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Let $X$ be a topological vector space. Assume that there exists a sequence $\phi_n:X\to X$ of finite range measurable functions with $\lim\phi_n(x)=x$ for every $x\in X$. Can we concluded there exists a sequence $\{X_n\}$ of subsets of $X$ with $X=\cup X_n$ such that $X_n$'s are all relatively second-countable?

Note that the answer will be negative if $X$ is only assumed to be a topological space.

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  • $\begingroup$ It took me a minute to see the difference from the previous question: now $X$ is a topological vector space. My first instinct would be to try the space $\mathbb{R}^{\kappa}$ for $\aleph_0 < \kappa \le \mathfrak{c}$ with the product topology, since this is separable but not even first countable. $\endgroup$
    – Nate Eldredge
    Commented May 21, 2018 at 15:16
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    $\begingroup$ @NateEldredge $\mathbb{R}^\mathfrak{c}$ cannot be a counterexample, since a counterexample must have cardinality at most $\mathfrak{c}$. So your idea cannot work in ZFC (but maybe under additional hypotheses?) However I didn't know these spaces were separable and I don't manage to see why, do you know where I can find a proof of it? $\endgroup$
    – N. de Rancourt
    Commented May 21, 2018 at 16:04
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    $\begingroup$ @N.deRancourt: You're right, I need more coffee :-) You can see a proof that such $\mathbb{R}^\kappa$ is separable here. As a concrete example, look at $\mathbb{R}^{\mathbb{R}}$, viewed as the set of all real-valued functions on $\mathbb{R}$. Basic open sets are of the form $U = U_{x_1, \dots, x_n, y_1, \dots, y_n, \epsilon} = \{f : |f(x_i) - y_i| < \epsilon, i = 1,\dots, n\}$. [...] $\endgroup$
    – Nate Eldredge
    Commented May 21, 2018 at 17:08
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    $\begingroup$ It's clear that you may find a polynomial $f$ with $f(x_i) = y_i$ for $i=1, \dots, n$, and you may find a polynomial with rational coefficients that is within $\epsilon$ of $f$ at the points $x_i$. So every basic open set contains a polynomial with rational coefficients, hence these are a countable dense set. $\endgroup$
    – Nate Eldredge
    Commented May 21, 2018 at 17:09
  • $\begingroup$ @AliBagheri do you know the book Counterexamples in topology by Seebach & Steen? If not, you should get it, there are a lot of pathological spaces that are counterexamples to many properties, and even if the properties you consider are not in this book, it could help you to have ideas! $\endgroup$
    – N. de Rancourt
    Commented May 21, 2018 at 21:31

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It seems that the space $X:=C_p(2^\omega)$ of real-valued continuous functions on the Cantor set is a counterexample to this question. The space $C_p(2^\omega)$ is endowed with the topology of pointwise convergence.

Claim 1. The space $X$ cannot be written as the countable union $X=\bigcup_{n\in\omega}X_n$ of first-countable subspaces.

Proof. To derive a contradiction, assume that $X=\bigcup_{n\in\omega}X_n$ for some first-countable subsets. For every $n\in\omega$ let $\bar X_n$ be the closeure of $X_n$ in the topology of the Banach space $C(2^\omega)$. By the Baire Theorem, some $\bar X_n$ contains some ball $B(f;\varepsilon):=\{g\in C(2^\omega):\|f-g\|<\varepsilon\}$. Replacing $B(f,\varepsilon)$ by a smaller ball in $B(f,\varepsilon)$, we can assume that the center $f$ of the ball belongs to $X_n$. Since $X_n$ is first-countable at $f$, there exists a countable neighborhood base $(U_k)_{k\in\omega}$ at $X_n$. We can assume that each $U_k$ is of the basic form $U_k=X_n\cap \{g\in C_p(2^\omega):\max_{x\in F_k}|g(x)-f(x)|<\varepsilon_k\}$ for some finite set $F_k\subset 2^\omega$ and some positive $\varepsilon_k<\frac12\varepsilon$. Choose any point $z\in 2^\omega\setminus\bigcup_{k\in\omega}F_k$ and consider the open neighborhood $V_z:=X_n\cap\{g\in C_p(2^\omega):|g(z)-f(z)|<\frac12\varepsilon\}$. We claim that $U_k\not \subset V_z$ for every $k\in\omega$. Indeed, for every $k\in\omega$, we can find a function $g\in B(f,\varepsilon)$ such that $|g(z)-f(z)|>\frac12\varepsilon$ and $|g(x)-f(x)|<\frac12\varepsilon$ for all $x\in F_k$. Since $X_n\cap B(f,\varepsilon)$ is dense in $B(f;\varepsilon)$, we can replace $g$ by a near function in $X_n$ and assume that $g\in X_n$. Then $g\not\in V_z$ but $g\in X_n\cap B(F_k,\varepsilon_k)=U_k$ and hence $U_k\not\subset V_z$. But this contradicts the choice of $(U_n)_{n\in\omega}$ as a neighborhood base of $X_n$ at $f$. $\square$

Claim 2. There exists a sequence $(\phi_n)_{n\in\omega}$ of measurable (in fact, constructible) finite range functions $\phi_n:C_p(2^\omega)\to C_p(2^\omega)$ that converges pointwisely to the identity function of $C_p(2^\omega)$.

Proof. Since the real line $\mathbb R$ is homeomorphic to the open interval $J:=(-1,1)$, the function space $C_p(2^\omega)$ is homeomorphic to the space $C_p(2^\omega,J)$ of continuous functions from $2^\omega$ to $J$. So, it suffices to construct th sequence $(\phi_n)$ for the function space $C_p(2^\omega,J)$.

The interval $J$ is a bit better than the real line as it admits a sequence of finite range measurable maps $\psi_n:J\to J$, which converges uniformly to the identity function of $J$.

Next, for every $n\in\omega$ denote by $\mathcal C_n$ the canonical finite disjoint cover of the Cantor set $2^\omega$ by basic closed-and-open sets of diameter $\frac1{3^n}$. In each basic set $C\in\mathcal C_n$ fix a point $x_C$ and for any function $f\in C_p(2^\omega,J)$ consider the finite range function $f_n:2^\omega\to C$ such that $f_n(C)=\{f(x_C)\}$ for every $C\in\mathcal C_n$. It is easy to see that the sequence $(f_n)$ converges to $f$ in $C_p(X,J)$.

So, $\varphi_n:C_p(2^\omega,J)\mapsto C_p(2^\omega,J)$, $\varphi_n:f\mapsto f_n$, is a sequence of continuous functions such that $\varphi_n(f)\to f$ for any $f\in C_p(2^\omega,J)$. Finally observe that the sequence of functions $\phi_n:=\varphi_n\circ\psi_n:C_p(2^\omega,J)\to C_p(2^\omega,J)$ has the desired property. They have finite range and converges pointwise to the identity function of $C_p(2^\omega,J)$.

Moreover, each function $\phi_n$ is not just measurable, but constructible. More percisely, the preimage of any open set can be written as the difference $U\setminus V$ of two open sets, because the half-intervals (= $\varphi^{-1}(y)$) on $J$ have this property.

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  • $\begingroup$ Thanks a lot. Just please let me know about the identification $C(2^\omega)\simeq C_p(2^{\omega},J)$ $\endgroup$
    – ABB
    Commented May 23, 2018 at 10:22
  • $\begingroup$ Actually I did not get how the argument shows that $C_p(2^\omega)$ can not be written as a countable union of second countable subsets! $\endgroup$
    – ABB
    Commented May 23, 2018 at 13:47
  • $\begingroup$ @AliBagheri I added detailed proofs. Is it sufficient and clear? And please make your own efforts to understand the proof, in particular, construct a homeomorphism between $C_p(2^\omega)$ and $C_p(2^\omega,J)$. This (almost) trivial and follows from the functoriality of the construction $C_p(\cdot,\cdot)$ (in this case by the second argument). $\endgroup$
    – Taras Banakh
    Commented May 23, 2018 at 14:45
  • $\begingroup$ Dear Taras, Could you please why $\phi_n:=\varphi_n\circ\psi_n:C_p(2^\omega,J)\to C_p(2^\omega,J)$ is finite range valued? I do not get it at all! $\endgroup$
    – ABB
    Commented May 28, 2018 at 10:11
  • $\begingroup$ Indeed, I have doubt it works. $\endgroup$
    – ABB
    Commented May 28, 2018 at 16:04

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