A topological space $X$ is called a [Fréchet–Urysohn space](https://en.wikipedia.org/wiki/Sequential_space#Fr%C3%A9chet%E2%80%93Urysohn_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?
> 
> 
> 



Fréchet–Urysohn subspaces in $[0,1]^{[0,1]}$
An affirmative answer to this problem is given by a Theorem on page 216 of [this book](https://books.google.com.ua/books/about/Sequences_and_series_in_Banach_spaces.html?id=UVmqAAAAIAAJ&redir_esc=y) 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.


