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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1If we remove the "each of which is homemorphic to $X$" condition, the long line is such a space. – Alex Becker May 17 '14 at 01:23
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Is $n$ fixed for this problem? – Cheerful Parsnip May 17 '14 at 01:24
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@GrumpyParsnip It suffices to take $n=1$. – Alex Becker May 17 '14 at 01:27
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1$(0,1) \times (0, 1)$ does not work, but I think $(0, 1) \times [0, 1]$ does. – Niels J. Diepeveen May 17 '14 at 01:29
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@NielsDiepeveen If we remove one point in the center, the result is connected. – Alex Becker May 17 '14 at 01:30
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@AlexBecker: I was not thinking of the usual topology but of the lexicographic order topology, as in the original question. – Niels J. Diepeveen May 17 '14 at 01:32
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@AlexBecker I agree that it suffices to solve it for $n=1$. Conceivably you could find a space where it works for $n=2$ and not $n=1$ though, or am I missing something? – Cheerful Parsnip May 17 '14 at 01:36
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@NielsDiepeveen: I agree. I think $(0,1)\times(0,1)$ works with the order topology. – Cheerful Parsnip May 17 '14 at 01:38
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2@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. – user642796 May 17 '14 at 01:42
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@NielsDiepeveen: oops, I forgot that $(0,1)\times (0,1)$ has infinitely many components! So, yeah, I guess neither of those examples works. – Cheerful Parsnip May 17 '14 at 01:45
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@ArthurFischer: Sorry, I forgot that condition. – Niels J. Diepeveen May 17 '14 at 01:47
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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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