# Counterexample to make the difference between weakly sequentially compact and sequentially compact explicit

Some background (all of this is from Barry Simon's "Real Analysis - A comprehensive course in analysis - vol 1"):

Reading both definitions I can see the difference between weakly sequentially compact and sequentially compact, but I can't find an example of a topological space which is weakly sequentially compact but not sequentially compact or at least one space where we have a sequence with at least one limit point but which doesn't have any convergent subsequence.

Here are some potential counter examples some friends and I thought of but didn't work:

Consider $$\mathbb{N} \times \{0, 1\}$$ where $$\mathbb{N}$$ has the discrete topology and the latter the indiscrete one. Now, the sequence:

$$(1, 0), (1, 1), (2, 0), (2, 1), \cdots$$

clearly has no limit point. It's also pretty clear that it doesn't have any convergent subsequence.

• Your example space is limit point compact but not strongly point compact. – Henno Brandsma Feb 8 at 22:42

$$\{0,1\}^{\mathbb{R}}$$ is weakly sequentially compact (it's even compact Hausdorff which implies that property), but it has sequences without convergent subsequences.

$$\beta \omega$$, the Cech-Stone compactification of the countable discrete space $$\omega$$ is another example, for the same reasons. The sequence $$x_n = n \in \omega$$ is a sequence without a convergent subsequence.

"Weakly sequentially compact" is a confusing name IMHO, just call it countably compact (every countable open cover has a finite subcover), to which it is equivalent. I show this fact in my answer here.

• Thanks for the answer and the suggestion at the end! Could you edit it to include at least one example of a sequence in $\{0, 1\}^{\mathbb{R}}$ which has no convergent subsequence (I haven't worked that much with function spaces yet)? – Matheus Andrade Feb 8 at 22:53
• I've managed to come up with an example of a sequence in $\{0, 1\}^{\mathbb{R}}$ with no convergent sub-sequence, but I've yet to show that any sequence has a limit point. I know compactness and Hausdorff together mean that it's true, but I can't use compactness yet so I would like to do it the usual way (proving it satisfies the definition) but it's been hard. Could you help? – Matheus Andrade Feb 9 at 6:37
• @MatheusAndrade you need the compactness of $\{0,1\}^\mathbb{R}$, so Tychonoff's theorem, you cannot show it directly from the definitions I think. You can deduce it from the compactness of the countable subproducts, if that helps. – Henno Brandsma Feb 9 at 6:44
• It does indeed. Thanks for taking the time! – Matheus Andrade Feb 9 at 7:29
• @MatheusAndrade I exhibit such a sequence here in the homeomorphic space $\{0,1\}^{(2^\mathbb{N})}$ – Henno Brandsma Feb 9 at 9:16

I would try the interval $$[-1,1]$$ with two origins. The sequence $$\{1/n\}$$ has two limit points (the two origins), but no convergent subsequence.

• That is actually not true, $\{1/n\}$ is itself convergent to both origins. – freakish Feb 8 at 23:43
• Ah! Yes, you are right. I retract my answer. – fauxefox Feb 8 at 23:46