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How can we prove $\mathbb R ^ 2$ is not homeomorphic to $\mathbb R ^3$ using Baire Category Theorem?

Here is a standard proof of this fact using algebraic topology. Note that $\mathbb{R}^{3}-\{x\}$ is homeomorphic to $S^{2}\times\mathbb{R}$ which has trivial fundamental group, but $\mathbb{R}^{2}-\{x\}$ is homeomorphic to $S^{1}\times\mathbb{R}$ which has the fundamental group $\mathbb{Z}$. Hence, $\mathbb{R}^2$ cannot be homeomorphic to $\mathbb{R}^3$.

So it would be interesting to know if Baire Category Theorem can provide another approach.

Martin Sleziak
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Babai
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    I edited the question to add some context. Perhaps it is good enough to reopen it? – Prism Oct 14 '14 at 02:09
  • There's a long list of applications of BCT at http://math.stackexchange.com/questions/165696/your-favourite-application-of-the-baire-category-theorem and this one isn't mentioned, so maybe it's not possible. – Gerry Myerson Oct 14 '14 at 05:09
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    @Prism: please don't do that. It leaves future people unaware what the OP was actually asking. The thing that actually needs clarification, from the OP, is why they believe that there is a solution using the Baire category theorem. That context would more helpful than the algebraic topology argument. – Carl Mummert Oct 14 '14 at 11:00
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    @Carl Mummert: Dear Carl, I definitely agree that we need clarification from OP. However, I have seen it encouraged elsewhere (meta) that "if you think the question is interesting, you can add some context." Perhaps, the question "What are some other ways to prove $\mathbb{R}^2\not\cong\mathbb{R}^3$" would be more welcoming than asking for specifically BCT solution. Hopefully, OP will tell us the source of this interesting problem. – Prism Oct 14 '14 at 12:24
  • @Prism: Form of submission shows respect, or lack thereof. The question prior to your edits looked like vomitted out, hence the closure. This is no longer asker's question, and I don't understand why some users feel compelled to do this. If you'd like the question addressed, re-post in the impeccable way that you formulated it in. – gnometorule Oct 14 '14 at 17:53
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    @Carl Mummert If you believe that Prism should have left the post alone, then why not roll it back? – user1729 Oct 15 '14 at 07:22
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    I did think about it, but I thought that a comment would be a more gentle way to raise the issue. @user1729 – Carl Mummert Oct 15 '14 at 11:09
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    Every proof that $\mathbb R^2$ and $\mathbb R^3$ are not homeomorphic uses some sort of algebraic topology. It cannot be proved in pure "point-set" topology. (How's that for a bold assertion?) – GEdgar Oct 15 '14 at 12:57
  • @GEdgar is there a proof for your statement? ("every proof uses some sort of algebraic topology") Maybe there are models of ZF+DC (Hence ZF + Baire) in wich R^2 are indeed homeomorphic to R^3. – Jonas Gomes Oct 15 '14 at 14:44
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    On the contrary, I imagine the usual algebraic topology works quite well in mere ZF. – GEdgar Oct 15 '14 at 14:48

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