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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 $[0,1]$]Answer below

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See e.g. math.stackexchange.com/q/152447 and the links and references there. –  user45931 Oct 24 '12 at 23:13
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2 Answers 2

up vote 2 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
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Spacebook, a searchable version of Steen and Seebach's Counterexamples in Topology, lists the following compact spaces that are not sequentially compact.

Stone-Cech Compactification of the Integers
Uncountable Cartesian Product of Unit Interval

The second is, of course, the one hinted at by Willard.

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please check my answer below –  Camilo Arosemena Oct 25 '12 at 0:00
You mean that the second is the one that Willard suggests. –  Brian M. Scott Oct 25 '12 at 12:20
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