I've seen both counterexamples and proofs to "compact implies sequentially compact", and I'm not sure what's going on.

Apparently there are compact spaces which are not sequentially compact; quick googling and wikipedia checks will turn up examples floating around; they tend to be variants of $[0,1]^{[0,1]}$ with the product topology. Here's a demonstration(?):

$[0,1]^{[0,1]}$ is compact by Tychonoff's theorem. So, we demonstrate failure of sequential compactness.

Choose a unique binary representation for each $x\in [0,1]$.

For each $n\in\mathbb{N}$, let $f_n : [0,1]\to[0,1]$ (an element of $[0,1]^{[0,1]}$) be the function which maps each $x$ to its $n$-th place digit in its binary expansion.

Let $f_{n_k}$ be a subsequence of this sequence.

Let $x'\in [0,1]$ be such that the $n_{2m}$-th digit is $0$ and the $n_{2m+1}$-th digit is $1$, for all $m\in\mathbb{N}$.

Then $f_{n_k} (x')$ does not converge (it alternates between $0$ and $1$), and hence $f_{n_k}$ cannot converge.

Thus we have found a sequence in $[0,1]^{[0,1]}$ without any convergent subsequence and so it is not sequentially compact.

(Aside: this is apparently based off of an example in Steen's Counterexamples in Topology, according to http://ncatlab.org/nlab/show/sequentially+compact+space )

Nevertheless, there's also some proofs(?) floating around that compactness of a space implies sequential compactness, along these lines (this proof(?) is of the contrapositive):

Suppose $X$ is not sequentially compact.

By definition, this means there is some sequence $(x_n)$ over $X$ with no convergent subsequence.

If any $x\in X$ had for each of its neighborhood $U$ infinitely many $n$ for which $x_n \in U$, then we could define a convergent subsequence of $(x_n)$, contradicting our assumption. (Presumably this is done by choosing for each neighborhood a sufficiently-large-indexed term in that neighborhood.)

Thus, for each $x \in X$ we can select an open set $U_x$ such that $x\in U_x$ but with $x_n \in U_x$ for only finitely many $n$.

The collection $\mathcal{U}=\{U_x : x\in X\}$ is clearly an open cover of $X$.

If $\mathcal{U}$ had a finite subcover $\{U_1, \dots, U_k\}$ then the union $U_1 \cup \cdots \cup U_k$ would contain all of $X$ but only contain $x_n$ for finitely many $n$, which is impossible.

Thus $X$ is not compact, since we have found an open cover without a finite subcover.

(Aside: this proof is essentially the same as the proof appearing in Rudin's Principles of Mathematical Analysis for Thm 2.37 that infinite subsets of compact spaces have limit points.)

So, what's going on here? It can't be that what both the counterexample and the proof are telling us is correct. Is there some subtle (or more embarassingly for me -- glaring) flaw in the proof?


The problem is here:

If any $x\in X$ had for each of its neighborhood $U$ infinitely many $n$ for which $x_n \in U$, then we could define a convergent subsequence of $(x_n)$, contradicting our assumption. (Presumably this is done by choosing for each neighborhood a sufficiently-large-indexed term in that neighborhood.)

In general topological spaces this only implies that we are able to construct a convergent net, not a convergent sequence. (A point $x$ is an accumulation point of a subset $S$ $\Leftrightarrow$ there exists a net of points of $S\setminus\{x\}$ converging to $x$.)

If $X$ is first countable at $x$ (the point $x$ has a countable base), then a sequence can be constructed. (This is more-or-less standard. We first construct a decreasing base $U_n$ at $x$ and then choose a point from each $U_n$). In particular, this works for metric spaces. Note that Rudin works only with compact subset of metric spaces in that chapter.

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    $\begingroup$ If compact implies every net has a convergent subnet, then that means every sequence would also have a convergent subnet but not necessarily convergent subsequences right? So there are sequences with convergent subnets but no convergent subsequences? $\endgroup$
    – Hrit Roy
    Sep 7 '20 at 7:16
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    $\begingroup$ @HritRoy Yes that is true. You can find some such examples also on this site. For example, this post: Example of converging subnet, when there is no converging subsequence. (The was the first one I was able to find, if you search a bit, you might find other related posts.) Or you can search for examples of space that are compact, but not sequentially compact - both here and in $\pi$-base. $\endgroup$ Sep 7 '20 at 7:21
  • $\begingroup$ Where can I find a good proof for X being first countable of this result? A book would be the best scenario. This result is going to be used in a journal of Applied Mathematics and I can't assume the reader has such background in Topology, since some of my readers might have graduated in other areas instead of Mathematics. Any help with it would be absolutely grateful. $\endgroup$ Jun 23 at 4:53
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    $\begingroup$ @RenanWillianPrado I have mentioned some references in the General Topology chatroom. (Maybe somebody will notice those messages and add some further references.) $\endgroup$ Jun 23 at 5:45
  • $\begingroup$ It's all good references for sure! I can't express my gratefulness for the Theorem 3.10.31 in Engelking's General Topology. $\endgroup$ Jun 23 at 18:50

Compactness is equivalent to sequential compactness for metric spaces; in general, neither implies the other. I believe that the point at which your proof requires that we be working in a metric space, is in

If any $x\in X$ had for each of its neighborhood $U$ infinitely many $n$ for which $x_n\in U$, then we could define a convergent subsequence of $(x_n)$

In a metric space, we can get this subsequence by taking balls of radius $\frac{1}{n}$; in general we can't.


In general spaces we have the problem, as pointed out by others, that the topology need not be "determined by sequences", like we have in metric spaces, or first countable spaces. A class of spaces where we do have that sequences suffice is the class of sequential spaces, which turn out to be the class of quotient spaces of metric spaces. Here we have that sequential closure always equals closure, and for Hausdorff sequential spaces we can prove (e.g. see this blog post) quite easily that sequential compactness is equivalent to countable compactness. The latter is always implied by compactness, so that for sequential spaces we have compact implies sequentially compact (but not reversely, as $\omega_1$ in the order topology shows).

Another classical example of a compact Hausdorff but not sequentially compact space is the Čech-Stone compactification of the integers. Such spaces fail sequential compactness because they are too big, as there is a so-called cardinal invariant $\omega_1 \le t \le c$ such that every compact Hausdorff space $X$ of size $\lt 2^t$ is always sequentially compact.

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    $\begingroup$ Out of sheer wonderment, in a model of ZF where there are only countably many ultrafilters over $\omega$, isn't the Stone-Cech compactification of $\omega$ at most continuum? $\endgroup$
    – Asaf Karagila
    Jun 12 '11 at 18:02
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    $\begingroup$ Quite. I gather that a lot of the proofs I mentioned assume choice in some way. I'm not even sure the whole Čech-Stone theory can be done without choice (like existence proofs). Topology without choice gets hairy pretty quickly. $\endgroup$ Jun 12 '11 at 18:54
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    $\begingroup$ Most things get hairy without choice, I should know, this is the main focus of my thesis. I am unfamiliar with choiceless topology, though this is a subject I will likely study sometime next year. Seems like a good place to start, I guess. $\endgroup$
    – Asaf Karagila
    Jun 12 '11 at 20:11

As others gave some explanations to the problem, I will address the problem of intuition.

When we think about [topological] spaces we usually think about well behaved spaces, such as $\mathbb R$ and other similar and familiar spaces, which usually happen to be metric spaces.

The problem, as pointed out in the other answers, is that you don't always have a countable base (at a point, or at all). Sometimes, our space is so big that sequences cannot describe convergence well enough, and we need a much stronger tool - nets.

Consider the compact space $\omega_1+1$ with the order topology, where $\omega_1$ is the least uncountable ordinal (namely $\aleph_1$). The open sets are intervals, and given a cover of $\omega_1+1$ by intervals we can find a decreasing sequence of ordinals which are endpoints of intervals forming a subcover. A decreasing sequence of ordinals is always finite. This space is compact.

Now consider the point $\omega_1$ in this space. It is not the limit of any sequence other than the (eventually) constant sequence. This is because you cannot get to $\omega_1$ from any countable ordinal with only $\aleph_0$ many steps.

This means that $\omega_1$ as a point does not have a countable base for its neighbourhoods, as each open set containing it is of the form $(\alpha,\omega_1]$ with $\alpha$ some countable ordinal. So this space is not even first countable, let alone second countable (it is not separable either).

This space is sequentially compact, since every infinite sequence of ordinals converges to its supremum.

If we remove the point $\omega_1$, and our space is now the space of all countable ordinals, then it is still sequentially compact but no longer compact (it is also first countable as well).

This is just a relatively simple example why general topological spaces are wild beasts and should be approached with caution, and not with Euclidean spaces in mind.

  • $\begingroup$ "every infinite sequence of ordinals converges to its supremum". That would be true of increasing sequences, but is not true in general. For example the sequence $\omega+1,1,2,3,...$ converges to $\omega$, which is not the supremum of the sequence. $\endgroup$
    – PatrickR
    Feb 3 '18 at 20:15
  • $\begingroup$ Yeah, and possibly the sequence is not convergent. I guess the right text should be "every infinite sequence of ordinals contains an increasing subsequence which converges to its supremum". But you should excuse me, I am not in a hurry to edit such an old answer. $\endgroup$
    – Asaf Karagila
    Feb 3 '18 at 20:42

Compactness does not imply sequential compactness. The part where you wrote "presumably" in the argument relies on first countability. In that case, where you wrote "for each neighborhood...", you could finish "...in a countable base at $x$."


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