcollection of an open cover, that subcollection might not cover $X$.

If you don’t care about the intermediate results of Exercises $23$ and $24$, you can combine the two arguments into one like this: Given an open cover, in step $k$ choose some element $G_k$ of the cover that contains a ball of the largest possible radius $r_k$ not contained in $G_1\cup\cdots\cup G_{k-1}$. (To avoid the problem that there might not be a largest radius, you can limit the radii, say, to reciprocals of integers.) If this process ends with a finite cover of $X$, you’re done. If it yields a countable cover of $X$, you can argue as in the second part of Exercise $26$. If it doesn’t yield a countable cover, $r_k$ doesn’t go to zero (since there are unused elements of the open cover with uncovered open balls of non-zero radius that would have been chosen at some point), so you can argue as in Exercise $24$.

Note that while Theorem $2.37$ holds for all topological spaces (the proof only uses the topology, not the metric), the converse does not. The first uncountable ordinal $\omega_1$, equipped with the order topology, is not compact (though it’s sequentially compact). But every infinite subset of $\omega_1$ has a limit point.

To see this, form the union of the initial segments defined by the first $\omega$ elements of the infinite subset $S$. Since this countable union of countable sets is countable, it cannot be all of $\omega_1$. Thus, there is an element of $\omega_1$ that is greater than infinitely many elements of $S$, and hence also a least such element $a$. If $a$ had an open neighbourhood that contains no element of $S$ other than $a$, some $b\lt a$ would also be greater than infinitely many elements of $S$. But $a$ is the least such element. Hence $a$ is a limit point of $S$.

Note that none of the results in the proofs sketched above hold in this case: $\omega_1$ (not being metrizable) isn't separable, has no countable base and isn’t compact.