# Using Baire Category Theorem to prove $\mathbb{R}^2\not\cong\mathbb{R}^3$.

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.

• I edited the question to add some context. Perhaps it is good enough to reopen it? – Prism Oct 14 '14 at 2:09
• @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
• @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
• @Carl Mummert If you believe that Prism should have left the post alone, then why not roll it back? – user1729 Oct 15 '14 at 7:22
• On the contrary, I imagine the usual algebraic topology works quite well in mere ZF. – GEdgar Oct 15 '14 at 14:48