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Let $\mathbb{P}_2(\mathbb{R})$ =( $\mathbb{R}^3$-{0})/~ where x~$r$x for any nonzero point x $\in \mathbb{R}^3$ and any nonzero $r \in \mathbb{R}$

I want to show that the $\mathbb{P}_2(\mathbb{R})$ is compact.. I used quotient map to solve, but it doesn't work. Would you inform me?

NNNN
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  • The image of a compact space by a continous map is compact. Here the projection $$\mathbb R^3-0\longrightarrow (\mathbb R^3-0)/\sim$ is obviously continous and surjective. – palio Oct 15 '13 at 14:35

2 Answers2

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Hint: $\mathbb{P}^2(\mathbb{R})$ is homeomorphic to the space $S^2/{\sim}$ where $x\sim -x$ for all $x\in S^2=\{x\in\mathbb{R}^3\mid \|x\|=1\}$. We know $S^2$ is compact by Heine-Borel.

Dan Rust
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  • By quotient mapping, the space $S^2/{\sim}$ is compact so is the projective plane. Right?? – NNNN Oct 15 '13 at 15:07
  • Yep, so in full $S^2$ is compact hence its image under a continuous map is compact (in this case the quotient map) and so $S^2/{\sim}$ is compact. Homeomorphisms preserve compact spaces and so $\mathbb{P}^2(\mathbb{R})$ is compact as it is homeomorphic to $S^2/{\sim}$ (you may need to prove this last fact if this is homework). – Dan Rust Oct 15 '13 at 15:16
  • In fact, I remembered (but not certainly) that my adviser said this was not compact. Maybe is my memory incorrect?? – NNNN Oct 15 '13 at 15:19
  • Sorry, what's not compact? I essentially gave a proof that $\mathbb{P}^2(\mathbb{R})$ is compact above. – Dan Rust Oct 15 '13 at 15:21
  • I said $\mathbb{P}^2(\mathbb{R})$. Maybe, my memory is incorrect as your reaction. – NNNN Oct 15 '13 at 15:25
  • Well the proof is above so unless I'm missing some detail in your question I think you may be mistaken :). – Dan Rust Oct 15 '13 at 15:26
  • I have one more question. I am distinguishable whether $\mathbb{P}^2(\mathbb{R})$ and $S^2$ are homeomorphic. I think the answer is no, not certainly. Is they are homeomorphic? – NNNN Oct 15 '13 at 15:29
  • They are not homeomorphic. This can be shown in a number of ways. Perhaps the easiest is to show that $S^2$ is an orientable surface and $\mathbb{P}^2(\mathbb{R})$ is a non-orientable surface. Another way would be to show that $\pi_1(S^2)=1$ whereas $\pi_1(\mathbb{P}^2(\mathbb{R}))=\mathbb{Z}/2\mathbb{Z}$ and yet another is to show that they have differing Euler characteristic. Depending on what you've been taught so far, some of these approaches may be more available to you than others. – Dan Rust Oct 15 '13 at 15:31
  • I see.. Thank you very much! – NNNN Oct 15 '13 at 15:33
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Hint: (1) The quotient map $\pi \colon \mathbb R^{3}-\{0\} \to \mathbb P^2(\mathbb R)$ is continuous. (2) The image $\pi[S^2]$ is all of $\mathbb P^2(\mathbb R)$.

martini
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