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A topological space $(X,\tau)$ is said to be homogeneous if for any $x,y\in X$ there is a homeomorphism $\varphi:X\to X$ such that $\varphi(x) = y$. Moreover, we say that $(X,\tau)$ is shrinkable if there is $S\subseteq X$ with $S\neq X$ and a homeomorphism $\psi: X\to S$ where $S$ is endowed with the subspace topology.

The real interval $[0,1]$ is shrinkable (it is homeomorphic to $[0,1/2]$), but not homogeneous. $S^1$ is homogeneous, but not shrinkable.

Is there an example of a shrinkable, homogeneous, connected, and compact $T_2$-space?

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    $\begingroup$ Just a remark: such a space can't have a open subset homeomorphic to $\mathbf{R}^n$. This would mean by homogeneity that it is a topological $n$-manifold. However compact connected topological manifold can't be homeomorphic to a proper subsbspace by Poincare-Lefschetz Duality (see Bredon corollary 8.4). $\endgroup$
    – Adam Chalumeau
    Commented Nov 17, 2020 at 8:32
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    $\begingroup$ The product $S^\mathbf{N}$, where $S$ is the circle. $\endgroup$
    – YCor
    Commented Nov 17, 2020 at 11:06
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    $\begingroup$ After @YCor, now we can ask about finite-dimensional shrinkable homogeneous Hausdorff continua, if there is any. $\endgroup$
    – Wlod AA
    Commented Nov 17, 2020 at 11:38
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    $\begingroup$ @WlodAA The pseudoarc makes the job. $\endgroup$
    – YCor
    Commented Nov 17, 2020 at 12:29
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    $\begingroup$ $[0,1]^\Bbb N$ also works, since it is (perhaps surprisingly) homogeneous $\endgroup$
    – Alessandro Codenotti
    Commented Nov 17, 2020 at 13:56

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While browsing the continuum-theory tag I noticed that this question was never answered, so I'm adding a CW answer just to remove it from the unanswered list.

As pointed out in the comments both the countable products of copies of $S^1$ (or any other homogeneous continuum) or of $[0,1]$ work. The pseudoarc does too for a finite dimensional example.

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