Let $(X,\tau)$ be a topological space. By $\text{End}(X)$ we denote the collection of all continuous functions $f:X\to X$. We say $f,g\in \text{End}(X)$ share a point if there is $x\in X$ such that $f(x) = g(x)$ and set $$E_{\text{End}(X)} = \big\{\{f,g\}: f\neq g\in \text{End}(X)\text{ and } f,g \text{ share a point}\big\}.$$ So $G_X := (\text{End}(X), E_{\text{End}(X)})$ is a simple, undirected graph. Is there a topological space $X$ such that no clique in $G_X$ has cardinality $\chi(G_X)$?

  • 1
    $\begingroup$ Thanks for your comment, Peter! For the discrete topology ${\cal P}(X)$ the answer is that $\chi(G_X)$ equals the cardinality of some clique in $G_X$, so I was wondering whether there is some "interesting" topology where this is not the case. $\endgroup$ – Dominic van der Zypen Feb 21 '18 at 14:30
  • 1
    $\begingroup$ Here is an idea that someone might turn into a proof. If f is an endomorphism of finite order with a fixed point (say on N, pick an integer n and let f be Max(0, (x mod n )-1) ), then the identity f and its powers form a finite clique, and this clique gets copied by acting by g on each of these maps. If every endomorphism has a finite order with fixed point, but the orders are bounded, this might be an example if there are enough different fixed points. Gerhard "How To Move Fixed Points?" Paseman, 2018.02.22. $\endgroup$ – Gerhard Paseman Feb 22 '18 at 18:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.