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Let $X$ be a space such that for any subset $S \subset X$ with finite cardinality $n$, the subspace $X \setminus S$ has exactly $n+1$ connected components, each of which is homeomorphic to $X$. Is there such a space $X$ which is not homeomorphic to $\mathbb{R}$?

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    $\begingroup$ If we remove the "each of which is homemorphic to $X$" condition, the long line is such a space. $\endgroup$ May 17, 2014 at 1:23
  • $\begingroup$ Is $n$ fixed for this problem? $\endgroup$ May 17, 2014 at 1:24
  • $\begingroup$ @GrumpyParsnip It suffices to take $n=1$. $\endgroup$ May 17, 2014 at 1:27
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    $\begingroup$ $(0,1) \times (0, 1)$ does not work, but I think $(0, 1) \times [0, 1]$ does. $\endgroup$ May 17, 2014 at 1:29
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    $\begingroup$ @NielsDiepeveen: Removing a point $\langle a, b \rangle$ (with $b> 0$) from $(0,1)\times[0,1]$ (with the lex order topology) results in one connected component being homeomorphic to $((0,1]\times[0,1]) \setminus\{\langle 1,1 \rangle\}$ (with the lex order topology). These are not homeomorphic, as you can remove a point from the latter resulting in one connected component homeomorphic to the real line, which is not possible in the former. $\endgroup$
    – user642796
    May 17, 2014 at 1:42

1 Answer 1

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Here's a bit of a cheap example. Consider the topology on $\mathbb{R}$ described in this previous answer. Since the topology is finer than the usual topology, the removal of any point disconnects the space, and it is relatively easy to see that there are two connected components, each of which is homeomorphic to the original space.

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