4
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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, 2019 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). $\endgroup$ Apr 15, 2019 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, 2019 at 9:26

1 Answer 1

1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.