## Compact space that is not sequentially compact

What is an example of a compact space that is not sequentially compact?

Let’s recall that: A space $X$ is compact if every open cover of $X$ has a finite subcover. And $X$ is sequentially compact if every sequence in $X$ has a convergent subsequence.

When $X$ is a metric space, these two conditions are equivalent (which is already interesting!). But I want to talk about a really cool example of a topological space which is compact but not sequentially compact.

Let $X=[0, 1]^{\mathcal{P}(\mathbb{N})}$ where $\mathcal{P}(\mathbb{N})$ is the power set of $\mathbb{N}$ (i.e. $\mathcal{P}(\mathbb{N})$ is the set of all subsets of the positive integers). One can think of the space $X$ in two ways: The formal way is to think of $X$ as the set of all functions $\mathcal{P}(\mathbb{N})\to [0, 1]$. Or, we can simply say $X$ is the space of all infinite tuples, all of whose entries are coming from $[0, 1]$. Of course, the infinite tuple will have uncountably many coordinates (namely the cardinality of $\mathcal{P}(\mathbb{N})$).

In any case, the space $X$ is compact by Tychonoff’s theorem. However, let’s show that $X$ is not sequentially compact. For this, we need to construct some sequence $\{a_n\}_{n=1}^{\infty}$ which does not have a convergent subsequence. The idea comes from the sequence $\{(-1)^{n}\}_{n=1}^{\infty}$ in $\mathbb{R}$ which does not have a convergent subsequence. In order to construct the sequence $\{a_n\}_{n=1}^{\infty}$ in $X$, we need to specify $\mathcal{P}(\mathbb{N})$-many coordinates for each $a_n$. Essentially we want to fill out a grid of dimension $|\mathbb{N}|\times|\mathcal{P}(\mathbb{N})|$: $a_{1} \ \ \ \ * * * * * * * * \cdots$ $a_{2} \ \ \ \ * * * * * * * * \cdots$ $a_{3} \ \ \ \ * * * * * * * * \cdots$ $\vdots \ \ \ \ \ \ \ \vdots \ \ \vdots \ \ \vdots \ \ \vdots \ \ \vdots \ \cdots$

The idea is to fill out the grid column by column! (It would seem more natural to fill the table row by row, since filling out $n$-th row exactly corresponds to specifying an element $a_{n}$ of the sequence). Recall that columns of the grid is indexed by elements of $\mathcal{P}(\mathbb{N})$. For every element $S\in\mathcal{P}(\mathbb{N})$, fill the $S$-th column by inserting $0, 1, 0, 1, 0, \cdots$ into the rows indexed by $a_{n_{1}}, a_{n_{2}}, a_{n_{3}}, a_{n_{4}}, \cdots$ respectively, where $S_{1}=\{n_{1}, n_{2}, n_{3},\cdots\}$, and entering $1/2$ (or anything arbitrary) to all other entries of this column. Let’s give an example. For example, if $S=\{2, 3, 4, 18\}$, then we would insert $0$ to the entries $(a_2, S), (a_4, S)$ and insert $1$ to the entries $(a_3, S), (a_{18}, S)$, and insert $1/2$ for $(a_j, S)$ for all $j\notin S$. Of course, $S$ could be infinite as well, but hopefully the idea is clear.

Now that the grid is filled, the elements of the sequence are ready. They are simply the rows of the grid. The claim is that the constructed sequence $\{a_n\}_{n=1}^{\infty}$ has no convergent subsequence. Indeed, assume to the contrary, that $\{a_{n}\}_{n=1}^{\infty}$ has some subsequence $\{a_{n_k}\}_{k=1}^{\infty}$ that converges. We take $S=\{n_1, n_2, n_3, \cdots\}$ and look at the $S$-th column of the grid. What do we see? We see $0, 1, 0, 1, 0, \cdots$ across the rows indexed by $\{a_{n_k}\}_{k=1}^{\infty}$. So this shows $\{a_{n_k}\}_{k=1}^{\infty}$ cannot possibly converge, because if it did, it would have to converge componentwise, but $S$-th component of $a_{n_{k}}$ alternates between $0$ and $1$. This is a contradiction, and the claim is proved.

So $X=[0, 1]^{\mathcal{P}(\mathbb{N})}$ is an example of compact space which is not sequentially compact.

Remark. Of course, the same idea goes on to show that $X=[0, 1]^{\mathbb{R}}$ is compact but not sequentially compact. The key is to write down a bijection (injection would suffice) $f:\mathcal{P}(\mathbb{N})\to\mathbb{R}$ and work from there…

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