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Let $(X,\tau)$ be a topological space. If $A\subseteq X$ we define the following equivalence relation on $X$: $$\sim_A = \{(x,y) \in X^2: x=y \text{ or }\{x,y\}\subseteq A \}.$$

Let $(X,\tau)$ be an infinite space with the following property:

If $A\subseteq X$ and $|A|<|X|$ then $X\cong X/\sim_A$.

Does this imply that $(X,\tau)$ is one of the following?

  1. $(X,\tau)$ is discrete;
  2. $(X,\tau)$ is indiscrete;
  3. $X$ can be well-ordered with a relation $\leq$ such that $\tau$ consists of the upper sets of $(X,\leq)$.
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  • $\begingroup$ Do you assume $|X|$ is a regular cardinal? I ask because for spaces with property 3 to have the property, one seems to need that $|X|$ is regular. $\endgroup$ – Joel David Hamkins Aug 6 '17 at 15:36
  • $\begingroup$ Thanks for your comment! The question is: if $|X|$ is singular, are there topologies with the desired property that are neither discrete nor indiscrete? $\endgroup$ – Dominic van der Zypen Aug 6 '17 at 15:51
  • $\begingroup$ Not sure, I don't have a good grasp of it yet. $\endgroup$ – Joel David Hamkins Aug 6 '17 at 15:54
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    $\begingroup$ What if you topologize $X$ so that the closed sets are exactly those $C\subseteq X$ satisfying $|C|<|X|$? $\endgroup$ – Keith Kearnes Aug 6 '17 at 18:59
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For a countable space $X$ the discussed property means that $X$ is homeomorphic to $X/A$ for any finite subset $A\subset X$. Let us call this property quotient-homogeneous.

There are many quotient-homogeneous countable spaces which do not fall into the categories (1)-(3) of Dominic van der Zypen:

4) any countable $T_1$-space with a unique non-isolated pont;

5) the space of rational numbers.

It seems that many (but not all) countable ordinals endowed with the interval topology are quotient-homogeneous.

Observe that spaces of form (3) do not satisfy the separation axiom $T_1$. Maybe Dominic van der Zypen had in mind the interval topology (not the upper-set topology)? But anyway the space of rationals is a counterexample.

The space pointed by @Keith Kearnes is another (non-metrizable) counterexample.

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