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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?

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  • $\begingroup$ Won't $X$ itself be in every family, since it is the zero set of the constant zero function? And in this case, we add it back in at each step. Could you explain? Also, in (1) you haven't described $H(0)$ as a family of sets, and in (2) you are taking the union of families, which would be a family, but I think you intend to take the union of the sets in that union family---could you clarify? $\endgroup$
    – Joel David Hamkins
    Commented Jan 29, 2011 at 9:14
  • $\begingroup$ For any normal space $X$ we have $\\{\\{x\\}: x\in X\\} \subset H(1)$ so $\bigcup_{\lambda\le1} H(\lambda)$ must contain $\\{\\{x\\}: x\in X\\}$ and every point of $X$. So the hierarchy is degenerate, and I don't think what you intended. $\endgroup$
    – Not Mike
    Commented Jan 29, 2011 at 9:17
  • $\begingroup$ Thanks both, right, I garbled my own question. Think I've improved it now. $\endgroup$
    – David Feldman
    Commented Jan 29, 2011 at 20:34

1 Answer 1

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

  1. $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 $f_V(x) = 0 \iff x \in V$.

  2. 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 $g_U(x) = 0 \iff x \in U$

  3. 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 $g_U$.

  4. 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 $F(x)=h_{UV}(x) + f_{V}(x)$.

  5. 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.

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  • $\begingroup$ You write: Then, for every $f \in C(X_1, [0,1])$, we have that $f \in C(X_0, [0,1])$ (as $f\circ i$ is continuous) and so $Z$ is a zero set for $X_0$ implies that $Z$ is a zero set for $X_1$, and every zero set of $X_1$ is contained in a zero set for $X_0$. Real-made-discrete maps ($X_0$) continuously to Real-standard-topology ($X_1$). The rationals are a zero set in $X_0$, but not in $X_1$. Of course it's true that zero set in $X_1$ is a zero set in $X_0$. $\endgroup$
    – David Feldman
    Commented Jan 30, 2011 at 0:10
  • $\begingroup$ Real-made-discrete? not sure I follow what you mean by that. $\endgroup$
    – Not Mike
    Commented Jan 30, 2011 at 0:16
  • $\begingroup$ spent about an hour trying to patch my proof, but I guess I should have thrown in the towl. Anyway, after playing around with it, I'd conjecture that $\nu(X)=1$ obtains, because the relative topology is always finer than the full spaces. $\endgroup$
    – Not Mike
    Commented Jan 30, 2011 at 0:20
  • $\begingroup$ Or something, idk... $\endgroup$
    – Not Mike
    Commented Jan 30, 2011 at 0:21
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    $\begingroup$ Indeed, I think so, because of how the tangent discs intermingle. So that would do it, that would be an example of a space with $\nu(X)=2$ $\endgroup$
    – Not Mike
    Commented Jan 30, 2011 at 2:15

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