$\mathbb{RP}^n$ is Hausdorff proof with $n$-sphere

Question

Problem:

Show that $\mathbb{RP}^n$ (the $n$-dimensional real projective space) is Hausdorff. We use the quotient topology: A set $X$ is open in $\mathbb{RP}^n$ if and only if $\{z\in \mathbb{R}^{n+1}\backslash \{0\} : [z]\in X\}$ is open in $\mathbb{R}^{n+1}\backslash \{0\}$.

Sketch of my proof:

Let $[x]$ denote the line spanned by $x$ in $\mathbb{R}^{n+1}$. To show that $\mathbb{RP}^n$ is Hausdorff, we show that for some lines $[x],[y]\in\mathbb{RP}^n$, there are open sets $X,Y$ in $\mathbb{RP}^n$ such that $[x]\in X, [y]\in Y$ and $X\cap Y=\emptyset$ where $x\neq y$ are some elements of $\mathbb{R}^{n+1}\backslash\{0\}$.

We project $x, y$ to the $n$-sphere. Let $x'\in[x]\cap \mathbb{S}^n, y'\in[y]\cap \mathbb{S}^n$. It is still true that $[x]=[x'], [y]=[y']$.

Now here comes the confusing part for me:

A set $X$ is open in $\mathbb{RP}^n$ if and only if $\{z\in \mathbb{R}^{n+1}\backslash \{0\} : [z]\in X\}$ is open in $\mathbb{R}^{n+1}\backslash\{0\}$. So I will try finding such disjoint $X$ and $Y$ and I will do it with the help of the $n$-sphere.

I claim that if we have some open sets $X',Y'\subset \mathbb{S}^n$ (open wrt. $\mathbb{S}^n$), then the set $\{z\in \mathbb{R}^{n+1}\backslash \{0\} : \exists w\in X': z\in [w]\}$ is open in $\mathbb{R}^{n+1}\backslash\{0\}$. Basically we have some open set $X'$ of the $n$-sphere and we take $X$ as the lines spanned by the points of $X'$. (I don't actually know if this is true or not, I would prove this as a lemma.)

Now I argue: Since the $n$-sphere is Hausdorff (was already proven in the book that it is a manifold), there are open disjoint sets $X', Y'\subset \mathbb{S}^n$ around $x', y'$. From the above claim also the sets $X''=\{z\in \mathbb{R}^{n+1}\backslash \{0\} : \exists w\in X': z\in [w]\}, Y''=\{z\in \mathbb{R}^{n+1}\backslash \{0\} : \exists w\in Y': z\in [w]\}$ are open in $\mathbb{R}^{n+1}\backslash\{0\}$. Then we intuitively construct $X,Y$ from $X'',Y''$.

Edit: Much appreciation to everyone for the help!

Answer 1

Your first claim is correct: linear extension of an open subset of $\mathbb{S}^n$ to $\mathbb{R}^{n+1}\backslash\{0\}$ is open. Why? You first consider the quotient map:

$$\pi:\mathbb{R}^{n+1}\backslash\{0\}\to\mathbb{S}^n$$ $$\pi(v)= \frac{1}{\lVert v\rVert}v$$

and then given an open subset $U\subseteq\mathbb{S}^n$ your set is nothing else than $\pi^{-1}(U)\cup\pi^{-1}(-U)$.

And this observation also shows that your last claim is false. Your sets $X''$ and $Y''$ are indeed open, but they don't have to be disjoint anymore. This happens whenever there is $x$ such that $x\in X'$ and $-x\in Y'$.

So how to fix that? Well, since $[x]\neq[y]$ then this means that $x\neq y$ and $-x\neq y$. The same for $y$. And so we can find small enough neighbourhoods $X'$ of $x$ and $Y'$ of $y$ (in $\mathbb{S}^n$) such that $(X'\cup -X')\cap (Y'\cup -Y')=\emptyset$. And then there rest of your argument stays the same.

However this might not be immediate and the devil is in the tedious details. There is another, arguably simpler way to do that, thanks to the following:

Lemma. Let $X$ be a Hausdorff space and $\sim\subseteq X\times X$ an equivalence relation. Then $X/\sim$ is Hausdorff if and only if $\sim$ is a closed subset of $X\times X$.

For proof see this.

So how do we apply this? Let $X=\mathbb{R}^{n+1}\backslash\{0\}$. Then our $\sim$ is defined as:

$$\sim=\{(x,y)\in X\times X\ |\ y=\lambda\cdot x\text{ for some }\lambda\in\mathbb{R}\backslash\{0\}\}$$

So why it is closed? Take a sequence $(x_n,y_n)\in \sim$ convergent to some $(x,y)\in X\times X$. Then $y_n=\lambda_n\cdot x_n$. We know that $y_n\to y$. Furthermore at least one coordinate of $y$ is nonzero, say $j$th. Then $y_{j,n}\to y_j$, i.e. $\lambda_n\cdot x_{j,n}\to y_j$. Since $y_j\neq 0$ and $\lambda_n\neq 0$ then $x_{n,j}$ eventually has to be nonzero. Therefore we can multiply by the convergent sequence $x_{j,n}^{-1}$ (note that it is convergent because $x_{j,n}$ converges to nonzero limit) to get $\lambda_n\to (\lim x_{j, n}^{-1})\cdot y_j$. Write down $\lambda$ for $\lim_n x_{j,n}^{-1}\cdot y_j$.

What we showed is that $\lambda_n\to \lambda$ and thus $(x,y)=(x,\lambda\cdot x)$ by continuity of multiplication. Which finally means that $(x,y)\in\sim$ and so $\sim$ is closed.