Consider an uncountable perfect $K_\sigma$ set $X\subseteq \omega^\omega$, where $K_\sigma$ means countable union of compact sets, perfect means that $X$ has no isolated points and $\omega^\omega$ is the Baire space. Suppose now that there exists a subset $A\subseteq X$ which is dense in $X$ (wrt the subspace topology).
My question is:
- Can we prove that there exists a Cantor set $\mathcal{C}\subseteq X$ such that $\mathcal{C}\cap A$ is dense in $\mathcal{C}$?
Since $X$ is uncountable it must contain a Cantor set, as all analytic sets have the perfect set property, but here we are looking for a Cantor set that preserves the property of $A$ of being dense.
Ideas?
Thanks!
Best Answer
Here's a counterexample. Let $X$ be the set of bounded sequences, and let $A$ be the set of sequences which have only finitely many nonzero terms and achieve a strict maximum at the last nonzero term. Clearly $A$ is a dense subset of $X.$
Let $P \subset \omega^{\omega}$ be a perfect set such that $P \cap A$ is dense in $P.$ We will show $P \not \subset X.$
Enumerate $P \cap A.$ Let $p_0$ be the first sequence, and recursively define $p_{i+1}$ to be the first sequence such that the nonzero part of $p_{i+1}$ strictly extends the nonzero part of $p_i$ (here we are using that $p_i$ is a limit point of $P$). Clearly $\langle p_i \rangle$ converges to an unbounded sequence $p,$ so $p \in P \setminus X.$