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](https://doi.org/10.1112/S0025579300009165) the answer to this problem is affirmative under MA$+\neg$CH. Maybe this fact is absolute? So does not depend on Axioms?




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**Added in Edit.** Since the [paper](https://doi.org/10.1112/S0025579300009165) 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.
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*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.



K-analytic spaces whose any compact subset is countable
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:
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> 
> 1) $\omega\_1<\mathfrak b$;
> 
> 
> 2) A K-analytic space $X$ is analytic if and only if every compact subset of $X$ is metrizable;
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> 
> 3) A K-analytic space $X$ is countable if and only if every compact subset of $X$ is at most countable.
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> 


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](http://%20%5Cnewcommand%7B%5CIZ%7D%7B%5Cmathbb%20Z%7D).


