Let $\mathcal{M}$ be a metric space, with $\mathbb{A}\subseteq\mathcal{M}$ compact. Suppose that $\ell\in\mathcal{M}$ is a limit point of $\mathbb{A}$, and that $\ell\notin\mathbb{A}$; further, let $\{\overline{\mathbb{B}}_{r_\alpha}(\ell)\}_{\alpha<\omega}$ be a sequence of closed balls centered at $\ell$ such that $\lim_{\alpha\rightarrow\omega}r_\alpha=0$, so $\lim_{\alpha\rightarrow\omega}\overline{\mathbb{B}}_{r_\alpha}=\{\ell\}$. We then have that $\{\mathcal{M}\sim\overline{\mathbb{B}}_{r_\alpha}(\ell)\}_{\alpha<\omega}$ is an open cover of $\mathbb{A}$ with no finite subcover, since $\forall\alpha\big(\mathbb{A}\cap\overline{\mathbb{B}}_{r_\alpha}(\ell)\neq0\big)$, a contradiction since $\mathbb{A}$ is compact. We thusly conclude that $\ell\in\mathbb{A}$, thus $\mathbb{A}$ is closed.

I was just wondering if this proof appeared in literature somewhere -- I liked it because it makes the interplay between the notion of a limit point and compactness in a metric space apparent.

EDIT: To be clear, $\omega$ is the first transfinite ordinal, and $\alpha<\omega\implies\alpha$ is a natural number.

  • $\begingroup$ I can feel your gladness :) +1. Not to extinguish it, but I guess it would have appeared in some textbooks, as far as I recall. Check especially those classic old analysis books. $\endgroup$ – Megadeth Jun 24 '17 at 7:16
  • $\begingroup$ I was reading through baby Rudin again when it occurred to me as a shorter proof than the one he had -- perhaps it's in another classic! $\endgroup$ – Alec Rhea Jun 24 '17 at 7:21
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    $\begingroup$ @AlexProvost : In a discrete metric space, those balls will have only one element but they will exist. As for $\alpha, \omega$, these notations usually refer to ordinals ($\omega$ being the smallest infinite one, and $\alpha<\omega$ implies that $\alpha$ is an integer). $\endgroup$ – Max Jun 24 '17 at 7:43
  • $\begingroup$ @AlexProvost : Actually here they will, since none of the balls centered at $l$ is a singleton, as $l$ is a limit point of $\Bbb{A}$. Unless you can pinpoint a step in the proof which is a mistake (maybe there is I havn't checked the details), the fact that there may exist singleton-balls isn't of any help in showing that the proof is invalid. $\endgroup$ – Max Jun 24 '17 at 8:14
  • $\begingroup$ @AlexProvost They will; there's nothing wrong in the argument for a generic metric space, so it applies also to discrete spaces. $\endgroup$ – egreg Jun 24 '17 at 8:17

I'm afraid this argument is not new.

Let me use a more standard notation and some simplification.

Let $X$ be a metric space and $A\subseteq X$. If $l\in X$ is a limit point of $A$ and $l\notin A$, we can consider the sequence of closed balls $\bar{B}_{1/n}(l)$, for $n>0$.

In a metric space it holds that $$ \bigcap_{n>0}\bar{B}_{1/n}(l)=\{l\} $$ and also $$ \bigcap_{n\in F}\bar{B}_{1/n}(l)=\bar{B}_{1/{\max F}}(l) $$ for every finite set $F$ of positive integers.

Thus the family $$ \bigl(X\setminus \bar{B}_{1/n}(l)\bigr)_{n>0} $$ is an open cover of $A$, which admits no finite subcover, because, for any finite set $F$ of positive integers, $$ \bigcup_{n\in F}\bigl(X\setminus \bar{B}_{1/n}(l)\bigr)= X\setminus \bar{B}_{1/{\max F}}(l) $$ and $\bar{B}_{1/{\max F}}(l)\cap A\ne\emptyset$. Therefore $A$ is not compact.

On the other hand, the proof can be given without reference to a metric. Suppose $X$ is a Hausdorff space; let $A\subseteq X$ and $l\in X$ be a limit point of $A$, $l\notin A$. In particular $A$ is not empty. It is known that a subset is closed if and only if it contains each of its limit points; we want to show that $A$ (which is not closed by assumption) is not compact.

For every $a\in A$, choose open neighborhoods $U_a$ of $a$ and $V_a$ of $l$ such that $U_a\cap V_a=\emptyset$. Then $(U_a)_{a\in A}$ is an open cover of $A$, but it fails to admit a finite subcover. Indeed, if $F$ is a finite subset of $A$, then $$ \biggl(\bigcup_{a\in F}U_a\biggr)\cap \biggl(\bigcap_{a\in F}V_a\biggr)=\emptyset $$ but $\bigcap_{a\in F}V_a$ is a neighborhood of $l$, so it intersects $A$. Therefore $A$ is not compact.

You may want to analyze the similarities and differences between the two arguments. In the argument for metric spaces, a further property is used, namely that every point has a basis of neighborhoods consisting of closed sets (that is, it is regular).

  • $\begingroup$ Much appreciated! $\endgroup$ – Alec Rhea Jun 24 '17 at 8:55

This argument is a combination of several facts, we can formulate it in general as:

In a sequential UL space $X$ a sequentially compact subspace $A$ is closed.


  • A space $X$ is sequential if all sequentially closed subsets are closed.

  • A subset $A \subset X$ of a space $X$ is sequentially closed iff for all sequences $(x_n)$ such that $\forall n: x_n \in A$ and $x_n \to x$ in $X$, we know that $x \in A$. (limits of sequences from $A$ are in $A$ too).

  • A space is UL if convergent sequences have unique limits. So for all sequences $(x_n)$: when $x_n \to x$ and $x_n \to y$ then $x=y$.
  • A space $X$ is sequentially compact when every sequence has a convergent subsequence.

Proof of the first statement basically writes itself: let $A$ be sequentially compact (as a subspace of $X$), and suppose for a contradiction that $A$ is not closed in $X$. As $X$ is sequential, $A$ cannot be sequentially closed, so there is a sequence $(x_n)$ such that $x_n \to x$ and all $x_n \in A$, and such that $x \notin A$. But as $(x_n)$ is a sequence in $A$, $A$ being sequentially compact means that there is an $a \in A$ and a subsequence $x_{n_k}$ of the sequence $(x_n)$ that converges to $a$. But in $X$, every subsequence of $(x_n)$ also converges to $x$, so on the one hand $x_{n_k} \to x$ but also $x_{n_k} \to a$. The UL property then forces $a = x$ while $a \in A, x \notin A$, contradiction. So $A$ is closed.

Compact implies closed in metric spaces follows from the above, because:

  • all metric spaces are sequential (from first countability, as you also use, by taking a countable collection of open balls) and also UL (from Hausdorff, which is why you can take the interesection of closed balls to get the singleton).

  • In a metric space compactness is equivalent to limit point compactness, which is also equivalent to sequential compactness (the last equivalence holds in all sequential Hausdorff spaces).

Your exact proof can also be generalised in general terms as follows:

A countably compact subset $A$ of a first countable Hausdorff space $X$ is closed in $X$.

For, suppose that $A$ is not closed, and pick a point $x \in \overline{A}\setminus A$. Let $U_n(x)$ be a countable local base at $x$. Then $$\cap_n \overline{U_n(x)} = \{x\}\text{:}$$

The right to left inclusion is trivial. Take any $y \neq x$ then $x$ and $y$ have disjoint neighbourhoods $U_x$ and $U_y$. For some $n$, $U_n(x) \subseteq U_x$ by definition of a local base. The fact that $U_y \cap U_n(x) \subseteq U_y \cap U_x = \emptyset$ shows that $y \notin \overline{U_n(x)}$, so $y \notin \cap_n \overline{U_n(x)}$, which shows the inclusion from left to right.

Then define $O_n = X \setminus \overline{U_n(x)}$, which are open sets. They cover $A$ as $$\cup_n (X \setminus \overline{U_n(x)}) = \text{ (de Morgan) } X \setminus \cap_n \overline{B_n(x)}= X \setminus \{x\} \supseteq A$$.

As $A$ is countably compact, we can find a finite subcover $O_{n_1}, \ldots, O_{n_k}$ of the $O_n$, and as $\cap_{i=1}^k B_{n_i}(x)$ is an open neighbourhood of $x$, it intersects $A$ in a point $a$ (as $x \in \overline{A}$). Then $a \notin O_{n_i}$ for any $i= 1,\ldots k$, contradicting the fact it's a subcover. This finally shows that $A$ is closed.

The above theorem is classical, and can even be generalised further I believe, and your proofs uses this exact argument, exploiting that compact implies countably compact and any metric space is first countable and Hausdorff. These latter fact also eneabled the sequence based proof as well.

  • $\begingroup$ This parses the various logical notions very nicely, muchas gracias. $\endgroup$ – Alec Rhea Jun 24 '17 at 11:04
  • $\begingroup$ @AlecRhea the notions of sequential compactness and being sequential were introduced to see in what classes of spaces the classic sequence based proofs from metric topology still worked, after all sorts of examples were found that show that general spaces need not be sequential, or that compact spaces don't need to be sequentially compact. $\endgroup$ – Henno Brandsma Jun 24 '17 at 11:07
  • $\begingroup$ @AlecRhea I added a second generalisation as well. $\endgroup$ – Henno Brandsma Jun 24 '17 at 11:34
  • $\begingroup$ Very cool -- my work is primarily surrounding ordinal-constructed number fields of arbitrarily high density, so I tend to take notions of sequencing etc for granted. It's nice to see all of the various implied notions laid bare like this. $\endgroup$ – Alec Rhea Jun 24 '17 at 18:36

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