# Set of homeomorphic fixed points that is dense, but not equal to whole space

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 $FH(X)\neq X$?

• Note: without the Hausdorff hypothesis, an example is a three-point space with one dense open point and two closed points. – R. van Dobben de Bruyn Sep 3 '18 at 23:51

Such an example can be constructed unifying two pathological examples of Cook and van Mill.

Example (Cook, 1967): 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): 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 $K$).

• Thanks @TarasBanakh for this effort, I am keen to see your new construction! – Dominic van der Zypen Sep 8 '18 at 16:09
• @DominicvanderZypen The construction presented in the answer is just the new (rewritten and correct) construction. – Taras Banakh Sep 8 '18 at 21:00