I'm looking at the proof of the following theorem by Hurewicz (7.10 in Kechris' Descriptive Set Theory):

Suppose $X$ is a Polish Space. Then one of the following holds:

  1. $X$ is $K_\sigma$ (i.e, $\sigma$-compact, i.e, a countable union of compact sets.

  2. $X$ has a closed subset homeomorphic to $\omega ^{\omega}$

My question is about a very small step in the proof of this. We build a Lusin scheme as expected,and something that comes up in that process is the following:

If $F \subseteq \omega ^ {\omega}$ is not $K_\sigma$, then $H:= \{x \in F: \forall U$ open nbhd of $x$, $\overline{U \cap F}$ is not $K_\sigma\}$ (this is what I called the perfect kernel analogue)

Now the text quickly says "$H$ is nonempty since $F$ is not $K_\sigma$, and similarly $F\setminus H$ is contained in a $K_\sigma$ set", but I don't see why this is true. I must be missing something obvious given that the book skips over it so I apologize if it's super dumb.

My guess is: suppose $H \not = \emptyset$. Then since $X$ is separable, there is a countable set of points $\{x_i\}$ dense in $F$ with open neighborhoods $U_i$ such that $\overline{U_i \cap F}$ is $K_\sigma$. These $U_i$ cover $F$ and thus $F$ is a countable union of $K_\sigma$ sets and thus $K_\sigma$, yielding a contradiction. A similar idea can be used to show $F\setminus H$ is $K_\sigma$. I don't think this is entirely right though becuase the $U_i$ don't have to cover $F$ necessarily.

What is the right approach here? I feel like I'm missing something super obvious.


Your idea seems correct to me, except for a typo at the beginning where it should be suppose $H=\varnothing$ rather than $H\neq\varnothing$ to reach a contradiction.

You say that the $U_i$ don't have to cover $F$, but in fact they do since the $\{x_i\}$ are a dense subset.

Similarly pick a countable dense subset $\{x_i\}$ in $F\setminus H$. By definition of $H$ each of those points has a neighbouhood $U_i$ such that $\overline{F\cap U_i}$ is $K_\sigma$, thus $F\setminus H\subseteq\bigcup_{i<\omega}\overline{F\cap U_i}$ is contained in a $K_\sigma$ set, which means that $H$ cannot be compact, otherwise the whole of $F$ would be $K_\sigma$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.