An ordinal invariant for spaces based on a hierarchy of closed sets from z-sets

2011-01-29T07:11:25Z

EDIT: I apologize for the confusion by which I originally framed this question for normal spaces, where it has an uninteresting answer (thanks to those who pointed this out). Hope I've got it right now.

Given a (Hausdorff/regular/Tychonoff) space $X$, attach to each ordinal a family of sets, as follows:

1) $H(0) = \{X\}$ ;

2) $H(\kappa + 1) = $ the family of all $Z$-sets (zero sets of continuous functions) of sets in $\bigcup_{\lambda \leq \kappa} H(\lambda) $;

3) for non-zero limit ordinal $\kappa$, $H(\kappa) = $ the family of all (nested? does it make a difference?) intersections of sets in $\bigcup_{\lambda < \kappa} H(\lambda) $

Associate to $X$, $\nu=\nu(X)$, the smallest ordinal such that $H(\nu)=H(\lambda)$ for all $\lambda > \nu$, i.e., the ordinal where the hierarchy collapses.

What ordinals occur as $\nu(X)$ for some (Hausdorff/regular/Tychonoff) space $X$? How does one build examples for those ordinal that do so occur?

Answer by Michael Blackmon for An ordinal invariant for spaces based on a hierarchy of closed sets from z-sets

2011-01-29T09:47:39Z

EDIT: This is for the normal case....

$V \in H(1) \cup H(0) $, if and only if $V$ is a zero set of $X$. (this is by definition) So in particular there exists some $f_V \in C(X,[0,1])$ such that $f_V(x) = 0 \iff x \in V$.
If $U \in H(2)$, then there exists some $V \in H(1) \cup H(0)$, and $g_U\in C(V,[0,1])$ such that $g_U(x) = 0 \iff x \in U$
Because $U$ a zero set of $V$, it will be closed, and so because $X$ is normal, we may apply Tietzes' Extension Theorem to produce a continuous map $h_{UV}:X\rightarrow[0,1]$ which extends $g_U$.
Both the functions $h_{UV}$ and $f_V$ are defined on all of $X$, so we may add them to produce the new function $F:X\rightarrow[0,2]$ given by $F(x)=h_{UV}(x) + f_{V}(x)$.
It follows that $U$ is the zero set of $F$, and we have that $U$ is a zero set of $X$.

By (1), (2), (3), (4) and (5), it follows that $U \in H(0) \cup H(1)$ and so $H(2) = H(1) \cup H(0)$. So noting that $X$ is always a zero set of $X$, we have that $H(2) = H(1)$.

Therefore, $\nu(X) = 1$

PS: If you are looking for a hard problem in this form that deals with topology, may I suggest Alan Dows' "Sequential order" chapter from "Open Problems in Topology II"

Edit: attempt at general case removed due to error.

An ordinal invariant for spaces based on a hierarchy of closed sets from z-sets - MathOverflow

An ordinal invariant for spaces based on a hierarchy of closed sets from z-sets

Answer by Michael Blackmon for An ordinal invariant for spaces based on a hierarchy of closed sets from z-sets