Compact space which is not sequentially compact

Question

Greets

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 1

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.

Thanks.

Answer 2

$\pi$-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.