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For topological spaces $X,Y,Z$ а function $f:X\times Y\to Z$ is called separately continuous if for any $(x,y)\in X\times Y$ the restrictions of $f$ to the sets $\{x\}\times Y$ and $X\times \{y\}$ are continuous.

It is well-known that for any zero-dimensional compact metrizable space $X$ the subset of injective functions is dense in the Banach space $C(X)$ of all continuous real-valued functions on $X$, endowed with the sup-norm $\|f\|=\sup_{x\in X}|f(x)|$. Is a similar approximation result true for separately continuous functions?

Problem: Let $f:X\times Y\to \mathbb R$ be a (bounded) separately continuous function defined on the product of two zero-dimensional compact metrizable spaces $X,Y$. Is there a separately continuous function $g:X\times Y\to\mathbb R$ such that the image $g(X\times Y)$ is countable (finite) and $\|f-g\|<1$?

In this problem we can assume (without loss of generality) that $X=Y$ is the Cantor cube $\{0,1\}^\omega$.

It should be mentioned that for functions of the first Baire class a similar problem has an affirmative answer, see the proof of Lebesgure-Hausdorff-Banach Theorem 24.10 in Kechris' book "The classical Descriptive Set Theory":

Theorem: For a bounded function $f:X\to\mathbb R$ of the first Baire class on a separable metrizable space $X$ there exists a function $g:X\to \mathbb R$ of the first Baire class such that $g(X)$ is finite and $\|f-g\|<1$.

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  • $\begingroup$ For the night I realized that the answer to this my question is affirmative. It follows from the fact that for each function $f:X\to\mathbb R$ of the first Baire class of a zero-dimensional Polish space $X$, each closed discrete subset $D\subset \mathbb R$ and each $\varepsilon>0$ there exists a continuous function $g:X\to[0,\varepsilon]$ such that $(f+g)(X)\cap D=\emptyset$. I will write down the proof and will post a link. $\endgroup$
    – Taras Banakh
    Sep 18, 2015 at 5:01
  • $\begingroup$ Ups! The proof of this fact on functions of the first Baire class does not work, so I will put it as another question on MathOverFlow. $\endgroup$
    – Taras Banakh
    Sep 18, 2015 at 8:38
  • $\begingroup$ Concerning the first comment, it turns out that there exists a function $f:2^\omega\to\mathbb R$ of the first Baire class such that for any continuous function $g:2^\omega\to\mathbb R$ there is $x\in 2^\omega$ with $f(x)+g(x)=0$, see mathoverflow.net/questions/218609/… $\endgroup$
    – Taras Banakh
    Sep 19, 2015 at 20:40
  • $\begingroup$ This problem is actually about the covering dimension of $2^\omega\times 2^\omega$ endowed with the topology of separate continuity. Is it equal to zero? $\endgroup$
    – Taras Banakh
    Sep 20, 2015 at 7:40

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