We consider a minimal compact metric flow $(X,T)$, where $T$ is a group, and a minimal idempotent $u^2=u\in E(X)$, where $E(X)$ is the Ellis semigroup (or enveloping semigroup) of the flow $(X,T)$. Using these, we define a subset of $X$ by $$ X_u:=\{x\in X: ux=x\}, $$ i.e. the set of all points in $X$ which are fixed by $u$.

Does $X_u$ contain any proximal points? (We say $x$ and $y$ are proximal points if $\inf_{t\in T} d(tx,ty)=0$.) My intuition would be 'no', since we have that a point is distal (i.e. proximal only to itself) if and only if all idempotents $u\in E(X)$ fix $x$, i.e. $ux=x \ \forall u^2=u\in E(X)$, and the condition used to define $X_u$ is related (though not the same). Any thoughts/suggestions/references?


The answer is no. According to http://ac.els-cdn.com/S0166864107001538/1-s2.0-S0166864107001538-main.pdf?_tid=281576e6-9489-11e6-bb98-00000aab0f01&acdnat=1476722928_757fc4c08fcbc966d5afdf7e0c9e0477 for $x,y$ to be proximal there must be $s$ in the enveloping semigroup with $sx=sy$. Then since $u$ is minimal, $su$ and $u$ generate the same left ideal. So $tsu=u$ for some $t$ in the semigroup. Then $x=ux=tsux=tsx=tsy=tsuy=uy=y$.

  • $\begingroup$ Can you elaborate on why the existence of a distal point shows that there are proximal points in $X_u$? If I understand correctly, you do not use the assumption that there are no distal points in $X_u$ anywhere in your argument which shows that if $x,y\in X_u$ are proximal then $x=y$. $\endgroup$ – Simon_Peterson Oct 17 '16 at 17:32
  • $\begingroup$ Sorry I misread the definition of distal. There are no proximal points. I will edit. $\endgroup$ – Benjamin Steinberg Oct 17 '16 at 22:40

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