Let $S$ be an uncountably infinite set (mainly interested in case that $S$ has same cardinality as $\mathbb R$) and look at the set $F$ of functions $f\colon S \to \mathbb R$. I equip $F$ with the topology of pointwise convergence. It is not hard to show that $F$ is not [Fréchet-Urysohn](https://en.wikipedia.org/wiki/Fr%C3%A9chet%E2%80%93Urysohn_space). On the other hand, define $F^f$ and $F^c \subseteq F$ to be the subsets of $F$ consisting of functions with finite, respectively countable, support. I'm trying to understand whether these spaces (equipped with the subspace topology from $F$, equivalently with the topology of pointwise convergence) are again Fréchet-Urysohn. My reason for expecting/hoping this is that my counterexample to the Fréchet-Urysohn property for $F$ makes essential use of functions with uncountable support; but this is of course not a proof. Thanks!



Space of functions with finite/countable support Fréchet-Urysohn?
Yes, these spaces are Fréchet-Urysohn.


Consider $F^c$; the proof for $F^f$ is the same. Let $A \subset F^c$ and suppose $f \in \overline{A}$; since $F^c$ is a topological vector space (as is $F^f$) we can suppose without loss of generality that $f = 0$. Note this means that for every finite set $S\_0 \subset S$ and every $\epsilon > 0$ we can find some $g \in A$ with $|g(x)| < \epsilon$ for every $x \in S\_0$.


We construct inductively a sequence $f\_n \in A$ which converges pointwise to $0$. Let $f\_1$ be arbitrary and enumerate its support as $x\_{1,1}, x\_{1,2}, \dots$. Then choose $f\_2$ such that $|f\_2(x\_{1,1})| < 1/2$, and enumerate the support of $f\_2$ as $x\_{2,1}, x\_{2,2}, \dots$. Then choose $f\_3$ such that $|f\_3(x\_{i,j})| < 1/3$ for $i,j = 1,2$, and continue in this way to produce a sequence $f\_n$ with the property that $|f\_n(x\_{i,j})| < 1/n$ for all $i,j = 1, \dots, n-1$, where $\{x\_{i,j}\}\_{j=1}^\infty$ enumerates the support of $f\_i$.


Now consider any arbitrary $x \in S$. If $x = x\_{i,j}$ for some $i,j$, then letting $N = \max(i,j)$, we have $|f\_n(x)| < 1/n$ for all $n \ge N$ and hence $f\_n(x) \to 0$. Otherwise, $x$ is outside the support of all the $f\_n$ and then $f\_n(x) \to 0$ trivially.


