This is an update to an older question.
Is there a contractible $T_2$-space $(X,\tau)$ on more than $1$ point such that no proper subspace of $X$ is homeomorphic to $X$?
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Sign up to join this communityThis is an update to an older question.
Is there a contractible $T_2$-space $(X,\tau)$ on more than $1$ point such that no proper subspace of $X$ is homeomorphic to $X$?
An example of such contractible (compact metrizable) space can be constructed as follows.
Let $K$ be the Cook continuum. It has the property that any continuous map $f:C\to K$ defined on a subcontinuum $C\subset K$ is either constant or the identity inclusion. Take any countable family $(K_{n,m})_{n,m\in\omega}$ of pairwise disjoint non-degenerate subcontinua of $K$. For every $n,m\in\omega$ let $\hat K_{n,m}=K_{n,m}\times [0,1]/(K_{n,m}\times\{0\})$ be the cone over $K_{n,m}$ with vertex $v_{n,m}$.
Let $X_0=\hat K_{0,0}$. For every $n\in\mathbb N$ we shall construct a compact metrizable contractible space $X_n$ that contain the space $X_{n-1}$ as a deformation retract. Assume that for some $n\in\mathbb N$ a space $X_{n-1}$ has been contructed. Fix any function $f_n:\omega\to X_{n-1}$ with dense image $f_n(\omega)$ in $X_{n-1}$. Let $X_n$ be the quotient space of the disjoint union $X_{n-1}\cup\bigcup_{m\in\omega}\hat K_{n,m}$ by the equivalence relation that identifies the vertex $v_{n,m}$ of each cone $\hat K_{n,m}$ with the point $f_n(m)$. The space $X_n$ has a natural compact metrizable topology such that $X_{n-1}$ is a reformation retract of $X_n$. This compeletes the inductive step.
Then the inverse limit $X=\varprojlim X_n$ is a contractible compact Hausdorff space such that every proper subspace $P$ of $X$ is not homeomorphic to $X$. This is because $P$ is either non-compact or not dense in $X$ and hence contains no subset homeomorphic to the continuum $K_{n,m}$ for suitable (large) numbers $n,m$.