Exercise 1.7 (Real projective space is compact)

Question

Show that $\mathbb{R}\mathbb{P}^n$ is compact. 
% [Hint: show that the restriction of $\pi$ to $\mathbb{S}^n$ is surjective.]

Elu Thingol · Accepted Answer

\begin{proof}
The trick is to notice that any element $[x] \in \mathbb{R}\mathbb{P}^n$, with an arbitrary representation $x \in \mathbb{R}^{n+1}$, has another, normalized representation
\begin{equation*}
\widetilde{x}:= \dfrac{x}{\|x\|}
\end{equation*}
since
\begin{equation*}
    [x] = \pi\left(xight) = \pi\left(\dfrac{1}{\|x\|} \cdot xight) = [\widetilde{x}],
\end{equation*}
and which lies on the unit sphere $\mathbb{S}^n$. Consider, therefore, the restriction $\pi:\mathbb{S}^n 	o \mathbb{R}\mathbb{P}^n$ of $\pi:\mathbb{R}^{n+1}\setminus\{0\} 	o \mathbb{R}\mathbb{P}^n$ to the unit sphere $\mathbb{S}^n$. By the previous argument, $\pi estriction_{\mathbb{S}^n}$ must be surjective, i.e, $\pi(\mathbb{S}^n)=\mathbb{R}\mathbb{P}^n$. By Proposition A.17., it is also continuous; therefore, the image of $\pi$ under compact set $\mathbb{S}^n$ must be compact as well.
\end{proof}

Exercise 1.7 (Real projective space is compact)

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Exercise 1.7 (Real projective space is compact)

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