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Let $X = [0,1]^{[0,1]}$ be equipped with the product topology. The overarching task here is to show that the set $K = \{f\in X : \exists Y-\text{countable }(\forall x\in[0,1]\backslash Y) f(x) = 0\}$ is sequentially compact. My general strategy is as follows:

Let $\{f_n\}_n$ be a sequence in $K$. Define $$A := \{x\in[0,1] : (\exists n\in\mathbb{N})(f_n(x) \neq 0)\}$$ $A$ is at most countable, so let $\{x_n\}_n$ be an enumeration of $A$. $\{f_n(x)\}_n$ is bounded for each $x\in A$, so (for each $x\in A$) there exists a subsequence $\{f_{n_i}\}_i$ such that $\{f_{n_i}(x)\}_i$ is convergent. From here I want to construct a convergent subsequence of $\{f_n\}_n$ using $\{x_n\}_n$ by doing something like the following:

Let $\{f_{n_i}\}^1_i$ be a subsequence of $\{f_n\}_n$ such that $\{f_{n_i}(x_1)\}_i$ is convergent.

Let $\{f_{n_i}\}^{k+1}_i$ be a subsequence of $\{f_{n_i}\}^k_i$ such that $\{f_{n_i}(x_{k+1})\}^k_i$ is convergent.

Define $\{f_{n_i}\}_i = \bigcap^\infty_{k=1}\{f_{n_i}\}^k_i$.

Hopefully the idea is clear even if my notation is messy. The obvious worry is that $\{f_{n_i}\}_i$ may be empty. So could the subsequence, as defined, be empty? If so, is there any way to carry out the strategy and get a nonempty subsequence that is pointwise convergent?

guest1908
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1 Answers1

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Your argument shows that the sequence $(f_n)$ essentially lives in the countable product $[0,1]^A$ (as all members of the sequence are $0$ outside of it).

Then use that the countable product of sequentially compact spaces is sequentially compact (e.g. this question) and this finishes the proof. The link I gave essentially makes your idea of successive subsequences per coordinate more precise. Or otherwise look at this book (by Joshi)) or Proposition 9 in Tao's notes for a detailed proof as well.

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Henno Brandsma
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