# K-analytic spaces whose any compact subset is countable

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.

• 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...!) – YCor Apr 15 at 9:15
• @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). – Taras Banakh Apr 15 at 9:24
• 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. – YCor Apr 15 at 9:26

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.