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.

  • $\begingroup$ 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 May 21 '18 at 15:16
  • 1
    $\begingroup$ @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 May 21 '18 at 16:04
  • 1
    $\begingroup$ @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 May 21 '18 at 17:08
  • 1
    $\begingroup$ 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 May 21 '18 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 May 21 '18 at 21:31

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.

| cite | improve this answer | |
  • $\begingroup$ Thanks a lot. Just please let me know about the identification $C(2^\omega)\simeq C_p(2^{\omega},J)$ $\endgroup$ – Ali Bagheri May 23 '18 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! $\endgroup$ – Ali Bagheri May 23 '18 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 May 23 '18 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! $\endgroup$ – Ali Bagheri May 28 '18 at 10:11
  • $\begingroup$ Indeed, I have doubt it works. $\endgroup$ – Ali Bagheri May 28 '18 at 16:04

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.