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.


1 Answer 1


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 $K_\sigma$.


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