Is an open subset of a rigid space rigid?

Question

Let $X$ be a locally compact Hausdorff space. Call $X$ rigid if its only autohomeomorphism is the identity, $\operatorname{Homeo}(X)=\{1\}$.

Questions:

1. Let $X$ be rigid. Is it true that every open subset $Y\subset X$ is rigid, i.e., $\operatorname{Homeo}(Y)=\{1\}$?

2. Let $Y\subset X$ be open. Is it true that every autohomeomorphism of $Y$ extends (perhaps non-uniquely) to an autohomeomorphism of $X$?

3. What can we assume about $X$ additionally to make sure that the answer to question 1 is positive? Metrizability, separability?

Discussion:

An affirmative answer to question 2 obviously implies an affirmative answer to question 1. Question 2 may be asking for too much, but there are simple cases where it is true. For instance, given a continuous bijective function $f:(0,1)\to(0,1)$, it can be continuously continued to a bijective function $F:\mathbb{R}\to\mathbb{R}$, hence every autohomeomorphism of $(0,1)$ can be continued to an autohomeomorphism of $\mathbb{R}$. A positive answer to question 1 seems more realistic to me. For instance, it is true for the Cook's curve, I believe. It is harder for me to check this condition for other examples of rigid spaces in the literature.

Thank you.

Answer 1

The answers to (1) and (2) are "no" in general for metrizable spaces.

Let $C$ be a connected compact metric space with more than one point, with the property that the only continuous self-maps of $C$ are constant maps and the identity. For example, Cook's curve has this property.

Let $C'$ be $C$ with one point removed. Since any autohomeomorphism of $C'$ extends to an autohomeomorphism of $C$ (its one-point compactification), $C'$ is also rigid.

Let $X$ be the disjoint union of $C$ and $C'$. Since $C$ is a connected component of $X$ and has no nonconstant continuous maps to $C'$, any autohomeomorphism $f$ of $X$ must satisfy $f(C)=C$ and hence $f(C')=C'$. Since both $C$ and $C'$ are rigid, $f$ is the identity, and hence $X$ is rigid.

But $X$ has the disjoint union of two copies of $C'$ as an open subspace, which is clearly not rigid, since there is an autohomeomorphism swapping the two copies.

Answer 2

For the third question: If you assume that $X$ is zero-dimensional (i.e., it has a basis consisting of clopen sets), then the answer to question 1 is yes.

Proof: Suppose $U$ is an open, non-rigid subset of $X$, and let $h: U \rightarrow U$ witness its non-rigidity, so that $h(x) = y$ for some $x \neq y \in U$. Using zero-dimensionality, we can find a clopen subset $W$ of $U$ such that $h(W) \cap W = \emptyset$ (let $V_x, V_y$ be disjoint clopen sets containing $x$ and $y$, and let $W = V_x \cap h^{-1}(V_y)$). Then you can define a non-identity homeomorphisms $h': X \rightarrow X$ by taking $h' = h$ on $W$, $h' = h^{-1}$ on $h(W)$, and $h' = \mathrm{id}$ everywhere else.

For example, every rigid subset of the real line is zero-dimensional, so a subset of $\mathbb R$ is rigid if and only if each of its open subsets is.

Answer 3

Here is a counterexample to the first question, a planar, non-separating one-dimensional locally connected continuum.

Let $A_1$ be the line segment in the $xy$-plane $\mathbb{R}^2$ with end points $(0,0)$ and $(1,0)$. Define set $A_2$ as the union of $A_1$ with a sequence of mutually disjoint trees (finite acyclic graphs) $T_i$, no two of them homeomorphic, each attached to $A_1$ at a single point other than the end point of $A_1$. The diameters of the trees should converge to zero, and the set of points of attachment should be dense in $A_1$ (see the first drawing). Also, $A_2\setminus A_1$ should be contained in the vertical open strip $0<y<1$. Observe now that every homeomorphism of $A_2$ onto itself must be the identity on $A_1$.

Next, to construct $A_3$, to each edge of every tree attached to $A_1$, attach a sequence of trees at a single point of the edge, in a similar manner as previously, and iterate this procedure, creating a nested sequence of sets $A_n$, each of which is a tree. Make sure that the diameters of the union $A=A_1\cup A_2\cup A_3\ldots$ is bounded, that the only points of $A$ lying on the boundary of the open strip $0<y<1$ are the end points of $A_1$. Also, make sure that the closure of $A$ (in $\mathbb{R}^2$), which we call $B$, does not contain any simple closed curve. This last requirement can be satisfied by designing the trees attached in the recursive procedure to be sufficiently small.

It is easy to see that every homeomorphism of $A_3$ onto itself must keep every point of $A_2$ fixed, and under any homeomorphism of $B$ onto itself, each point of $A_n$ must be fixed. Hence $B$ is rigid. Let $B’$ be the reflection of $B$ in the $y$-axis, let $X=B\cup(B’)\cup(B+(1,0))$, and let $U=[B\cup(B’)]\setminus\{(-1,0),(1,0)\}$ (see the second drawing). Here is the open set $U$, in the rigid space $X$, and it is not rigid, since it is symmetric about the $y$-axis.

Note 1. The set $B$ can also be described as the inverse limit of the nested sequence $\{A_n\}\ (n=1,2,\ldots)$ with the obvious retractions $r_n$ from $A_{n+1}$ to $A_n$, collapsing each tree in $A_{n+1}$ attached to $A_n$ to its point of attachment, as the bonding maps.

Note 2. In each stage of the construction a sequence of trees is used. In all, a countable number of trees appear; we may assume that all of them are mutually non-homeomorphic, though it is not quite necessary.