Distinguishing perfect set properties in separable metrizable spaces

TL;DR

This paper explores the perfect set property in separable metrizable spaces, linking it to set-theoretic assumptions like $rak{b}>\omega_1$, and introduces the Grinzing property to analyze pointclass complexities.

Contribution

It establishes the independence of certain perfect set property statements from ZFC and introduces the Grinzing property for subsets of $2^\omega$.

Findings

The statement about perfect set properties is equivalent to $rak{b}>\omega_1$.

Existence of subsets with the Grinzing property under certain set-theoretic assumptions.

Characterization of the Grinzing property's presence under $ ext{MA}+ eg ext{CH}$ and $ ext{CH}$.

Abstract

All spaces are assumed to be separable and metrizable. Our main result is that the statement "For every space $X$, every closed subset of $X$ has the perfect set property if and only if every analytic subset of $X$ has the perfect set property" is equivalent to $\mathfrak{b}>\omega_1$ (hence, in particular, it is independent of $\mathsf{ZFC}$). This, together with a theorem of Solecki and an example of Miller, will allow us to determine the status of the statement "For every space $X$, if every $\mathbf{\Gamma}$ subset of $X$ has the perfect set property then every $\mathbf{\Gamma}'$ subset of $X$ has the perfect set property" as $\mathbf{\Gamma},\mathbf{\Gamma}'$ range over all pointclasses of complexity at most analytic or coanalytic. Along the way, we define and investigate a property of independent interest. We will say that a subset $W$ of $2^\omega$ has the Grinzing property if it is uncountable and for every uncountable $Y\subseteq W$ there exists an uncountable collection consisting of uncountable subsets of $Y$ with pairwise disjoint closures in $2^\omega$. The following theorems hold. (1) There exists a subset of $2^\omega$ with the Grinzing property. (2) Assume $\mathsf{MA}+\neg\mathsf{CH}$. Then $2^\omega$ has the Grinzing property. (3) Assume $\mathsf{CH}$. Then $2^\omega$ does not have the Grinzing property. The first result was obtained by Miller using a theorem of Todor\v{c}evi\'c, and is needed in the proof of our main result.