Polish spaces are continuous images of the Baire space

Question

I'm having some troubles understanding the proof of Theorem 7.9 (pag. 39) in Kechris' "Classical Descriptive Set Theory":

There are two points of the proof proposed that I don't quite understand. First of all, when we define the Lusin scheme, why do we need to specify that every $F_s$ is going to be a $F_\sigma$ if right afterwards we set $F_s = \bigcup_i \overline{F_{s^\smallfrown i}}$, making it a $F_\sigma$ set by definition. Moreover I truly don't get how does he manage, at the end of the proof, to write $C_{i+1} \setminus C_i = \bigcup_j E_j^{(i)}$ with $E_j^{(i)}$ being pairwise disjoint $F_\sigma$ sets of diameter $< \epsilon$. Where do $(E_j^{(i)})_j$ come from? How do we know that $C_{i+1} \setminus C_i$ can be covered by pairwise disjoint $F_\sigma$ sets of diameter $< \epsilon$?

Answer 1

(ii) is somewhat superfluous. (iii) is already the intended construction: each $F_s$ is partitioned by its successors $F_{s\smallfrown i}$ and these sets are all $F_\sigma$ too. (ii) is just to reinforce and anticipate (iii), I think. (iii) is a statement of intent, not the definition of $F_s$.

The final point is more subtle: $C_{i+1}\setminus C_i$ is a relatively open set of the Polish space $C_{i+1}$. Every open subset of a Polish space can be written as a pairwise disjoint countable union of small diameter $F_\sigma$ sets. This follows as we can take a cover by small (diameter) open sets and as the set is hereditarily Lindelöf (being second countable) we can find a countable subcover of it, which we enumerate. After that, taking the standard trick of subtracting all previous sets, we get a countable disjoint family of sets that are all $F_\sigma$ (the difference of open sets in a metric space, using open sets are $F_\sigma$) and still small.

Answer 2

Ok, I just read the same proof on these notes, and after a while I realized that this proof actually answers my doubts. The fact is that given a $F_\sigma$ set $F$ and having $F = \bigcup_i (C_{i}\setminus C_{i-1})$, we can always find a countable open cover (since Polish spaces are second-countable and therefore Lindelöf) $(U_n^{(i)})$ of $D_i = C_{i}\setminus C_{i-1}$ of diameter $< \epsilon$. Then we can rearrange it in: $$D_i = \bigcup_n [D_i \cap (U_n^{(i)} \setminus \bigcup_{k=1}^{n-1} U_k^{(i)})]$$ If we call $F_{i,n} = D_i \cap (U_n^{(i)} \setminus \bigcup_{k=1}^{n-1} U_k^{(i)})$ then it is easy to see that these new pairwise disjoint sets $(F_{i,n})_{i,n}$ are $F_\sigma$ sets (since we are in a metrizable space) of diameter $< \epsilon$, and also we have that $$\forall i,n \quad \overline{F_{i,n}} \subseteq \bigcup_{j=1}^i D_j = C_i \subseteq F$$. So this proves the claim.