If $(X,\tau)$ is a topological space, let $FH(X)$ denote the collection of $x\in X$ such that there is a non-identity homeomorphism $\varphi:X\to X$ with $\varphi(x) = x$.


What is an example of a $T\_2$-space $(X,\tau)$ such that $FH(X)$ is dense in $X$, but $FH(X)\neq X$?



Set of homeomorphic fixed points that is dense, but not equal to whole space
Such an example can be constructed unifying two pathological examples of Cook and van Mill.


**Example** ([Cook, 1967](http://matwbn.icm.edu.pl/ksiazki/fm/fm60/fm60123.pdf)): There exists a non-degenerated metric continuum $K$ such that any continuous map $f:K\to K$ is either constant or the identity.


**Example** ([van Mill, 1983](http://www.ams.org/journals/tran/1983-280-02/S0002-9947-1983-0716833-2/S0002-9947-1983-0716833-2.pdf)): There exists a metrizable separable Boolean group $G$ that admits no homeomorphism of $G$ has no fixed points.


The van Mill group $G$, being separable, contains a dense countable subgroup $G\_\omega$. The countable subgroup $G\_\omega$, being Boolean, can be written as the countable union $G\_\omega=\bigcup\_{n\in\omega}G\_n$ of an increasing sequence $(G\_n)\_{n\in\omega}$ of finite Boolean subgroups. Since each subgroup $G\_n$ is complemented in $G\_{n+1}$, we can fix a homomorphism $r\_n:G\_{n+1}\to G\_n$ such that $r\_n(x)=x$ for all $x\in G\_n$. 


Now take the Cook continuum $K$ and fix two distinct points $a,b\in K$.


Consider the space $$U:=(K\times G\times\{\omega\})\cup\bigcup\_{n\in\omega}K\times G\_n\times\{n\}\subset K\times G\times(\omega+1)$$where the ordinal $\omega+1=\omega\cup\{\omega\}$ is endowed with the order topology (which is compact and metrizable).


On the space $U$ consider the equivalence relation $$
\begin{aligned}
\sim:&=\{((x,g,\omega),(y,g,\omega)):x,y\in K,\;g\in G\}\cup\\
&\cup\bigcup\_{n\in\omega}\bigcup\_{g\in G\_{n+1}}\{((a,g,n+1),(b,r\_n(g),n)),((b,r\_n(g),n),(a,g,n+1))\}.
\end{aligned}
$$


Consider the quitient space $X:=U/\_\sim$ and the corresponding quotient map $q:U\to X$. The space $X$ contains a closed copy $G':=q(K\times G\times \{\omega\})$ of the van Mill's group $G$ and the complement $X\setminus G'$ is a connected dense set, which is the union of countably many copies $K\_{n,g}:=q(K\times \{(g,n)\})$, $g\in G\_n$, of the Cook continuum $K$. Each $K\_{n,g}$ intersects the continua $K\_{n+1,p}$ where $p\in r\_n^{-1}(g)$.


It can be shown that the space $X$ has the required property:
the set $FH(X)$ is dense not not coincide with $X$. More precisely, $FH(X)$ is equal to the complement $X\setminus G'$ of the copy $G'$ of the group $G$ in $X$. Indeed, for any $s\in G\_{n+1}$ with $r\_n(s)=0$ the translation $x\mapsto x+s$ of the group $G$ extends to a homeomorphism of $X$ that does no move points of the union $\bigcup\_{i\le n}\bigcup\_{g\in G\_i}K\_{i,h}$. So, $FH(X)\supset X\setminus G'$. On the other hand, any homeomorphism of $X$ that fixes some point of the group $G'$ is identity on $G'$ and extends unquely to the identity homeomorphism of $X$ (because of the rigidity properties of the Cook continuum $K$). 


