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As the title said, I want to know how to prove that $\mathbb{R}P^2$ is a closed surface.

My trial：

$\mathbb{R}P^2$ can be defined as: $D^2/\sim$, which is homeomorphic to $I\times I/\sim$, here $I=[0,1], \quad(s,0)\sim (1-s,1),\quad (0,t)\sim (1,1-t)$.

then if $\mathbb{R}P^2$ is a closed surface, firstly I have to prove it is a 2-dimension compact manifold without boundary, which means $\forall x \in I\times I/\sim$ ,there is a open neiborhood $U\subset I\times I/\sim$ and a homeomorphic map $\varphi :U\longrightarrow \varphi (U)$, where $\varphi (U)\subset \mathbb{R}^2_{+}$ is a open set, and every $x\in I\times I/\sim , \varphi(x)$ can not be the form $(a,0)\in \mathbb{R}^2 $, but I do not know how find the map.

Harish Chandra Rajpoot
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F.Liu
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  • Not really, i still don't know why $\mathbb{R}P^2$ has no boundary. – F.Liu Apr 15 '23 at 17:57
  • Do you know that every point of a manifold is either a boundary point or an inner point? If you can find for every point $P$ a map $\psi_P:U_P\rightarrow V_P$ to some open subset of $\mathbb{R}^2$ you know that all points are inner points. It may help to draw $I\times I$ with the according equivalence relation and trying to find the according open subset $U_P$. – WhenYouHaveNoClue Apr 15 '23 at 19:32
  • Compactness is an immediate consequence of the definition of the quotient topology, the fact that the projection map to the quotient is continuous and the fact that images of compact subsets under continuous maps are again compact. – WhenYouHaveNoClue Apr 15 '23 at 19:34
  • So, the proposed duplicate does answer your question. Actually, my comment was generated automatically when I voted to close this post as a duplicate. I did not intend to ask a such a question. – Anne Bauval Apr 15 '23 at 21:25
  • The definition $\mathbb{R}P^2 = D^2/\sim$ is a bit unusual (but is a valid approach). The standard approach is described for example in https://math.stackexchange.com/q/1085706. Also see https://math.stackexchange.com/q/3999019 and https://math.stackexchange.com/q/3445764. – Paul Frost Apr 16 '23 at 09:51

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