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A subset $A$ of a topological space $X$ is called projectively countable if for any continuous map $f:X\to\mathbb R$ the image $f(A)$ is countable. It is easy to see that each projectively countable set in a finite-dimensional cube $[0,1]^n$ is countable.

Problem. Is each projectively countable set in the Hilbert cube $[0,1]^\omega$ countable?

Added in Edit. The answer is affirmative under $\omega_1<\mathfrak c$ as shown by the following

Theorem 1. For any subset $X\subset K$ of cardinality $|X|<\mathfrak c$ in a compact metrizable space $K$ there exists a continuous function $f:K\to\mathbb R$ such that $f{\restriction}X$ is injective.

Proof. Consider the family $\mathcal H=\{H_{x,y}:x,y\in X,\;x\ne y\}$ of hyperplanes $H_{x,y}=\{f\in C(K):f(x)=f(y)\}$ in the Banach space $C(K)$. By the comment of @fedja to this question, the Banach space $C(K)$ cannot be covered by $<\mathfrak c$ hyperplanes. Consequently, there exists a function $f\in C(K)\setminus \bigcup \mathcal H$. This function has injective restriction $f{\restriction}X$. $\quad\square$

By a similar argument one can prove a linear version of the above theorem.

Theorem 2. For any subset $A\subset X$ of cardinality $|A|<\mathfrak c$ in a separable Banach space $X$ there exists a linear continuous functional $f:X\to\mathbb R$ such that the restriction $f{\restriction}A$ is injective.

Remark. Under CH the above problem has negative answer if this MO-problem has negative answer.

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  • $\begingroup$ I think I know how to show that every projectively-countably subset of the Hilbert Cube is countably-dimensional (so in particular, if there is an uncountable one, then there is an uncountable zero-dimensional projectively countable subset). Would this help? $\endgroup$
    – Arno
    May 1, 2019 at 3:34
  • $\begingroup$ @Arno Each projectively countable subset $A$ of the Hilbert cube $[0,1]^\omega$ is contained in the countable power $C^\omega$ of some countable set $C\subset[0,1]$ and consequently, $A$ is zero-dimensional (not only countably-dimensional). On the other hand, for any compact finite-dimensional subset $K\subset [0,1]^\omega$ the intersection $K\cap A$ is countable. But all this information is insufficient for concluding that $A$ is countable. Moreover, I strongly suspect that under CH the Hilbert cube does contain an uncountable projectively countable subset, put the proof escapes. $\endgroup$ May 1, 2019 at 5:21
  • $\begingroup$ Thanks for pointing out the obvious that I missed - I had taken the scenic route there. I suspect that the computability-theoretic techniques I'd use for this will not work then, but this certainly is an intriguing question. $\endgroup$
    – Arno
    May 1, 2019 at 6:31

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