A topological space $X$ is called a Fréchet–Urysohn space if for each subset $A\subseteq X$ and for each point in its closure, $x\in\overline{A}$, there is a sequence (not just a net, but a sequence) $\{a_n\}\subseteq A$ that converges to $x$: $$ a_n\underset{n\to\infty}{\longrightarrow}x. $$

Let $\varPhi$ be a set of continuous maps $\varphi:[0,1]\to[0,1]$ with the following property:

each sequence $\{\varphi_n\}\subseteq\varPhi$ has a subsequence $\{\varphi_{n_k}\}$ that converges to some map $f:[0,1]\to[0,1]$ pointwisely: $$ \forall t\in [0,1]\quad \varphi_{n_k}(t)\underset{k\to\infty}{\longrightarrow}f(t) $$ ($f$ is not necessarily continuous).

We consider $\varPhi$ as a set in the space $[0,1]^{[0,1]}$ of all maps $f:[0,1]\to[0,1]$ with the topology of pointwise convergence (in other words, we treat $[0,1]^{[0,1]}$ as the direct product of $\mathfrak{c}=\operatorname{card}([0,1])$ copies of the interval $[0,1]$).

Question: is the closure $\overline{\varPhi}$ of the set $\varPhi$ in the space $[0,1]^{[0,1]}$ a Fréchet-Urysohn space?


An affirmative answer to this problem is given by a Theorem on page 216 of this book of Diestel. This theorem says that the space $B_1[0,1]$ of functions of the first Baire class is angelic in the topology of pointwise convergence. This means that each relatively countably compact subset of $B_1[0,1]$ is relatively compact and the closure of a relatively compact set is precisely the set of limits of its sequences.

By the assumption, the set $\Phi$ is relatively countably compact in $B_1[0,1]$, so its closure $\bar\Phi$ is compact and Frechet-Urysohn.

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  • $\begingroup$ Taras, does this mean, in particular, that the closure of $\varPhi$ in $B_1[0,1]$ is already compact (so that it coincides with the closure of $\varPhi$ in $[0,1]^{[0,1]}$)? $\endgroup$ – Sergei Akbarov Jun 12 '18 at 15:24
  • $\begingroup$ @SergeiAkbarov I think so. The relative compactness means that the closure of the set in $B_1[0,1]$ is compact (if I understood correctly what was written in Diestel's book). $\endgroup$ – Taras Banakh Jun 12 '18 at 15:37
  • $\begingroup$ Interesting... OK, I'll check it. Thank you, Taras! $\endgroup$ – Sergei Akbarov Jun 12 '18 at 15:39
  • $\begingroup$ Taras, excuse me, if we replace $[0,1]$ by a metric compact space $X$, then Diestel's argument can't be applied directly. Is it possible that nevertheless the same is true in this general situation? $\endgroup$ – Sergei Akbarov Jun 14 '18 at 16:57
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    $\begingroup$ @SergeiAkbarov Because $B_1(X,[0,1]^\omega)$ is homeomorphic to $B_1(X\times\omega, [0,1])$. $\endgroup$ – Taras Banakh Jun 14 '18 at 18:29

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