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According to [2]: Let $X$ be a space. We call a system $(X_s)_{s\in T}$ a Sierpinski stratification of $X$ if $T$ is a nonempty tree over a countable alphabet and $X_s$ is a closed subset of $X$ for each $s \in T$ such that:

(i) $X_{\varnothing} = X$ and $X_s =\bigcup \{X_t : t \in \text{succ}(s)\}$ for all $s \in T$, and

(ii) if $\sigma\in [T]$ then the sequence $X_{σ\restriction 0},X_{σ\restriction 1},...$ converges to a point $x_{\sigma} \in X$.

Here $[T]$ is the set of all infinite branches of $T$.

The authors of [2] state that in [1] Sierpinski proved: A (separable metrizable) space $X$ is absolute $F_{\sigma\delta}$ if and only if $X$ has a Sierpinski stratification.

My question concerns replacing condition (ii) with:

(ii') if $\sigma\in [T]$ then $\bigcap_{n<\omega} X_{σ\restriction n}$ is a singleton.

Apparently, if $(X_s)_{s\in T}$ is a Sierpinski stratification of $X$ then (ii') holds. My main question is whether the definition above is equivalent to the definition with (ii') in the place of (ii). Can this be proved simply by refining the stratification so that at level $n$ of the tree, we intersect each $X_s$ with sets from a countable cover of $X$, such that each set in the cover is a closed neighborhood of diameter $<1/n$?

Also I noticed that neither of these definitions will be identical to what Sierpinski describes in [1]. Is there a more recent reference where Sierpinski's theorem and its variants are proved? Surely there have been more proofs since 1924!

[1] Sierpiński, W., Sur une définition topologique des ensembles (F_{\sigma \delta})., Fund. math. 6, 24-29 (1924). ZBL50.0141.03.

[2] Dijkstra, Jan J.; van Mill, Jan, Erdős space and homeomorphism groups of manifolds, Mem. Am. Math. Soc. 979, v, 62 p. (2010). ZBL1204.57041.

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  • $\begingroup$ What do you mean that the sequence of sets converges to a point? I would interpret that as meaning that their intersection is the singleton. So then (ii) and (ii)’ are reformulations. $\endgroup$ Commented Sep 12, 2020 at 17:27
  • $\begingroup$ Suerpiński himself states it as: if we pick a point $p_n$ from each $X_{\sigma\restriction n}$ then $(p_n)$ converges to some point $x$ in the space. $\endgroup$ Commented Sep 12, 2020 at 17:31
  • $\begingroup$ @HennoBrandsma I think that converging to a point means that every neighborhood of the point contains all but finitely many of the sets $X_{\sigma\restriction n}$. So this condition is stronger than Sierpinski's. I think they are likely equivalent, whereas the singleton intersection property might be too weak. $\endgroup$ Commented Sep 12, 2020 at 18:00
  • $\begingroup$ The answer seems to be no, even assuming each $X_s$ is non-empty. Take any $F_{\sigma\delta}$-space $X$ and any non-$F_{\sigma\delta}$-space $Y$. Then $X\times Y$ has a stratification (of non-empty sets) satisfying (ii'), but $X\times Y$ is not $F_{\sigma\delta}$ and therefore does not have a Sierpinski stratification. $\endgroup$ Commented Oct 27, 2020 at 18:16

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