Invariants of category in Polish spaces

Question

Consider the invariants appearing in the Cichoń's diagram: $add(\mathcal I)$, $cov(\mathcal I)$, $non(\mathcal I)$, $cof(\mathcal I)$, where $\mathcal I$ is either the ideal of null sets for the Lebesgue measure or the ideal of meager sets.

By theorem 17.41 of Kechris' book Classical Descriptive Set Theory, if $X$ is any Polish space and $\mu$ is any continuous Borel measure on $X$ (continuous = points are null sets), there exists a Borel isomorphism $f:X\longrightarrow [0,1]$ through which $\mu$ corresponds to the Lebesgue measure.

This result implies that the cardinal invariants corresponding to the ideal of null sets are the same for every Polish space and every continuous Borel measure on it.

I would like to know if there is an analogous result for the ideal of meager sets. Specifically: Are the four invariants the same for the ideal of meager sets of any perfect Polish space?

I suspect that the answer should be "yes", but I have not seen any result in this direction.

Of course, the answer is positive when restricted to the "usual" Polish spaces: $[0,1]$, $\mathbb R$, $^\omega\omega$, $^\omega2$, etc.

Answer 1

Your suspicion is correct -- the answer is yes.

To see this, first observe that if $Y$ is a dense $G_\delta$ subset of (a perfect Polish space) $X$, then $X$ and $Y$ must agree on the four cardinal invariants you mention. Second, I claim that that the Baire space $\omega^\omega$ is a dense $G_\delta$ subspace of every perfect Polish space. Therefore all perfect Polish spaces agree with the Baire space (and hence with each other) on the value of these four invariants.

To prove this claim, begin with any perfect Polish space $X$. If $B$ is a countable basis for $X$, then $Y = X - \bigcup_{U \in B}\partial U$ is a dense $G_\delta$ subspace of $X$, and is also zero-dimensional. If $D$ is a countable dense subset of $Y$, then $Z = Y - D$ is a dense $G_\delta$ subset of $X$ that is zero-dimensional, and in which no nonempty open subset is compact (because no open subset is closed in $Y$). By a theorem of Alexandrov and Urysohn (Theorem 7.7 in Kechris's book), $Z$ is homeomorphic to the Baire space.