2
$\begingroup$

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!

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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! $\endgroup$ – David Holmes Dec 17 '20 at 22:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.