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If $(X,\tau)$ is a topological space, we say $Y\subseteq X$ is homeomorphism-fixing if the only homeomorphism $\varphi:X\to X$ such that for all $y\in Y$ we have $\varphi(y)=y$ is the identity map. Moreover we say that $Y$ is minimally homeomorphism-fixing if for all $y\in Y$ the set $Y\setminus \{y\}$ is not homemorphism-fixing.

What is an example of an infinite Hausdorff space without minimally homeomorphism-fixing subsets?

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The real numbers.

Any dense subset, like the rationals is homeomorphism fixing. Conversely if that subset $Y$ is not dense, we can find a point $x\in X$ and a small interval around $x$ that does not contain any element of $Y$. Now it is easy to construct a homeomorphism that is the identity outside of that interval (and thus fixes $Y$) and does something on that interval (reparametrizing).

If you remove a single point from a dense subset, that subset is still dense and thus homeomorphism fixing. Thus there are no minimal homeomorphism fixing subsets.

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    $\begingroup$ This argument also shows that the class of examples is enormous... $\endgroup$ – Todd Trimble May 22 '18 at 14:26
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    $\begingroup$ Yes. It seems more interesting to ask for reasonable spaces for which the notions of denseness and homeomrohism fixing don't agree. A non-reasonable example is the Sierpinski- space. (The one with two points, where only one of them is closed). I don't see an Hausdorff example yet. $\endgroup$ – HenrikRüping May 22 '18 at 14:41
  • $\begingroup$ Is dense the equivalent of homeomorphism fixing for weaker separation axioms? Gerhard "Don't Keep Me From Knowing!" Paseman, 2018.05.22. $\endgroup$ – Gerhard Paseman May 22 '18 at 14:43
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    $\begingroup$ Any rigid space has every set as a homeomorphism fixing set. So in general, it isn't equivalent to density. $\endgroup$ – Joel David Hamkins May 22 '18 at 14:59

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