Let $X$ be a second countable topological vector space. Does there exist any sequence of finite valued functions $f_n\colon X\to X$ converging pointwise to the identity mapping on $X$?

$\begingroup$ Does "finite valued" just mean "takes values in $X$"? $\endgroup$ – LSpice May 5 '18 at 0:02

3$\begingroup$ @LSpice: I think it means that the image of $f_n$ is a finite set. $\endgroup$ – Nate Eldredge May 5 '18 at 0:38
To answer a question raised in comments: if $X$ is merely separable then this can fail. Let $X = \mathbb{R}^\mathbb{R}$ with the product topology, which is a Hausdorff topological vector space. Then $X$ is separable. (Any product of continuum many separable spaces is separable. Alternatively, viewing $X$ as the space of all functions from $\mathbb{R}$ to $\mathbb{R}$, one can show that the set of polynomials with rational coefficients is dense.)
Now let $f_n : X \to X$ be a sequence of finiterange functions converging pointwise to some $f$. Let $A = \bigcup_n f_n(X)$ be the union of the images of the $f_n$, so that $A$ is countable. Then the number of sequences of elements of $A$ has cardinality at most $(\aleph_0)^{\aleph_0} = \mathfrak{c}$, so that the set $B$ of possible limits of sequences from $A$ likewise has cardinality at most $\mathfrak{c}$. But the image of $f$ is contained in $B$. Since $X$ has cardinality $\mathfrak{c}^\mathfrak{c} > \mathfrak{c}$, the function $f$ cannot be surjective.


1$\begingroup$ Let $X$ be a separable topological space and $A$ be a countable dense subset in $X$. Any point $x$ of $X$ is the limit of a sequence of elements contained in $A$. Does not it show that size of $X$ is at most $\mathfrak{c}$? $\endgroup$ – Ali Bagheri May 21 '18 at 16:27

$\begingroup$ @AliBagheri: Nope, your second sentence is false in general. We can say that every neighborhood of $x$ contains a point of $A$. This does not imply that there is a sequence of points of $A$ which converge to $x$, if $X$ is not first countable (or more generally, if $X$ is not sequential). Indeed in this example, if you take the polynomials with rational coefficients, the limit of any sequence of them is Baire class one and in particular Borel. Yet their closure is all of $X$ which contains lots of nonBorel functions. $\endgroup$ – Nate Eldredge May 21 '18 at 17:01
I think it is true (even in the more general setting of second countable topological spaces). Let $\{A_n\}_{n\in\mathbb{N}}$ be a countable basis of the topology of $X$. For ${n\in\mathbb{N}}$ let $\Pi_n$ be the finite partition of $X$ generated by the sets $A_1,\dots, A_n$, and let $f_n:X\to X$ be constant on each $E\in\Pi_n$, with $f_n(E)\in E$. Then $f_n(x)\to x$ for any $x\in X$. (Indeed, for any $x\in X$ and for any nbd $U$ of $x$ there is $n_0\in\mathbb{N}$ such that $x\in A_{n_0}\subset U$. Then by construction, for any $n\ge n_0$ there holds $f_n(x)\in U$).

$\begingroup$ Thanks. What do you mean "even in the more ..."? $\endgroup$ – Ali Bagheri May 4 '18 at 21:48

1$\begingroup$ I mean, not only topological vector spaces, but just topological spaces. $\endgroup$ – Pietro Majer May 4 '18 at 21:50

1$\begingroup$ It seems that this is a characterization of second countability: a Hausdorff (?) topological space is second countable if and only if there is a sequence of finitely valued functions which converge to the identity pointwise. $\endgroup$ – erz May 4 '18 at 21:51

3$\begingroup$ @erz: I don't think so. For instance, if $X$ has this property, then so does any weaker topology on $X$. But it can be possible to take a second countable topology and weaken it to one which is not second countable (but still Hausdorff). For instance, the norm and weak topologies on a separable Banach space. $\endgroup$ – Nate Eldredge May 4 '18 at 22:02

1$\begingroup$ @PietroMajer: I posted another answer with a separable TVS which fails this property. $\endgroup$ – Nate Eldredge May 5 '18 at 0:50
I'll assume that $X$ is Hausdorff.
A Hausdorff topological vector space is metrizable iff it is first countable. Hence, $X$ is metrizable and has a dense subset $\{x_n\}_{n=1}^{\infty}$. Define $f_n(x)$ as an arbitrary closest of $x_1,...,x_n$ to $x$. Then $f_n(x)\to x$. Indeed, $d(x,f_n(x))$ does not increase, and for every $\varepsilon>0$ there is $n$ such that $d(x,f_n(x))\le d(x,x_n)<\varepsilon$.

$\begingroup$ I find amusing that we had two different proofs with very similar wording :) $\endgroup$ – Pietro Majer May 4 '18 at 21:21

$\begingroup$ @PietroMajer I think, they are similar: in my proof I just use a particular base of topology (balls). Mine is probably a bit easier to come up with, and yours is more general. $\endgroup$ – erz May 4 '18 at 21:49

1$\begingroup$ In fact your proof also works without assuming Hausdorff  then the topology is induced by a semidistance, but the conclusion holds true. $\endgroup$ – Pietro Majer May 4 '18 at 22:00

$\begingroup$ @PietroMajer you are right, although it still uses the algebraic structure unlike your proof. $\endgroup$ – erz May 5 '18 at 6:12