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A regular topological space $X$ is called

$\bullet$ analytic if $X$ is a continuous image of a Polish space;

$\bullet$ $K$-analytic if $X$ is the image of a Polish space $P$ under an upper semicontinuous compact-valued map $\Phi:P\multimap X$.

It is well-known that an analytic space $X$ is countable if and only if every compact subset of $X$ is countable. Is the same fact true for $K$-analytic spaces?

Problem. Is a $K$-analytic space $X$ countable if every compact subset of $X$ is countable?

Remark. By an old result of Fremlin the answer to this problem is affirmative under MA$+\neg$CH. Maybe this fact is absolute? So does not depend on Axioms?


Added in Edit. Since the paper of Fremlin is under payball, I give a simple proof (I do not know if it coincides with the original proof of Fremlin).

Theorem. Under $\omega_1<\mathfrak b$ a $K$-analytic space $X$ is countable if and only if every compact subset of $X$ is countable.

Proof. Write $X$ as the image of the Polish space $P=\omega^\omega$ under an usco map $\Phi:P\multimap X$. Assume that $X$ is uncountable but every compact subset in $X$ is countable. Then we can construct a transfinite sequence of points $\{x_\alpha\}_{\alpha<\omega_1}\subset P$ such that for every $\alpha<\omega_1$ the compact set $\Phi(x_\alpha)$ is not contained in the countable set $\bigcup_{\beta<\alpha}\Phi(x_\beta)$. By the definition of $\mathfrak b$, the set $\{x_\alpha\}_{\alpha<\omega_1}$ is contained in some $\sigma$-compact set and hence there exists a compact subset $K\subset P$ such that $K\cap\{x_\alpha\}_{\alpha<\omega_1}$ is uncountable. Then the compact set $\Phi(K)$ is uncountable, too. This is a contradiction, completing the proof.

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  • $\begingroup$ maybe useful to say that "compact-valued map" is not a map, but a multi-valued map. (It should be part of the Universal Constitution that an attributing adjective shouldn't be disattributing...!) $\endgroup$ – YCor Apr 15 at 9:15
  • $\begingroup$ @YCor This is standard terminology (I mean ``usco" --- "upper semicontinuous compact-valued"). Also "compact-valued multi-valued map" also does not sound good. In fact, a basic notion should be (and is) a multi-valued map (i.e., a relation) and a function is just a special type of relation. Exactly this way it is defined in Set Theory. But I think that the question itself is more interesting than all these formalities around notations and terminology (especially when it is clear what is going on). $\endgroup$ – Taras Banakh Apr 15 at 9:24
  • $\begingroup$ I know it's standard, and that the standard terminology is bad. Anyway most readers (including myself) don't know what a compact-valued map is and would guess it's a kind of map, so I added a comment so they don't also have to make a search. $\endgroup$ – YCor Apr 15 at 9:26
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I looked at the paper of Fremlin and have seen that a minor modification of his example yields the following theorem showing that my question is independent of ZFC.

Theorem. The following statements are equivalent:

1) $\omega_1<\mathfrak b$;

2) A K-analytic space $X$ is analytic if and only if every compact subset of $X$ is metrizable;

3) A K-analytic space $X$ is countable if and only if every compact subset of $X$ is at most countable.

The implication $(3)\Rightarrow (1)$ can be proved as follows. Assuming that $\omega_1=\mathfrak b$, we can find an uncountable subset $B\subset\omega^\omega$ such that for every compact subset $K\subset\omega^\omega$ the intersection $K\cap B$ is at most countable. Consider the space $X=B\cup\{\infty\}$ where $\infty\notin B$ is any point. The topology of the space $X$ is generated by the base

$$\mbox{$\{\{x\}:x\in B\}\cup\{X\setminus D:D\subset B$ is closed and discrete in $X\}$.}$$

The space $X$ is $K$-analytic, being the image of $\omega^\omega$ the upper semicontinuous map $\Phi:\omega^\omega\multimap X$ defined by $$\Phi(x)=\begin{cases} \{\infty,x\}&\mbox{if $x\in B$};\\ \{\infty\}&\mbox{otherwise}. \end{cases} $$ More details can be found in Theorem 4 of this paper.

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