Does $X \subseteq \mathbb N^{\mathbb N}$ non-countable and $F_{\sigma}$ imply that $X$ contains a perfect set?

Question

I think that the claim below is true, but whenever I try to prove it, I find myself using the continuum hypothesis ($\aleph_1 = \mathfrak c$).

My question: Can the following statment be proved without using CH (continuum hypothesis)?

Let $X \subseteq \mathbb N^{\mathbb N}$ be a non-countable set which is $F_{\sigma}$. Does this imply that $X$ contains a perfect set?

Thank you!

Answer 1

Yes. If $X$ is an uncountable Borel set of a standard Borel space (e.g. Baire space), then $X$ includes a perfect set.

You should be able to find the proof in any standard descriptive set theory book.