# What does the perfect set property really mean?

In the book "Classical Descriptive Set Theory" by Kechris, we have the following definitions:

A topological space $$X$$ is called perfect if it has no isolated points. A subset $$P \subseteq X$$ is called perfect in $$X$$ if it is closed in $$X$$ and is a perfect space with respect to its subspace topology.

Now I looked up in the internet that the perfect set property means the following:

A subset $$A \subseteq X$$ satisfies the perfect set property if it is either countable or contains a non-empty perfect subset.

Now, what is really meant hear? Does it mean $$A$$ has a subset $$P$$ that is closed in $$A$$ and perfect, or does $$P$$ even have to be closed in $$X$$?

Moreover, Kechris calls it the "perfect set theorem for analytic sets" that every uncountable analytic set in a polish space contains a homeomorphic image of the Cantor space. In which of the above sences is a homeomorphic image of the Cantor space a perfect subset?

Thank you for your help!

## 1 Answer

(1) It requires that $$P$$ is closed in the original space $$X$$.

(Actually, say the subspace-psp is the "perfect set property" in the other sense you asked about. Then if $$X$$ has a countable base, every $$A\subseteq X$$ has the subspace-psp. For via the usual Cantor-Bendixson analysis, iteratively remove isolated points from $$A$$, leading to eventual set $$A_\infty$$. That is, set $$A_0=A$$, and given $$A_\alpha$$, set $$A_{\alpha+1}=A_\alpha\backslash I_\alpha$$ where $$I_\alpha\subseteq A_\alpha$$ is the set of elements of $$A_\alpha$$ which are isolated in $$A_\alpha$$, and $$A_\beta=\bigcap_{\alpha<\beta}A_\alpha$$ for limit $$\beta$$. Note for all countable ordinals $$\alpha<\beta$$, $$A_{\alpha}\backslash A_\beta$$ is countable (using that there is a countable base for the topology), and there is some countable $$\alpha$$ such that $$A_\alpha=A_{\alpha+1}$$. Set $$A_\infty=$$ this $$A_\alpha$$.

Now if $$A_\infty=\emptyset$$ then $$A$$ is countable. If $$A_\infty\neq\emptyset$$ then every point in $$A_\infty$$ is non-isolated in $$A_\infty$$, and $$A_\infty$$ is closed in $$A$$. For if $$x\in A$$ is in the closure of $$A_\infty$$ but $$x\notin A_\infty$$, then since $$x\in A=A_0$$, there is some stage $$\alpha$$ such that $$x\in A_\alpha\backslash A_{\alpha+1}$$. So $$x$$ was isolated in $$A_\alpha$$. But $$A_\infty\subseteq A_\alpha$$, so $$x$$ is in the closure of $$A_\alpha\backslash\{x\}$$, a contradiction. Therefore $$A_\infty$$ is perfect in the subspace topology induced by $$A$$, so $$A$$ has the subspace-psp.)

(2) The homeomorphic image of Cantor space is a perfect subset in both senses (but one just implies the other: note that if $$P\subseteq A\subseteq X$$ and $$P$$ is a perfect subset of $$X$$ w.r.t. the $$X$$-topology, then $$P$$ is also a perfect subset of $$A$$ w.r.t. the subspace topology).