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Fix a set $X$. For a topology $\tau$ on $X$, let $F_\tau = \{f: X \to X \mid f \text{ continuous w.r.t. } \tau\}$. What would be an example of a set $X$ and two Hausdorff topologies $\tau$ and $\sigma$ on $X$ such that $\tau \neq \sigma$ but $F_\tau = F_\sigma$?

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  • $\begingroup$ Maybe explore the very limited world of topologies on finite sets, with maybe 4 or 5 elements. $\endgroup$
    – Dan Asimov
    Mar 1 at 14:54
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    $\begingroup$ @DanAsimov If the set is finite there is only one Hausdorff topology: the discrete one. So, this doesn't work. $\endgroup$
    – psl2Z
    Mar 1 at 14:55
  • $\begingroup$ Well, so much for that idea. Here's another possibly bad idea: What if we define an alternative topology on the real line with a base of half-open intervals [a, b) instead of (a, b). Will this have the same continuous maps ℝ → ℝ as the standard topology? $\endgroup$
    – Dan Asimov
    Mar 1 at 14:57
  • $\begingroup$ @DanAsimov, I think that one is disconnected, so for example there is a continuous map whose image is exactly $\{0, 1\}$. $\endgroup$ Mar 1 at 15:07

1 Answer 1

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One way to find an example is via this fact.

In more detail: suppose $(X, \tau)$ is a nontrivial Hausdorff space with the property that any $\tau$-continuous function $X \to X$ is either the identity or constant. Let $f: X \to X$ be any non-identity bijection of $X$ (eg one that swaps two elements). Let $\sigma$ be the topology on $X$ that makes $f: (X, \tau) \to (X, \sigma)$ into a homeomorphism (ie "transport the structure along $f$"). Then $F_\tau$ and $F_\sigma$ both consist only of the identity function and the constant functions. But $\tau$ and $\sigma$ are not equal, because otherwise $f$ would be a continuous function from $(X, \tau) \to (X, \tau)$.

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    $\begingroup$ But they are still homeomorphic. Is there an example where the topologies are not homeomorphic? $\endgroup$
    – psl2Z
    Mar 1 at 15:09
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    $\begingroup$ @psl2Z, the question didn't say anything about non-homeomorphic. But yes, you can get that. For example, any proper subcontinuum of the Cook continuum can't be homeomorphic to the Cook continuum, so this gives two non-homeomorphic Hausdorff spaces of the same cardinality whose only continuous self-maps are the identity of constant. $\endgroup$ Mar 1 at 15:14
  • $\begingroup$ @Soma, that's not so hard - $\sigma$ is just the collection of all sets of the form $f(U)$ where $U \in \tau$. The idea is to use $f$ to "relabel" the points of $X$. See also here. $\endgroup$ Mar 1 at 15:16
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    $\begingroup$ Thanks!! Very nice example $\endgroup$
    – Soma
    Mar 1 at 15:19

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