Real projective space is Hausdorff: is this proof correct?

Let $\mathbb{RP}^n=\frac{\mathbb{R}^{n+1}\setminus\{0\}}{\sim}$ be the real projective space of dimension $n$, where $\sim$ is the proportionality relation.

Then $\mathbb{RP}^n$ has the quotient topology induced by the canonical surjection

$$\pi\colon\mathbb{R}^{n+1}\setminus\{0\}\to \mathbb{RP}^n, \quad x\mapsto [x]_{\sim}$$

Let $p$ and $q$ be distinct element of $\mathbb{RP}^n$, and let be $\pi_{|S^n}^{-1}(\{p\})=\{x,-x\}$ and $\pi_{|S^n}^{-1}(\{q\})=\{y,-y\}$.

Let be $\varepsilon\leq \frac{1}{2}\text{min}\{\lVert x-y\rVert,\lVert x+y\rVert\}$.

Let be $U=B_\varepsilon^{n+1}(x)\cup B_\varepsilon^{n+1}(-x) $ and $V=B_\varepsilon^{n+1}(y)\cup B_\varepsilon^{n+1}(-y) $. Then $U$ and $V$ are open and disjoint.

Let be $U'=\pi(U)$ and $V'=\pi(V)$. Since $\pi$ is an open map, then $U'$ and $V'$ are open neighborhood of $p$ and $q$ respectively.

If I show that $U'$ and $V'$ are disjoint, then I have completed the proof that $\mathbb{RP}^n$ is Hausdorff.

Suppose there are $u\in U$ and $v\in V$ such that $\pi(u)=\pi(v)$.

My book says that I can consider $u$ and $v$ of norm${}=1$.

Then, by this assumption, from $\pi(u)=\pi(v)$ follows that $u=\pm v \in U\cap V$ which is absurd.

My question is: why can I consider $u$ and $v$ of norm${}=1$? For example, suppose that $u\in B_\varepsilon^{n+1}(x)$. According to my book, I can replace $u$ with $\frac{u}{\lVert u\rVert}$ since $\pi(u)=\pi(\frac{u}{\lVert u\rVert})$.

But form $u\in B_\varepsilon^{n+1}(x)$ doesn't follow that $\frac{u}{\lVert u\rVert}\in B_\varepsilon^{n+1}(x)$. So $\frac{u}{\lVert u\rVert}$ could not be in $U$.

You are right: $\frac{u}{\lVert u \rVert}$ is in general not in $U$. To verify this, let $S_\epsilon(x)$ denote the sphere with center $x$ and radius $\epsilon$. Choose any point $z \in S_\epsilon(x) \cap S^n$ and let $E$ be the two-dimensional subspace of $ \mathbb{R}^{n+1}$ generated by $x, z$. With the induced norm it is nothing else than a Euclidean plane, $S = S^n \cap E$ is the ordinary unit circle in $E$ and $S' = S_\epsilon(x) \cap E$ is a circle with radius $\epsilon$ and center $x$. There are two points $z_1, z_2 \in S'$ such that the line $l_i$ through $0$ and $z_i$ is tangent to $S'$. The points $0, x, z_1$ form a right triangle and we conclude $\lVert z_1 \rVert = \sqrt{1- \epsilon^2} < 1$. The line $l_1$ intersects $S$ in $\frac{z_1}{\lVert z_1 \rVert}$ and it is geometrically obvious that $\lVert \frac{z_1}{\lVert z_1 \rVert} - x \rVert > \lVert z_1 - x \rVert = \epsilon$. If you want you can formalize these arguments a little bit, but I believe it is okay. This shows that $\lVert \frac{u}{\lVert u \rVert} - x \rVert > \epsilon$ for all $u \in \mathbb{R}^{n+1}$ which are sufficiently close to $z_1 \in S_\epsilon(x)$, in particular for suitable $u \in B^{n+1}_\epsilon (x)$.

This leaves the question what to do. We redefine $U, V$ as follows:

Choose $x \in \pi^{-1}(p) \cap S^n$ and $y \in \pi^{-1}(q) \cap S^n$ and define $\epsilon$ as in your question. Let $$U^\ast = S^n \cap B^{n+1}_\epsilon (x), V^\ast = S^n \cap B^{n+1}_\epsilon (y) .$$

These are open subsets of $S^n$. Define

$$U = \lbrace t \cdot y \mid y \in U^\ast, t \in \mathbb{R}\backslash \lbrace 0 \rbrace \rbrace, V = \lbrace t \cdot y \mid y \in V^\ast, t \in \mathbb{R}\backslash \lbrace 0 \rbrace \rbrace .$$

It is easy to see that they are open in $\mathbb{R}^{n+1} \backslash \lbrace 0 \rbrace$.

The "new" $U, V$ are disjoint and have the property $\pi^{-1}(\pi(U)) = U, \pi^{-1}(\pi(V)) = V$. This shows that $\pi(U)$ and $\pi(V)$ are disjoint open neighborhoods of $p $ and $q$. Note that $U$ is the union of all punctured lines going through $U^\ast$ (i.e. a punctured "double cone"), similarly $V$.

The core of this proof is this:

Consider the restriction $\hat{\pi} = \pi \mid_{S^n} : S^n \to \mathbb{RP}^n$. We have $\hat{\pi}^{-1}([x]) = \{ x, -x \}$. The following are equivalent for $W \subset \mathbb{RP}^n$:

(1) $W$ is open in $\mathbb{RP}^n$.

(2) $\pi^{-1}(W)$ is open in $\mathbb{R}^{n+1} \backslash \lbrace 0 \rbrace$.

(3) $\hat{\pi}^{-1}(W) = \pi^{-1}(W) \cap S^n$ is open in $S^n$.

(1) $\Leftrightarrow$ (2) is the definition of the quotient topology and (2) $\Rightarrow$ (3) is trivial. To verify (3) $\Rightarrow$ (2) note that the map $\rho : \mathbb{R}^{n+1} \backslash \lbrace 0 \rbrace \to \mathbb{R}^{n+1}, \rho(x) = \frac{x}{\lVert x \rVert}$ is continuous. Its image is $S^n$ so that it induces a continuous map $r : \mathbb{R}^{n+1} \backslash \lbrace 0 \rbrace \to S^n$. But now $\pi^{-1}(W) = r^{-1}(\pi^{-1}(W) \cap S^n)$.

This immediately implies that $\hat{\pi} : S^n \to \mathbb{RP}^n$ is a quotient map which gives a nice alternative representation of $\mathbb{RP}^n$. In fact, we can define $\mathbb{RP}^n = S^n/ \equiv$ where the equivalence relation "$\equiv$" is defined by $x \equiv y \Leftrightarrow x = \pm y$, the latter being the explicit meaning of $\hat{\pi}(x) = \hat{\pi}(y)$.

Using $\hat{\pi}$ instead of $\pi$ is the best approach to prove that $\mathbb{RP}^n$ is Hausdorff. If you check the above proof, you will see that it exactly comes down to that.

Another alternative is to define $\mathbb{RP}^n$ as the set of all lines through $0$, i.e. the set set of all one-dimensional subspaces of $\mathbb{R}^{n+1}$. The map $\pi : \mathbb{R}^{n+1} \backslash \lbrace 0 \rbrace \to \mathbb{RP}^n$ is then defined by $\pi(x) =$ line through $x$. Note that it is impossible to define $\pi(0)$.