# Space of functions with finite/countable support Fréchet-Urysohn?

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. 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!

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.
• Thank you so much! This has been driving me nuts... I missed the idea of including some of the support of $f_2$ in the set on which I make $f_3$ small. I’d somehow convinced myself it should really be false and was trying to construct counterexamples. I’ll sleep better now! Dec 17, 2020 at 22:17