3
$\begingroup$

Let $X$ be a separable topological space of size $\mathfrak{c}$. By a simple function $\phi:X\to X$, we mean a finite range valued measurable function.

Q. Is it possible to find a sequence of simple functions $\{\phi_n\}$ with $\lim\phi_n(x)=x$ for every $x\in X$?

Improve this question
$\endgroup$
1
  • $\begingroup$ Maybe ask this question for (paracompact) Frechet-Urysohn (or even first-countable) spaces? By the way, for the Sorgenfrey line such a function sequence does exist. $\endgroup$
    – Taras Banakh
    May 19, 2018 at 6:52

1 Answer 1

Reset to default
3
$\begingroup$

No, take any $\mathbb{N} \subseteq X \subseteq \beta\mathbb{N}$ with $|X|=\mathfrak{c}$. If ${\phi_n}$ is any sequence of functions with finite range (measurable or not) and $x\in X \setminus \bigcup_n \phi_n(X)$, then $\lim \phi_n(x) \neq x$.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .