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So this is exercise 17G.1 of Stephen Willard's "general topology", and it's stated:

Show that there is a compact space that is not sequentially compact

[Hint: Consider an uncountable product of copies of $[0,1]$]

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See e.g. and the links and references there. – user45931 Oct 24 '12 at 23:13
up vote 3 down vote accepted

So here is my answer

Let $S$ be the set of all strict increasing sequences of natural numbers, and for each $s\in{S}$ let $X_s=\left\{{0,1}\right\}$ with the discrete topology, put $X=\prod_{s\in{S}}X_s$. Then $X$ is compact by Tychonoff's theorem. Let us see that $X$ is not sequentially compact. Define $\left\{{x_n}\right\}_{n\in{\mathbb{N}}}$ as follows: let $s\in{S}$ with $s=\left\{{n_k}\right\}_{n\in{\mathbb{N}}}$, then define $(x_n)_s=0$ if $n=n_k$ for some $k$ even and define $(x_n)_s=1$ otherwise. Let us see that $\left\{{x_n}\right\}_{n\in{\mathbb{N}}}$ has no convergent subsequence in $X$. Let $s\in{S}$ with $s=\left\{{n_k}\right\}_{n\in{\mathbb{N}}}$, then $(x_{n_k})_s=0$ for $k$ even and $(x_{n_k})_s=1$ for $k$ odd, thus $\left\{{x_{n_k}}\right\}_{k\in{\mathbb{N}}}$ does not converge in $X=\prod_{s\in{S}}X_s$ since it doesn't converge componentwise.

As you can see this example is too straightforward, and I would like to see other examples.


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You want $s=\{n_k\}_{k\in\Bbb N}$, not $s=\{n_k\}_{n\in\Bbb N}$. Apart from that the argument is correct, though there are a couple of places where it could be made a little clearer. I don’t think that I’d call it too straightforward; it’s one of the two most obvious examples of a compact space that isn’t sequentially compact, the other being $\beta\Bbb N$, the Čech-Stone compactification of $\Bbb N$. – Brian M. Scott Oct 24 '12 at 23:48

$\pi$-Base, a searchable version of Steen and Seebach's Counterexamples in Topology, lists the following compact spaces that are not sequentially compact. You can learn more about these spaces by viewing the search result.

Stone-Cech Compactification of the Integers

Uncountable Cartesian Product of Unit Interval ($I^I$)

The second is the one hinted at by Willard.

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You mean that the second is the one that Willard suggests. – Brian M. Scott Oct 25 '12 at 12:20

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