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.