If $(X,\tau)$ is a topological space and $x,y\in X$ we say that $x$ is mappable to $y$ if there is a non-constant continuous map $f:X\to X$ with $f(x) = y$.

Is there a Hausdorff space $X$ with more than one point such that whenever $x\neq y\in X$ and $x$ is mappable to $y$, then $y$ is not mappable to $x$?

  • $\begingroup$ Do you mean a non-constant continuous map? $\endgroup$ – user1688 Feb 2 '17 at 8:09
  • $\begingroup$ mappable?--how about "$x$ pays attention to $y$" or "$x$ appreciates $y$" or .... $\endgroup$ – Włodzimierz Holsztyński Feb 2 '17 at 8:14
  • $\begingroup$ @Anton, in topology it's common to assume that maps or mappings stand for continuous functions. $\endgroup$ – Włodzimierz Holsztyński Feb 2 '17 at 8:17
  • $\begingroup$ The question seems about quasi-ordering not being a partial ordering(?). $\endgroup$ – Włodzimierz Holsztyński Feb 2 '17 at 8:22
  • $\begingroup$ Is there a Hausdorff space (with more than one point) having no non-constant non-identity continuous self-maps? So every continuous self-map is either constant or the identity. Such a space would be an example. $\endgroup$ – Joel David Hamkins Feb 2 '17 at 14:09

Yes, there are such spaces. One example is referenced by Ramiro de la Vega in his answer to this related question.

As Joel points out in the comments, the property you describe, namely "$X$ has an anti-symmetric mappability relation", is strictly weaker than the property "$X$ is strongly rigid", where the latter is defined to mean that the only continuous functions from $X$ to itself are the identity map and the constant functions.

The paper mentioned in Ramiro's answer (available here) gives a hereditarily indecomposable continuum with this property. The construction is due to Howard Cook, and the fact that it has the desired property is expressed in Theorem 11 of his paper. (The theorem states that the identity is the only mapping of $M_2$ onto a non-degenerate subcontinuum of $M_2$ (where $M_2$ is a continuum constructed earlier). Here mapping means continuous function and non-degenerate subcontinuum means a closed connected set having more than one point. Since the continuous image of $M_2$ is always closed and connected, this theorem implies immediately that the only non-identity continuous functions from $M_2$ to itself are the constant maps.)

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  • $\begingroup$ Thanks a lot - I didn't think of the question you referenced to, which I had asked a while ago. Great you made the link! $\endgroup$ – Dominic van der Zypen Feb 2 '17 at 16:26

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