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Let $\mathcal{F}_0=[0,1]^{\mathbb{R}^J}$, i.e. $\mathcal{F}_0=$ the set of functions $q:\mathbb{R}^J \rightarrow [0,1]$. Note that the $q$'s are not necessarily continuous. My question is: Is $\mathcal{F}_0$ compact? I'm reading something which uses a proof which seems to rely on compactness of $\mathcal{F}_0$.

My approach: Consider a sequence $\{q_n\}_n\in \mathcal{F}_0$. For every $x \in \mathbb{R}^J$, the sequence of real numbers $\{q_n(x)\}_n \in [0,1]$ has a convergent subsequence. After this I would have applied the "subsequence of subsequence" approach, but the set of $x$'s, $\mathbb{R}^J$, is not countable, so I don't think that approach works.

Any help is most appreciated.

Canine360
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    It is compact wrt the product topology (by Tychonoff's thereom). Moreover, it does not suffice to show that every sequence has a convergent subsequence, since the space might not be sequentially compact. – Evangelopoulos Foivos Oct 12 '21 at 09:54
  • I was not familiar with Tychonoff's theorem. I looked it up and it makes perfect sense. Thanks! But I didn't fully understand the second part. Why can the space be not sequentially compact? – Canine360 Oct 12 '21 at 11:58
  • Also, I'm confused about what the product topology is in this case. – Canine360 Oct 12 '21 at 11:59
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    Convergence in product topology is identical to pointwise convergence. If $(u_n)$ is a sequence (or, more generally, a net) in the product space, then $u_n\to u$ if and only if $u_n(x) \to u(x)$ for every $x \in \mathbb R^J$. See this for a counter example of a product that is not sequentially compact. – Evangelopoulos Foivos Oct 12 '21 at 13:30
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    thanks so much. – Canine360 Oct 13 '21 at 00:07

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