I am trying to understand and complete the proof that the real projective space ${\mathbb{P}}^n$ is Hausdorff.In my notes it is modeled as${\mathbb{R}}^{n+1}\setminus \{0\}/\sim $ and it goes like this:
It is enough to construct, given two different points $[a]$ and $[b]$ a continuous function $f:{\mathbb{P}}^n \rightarrow {\mathbb{R}} $ such that $f[a] \neq f[b] $ (why ?.......(1))
We fix $\omega$ in ${\mathbb{R}}^{n+1}\setminus \{0\}$ and define $f[\nu]$ as the squared of the distance from $\omega$ to the vector line $R\nu$ generated by $\nu$. Since $f\circ \pi(\nu) = f[\nu] = |\omega|^2-(\omega . \nu)^2/|\nu|^2$, it follows that $f \circ \pi $ is continuous and hence $f$ is continuous...(why?........(2)) ($\pi$ is the canonical projection $\pi : {\mathbb{R}}^{n+1}\setminus \{0\} \rightarrow {\mathbb{P}}^{n} $) It is then enough to take $w \sim a$ to have $f[a] = 0 \neq f[b]$ ...(*)
I have two questions:
1)why is (1) true ?
At first I thought I could justify (1) like this: Since ${\mathbb{R}}$ is Hausdorff, $\exists $ open sets $A,B$ such that for $f[a] \neq f[b] $, $f[a] \in A$ and $f[b] \in B$, with $A \cap B = \emptyset$. By hypothesis then $[a] \neq [b] $ and since $f$ is continuous, $f^{-1}(A) $ and $f^{-1}(B) $ are open sets of ${\mathbb{P}}^n$ containing $[a]$ and $[b]$ respectively such that $f^{-1}(A) \cap f^{-1}(B) = f^{-1}(A\cap B)=f^{-1}(\emptyset)=\emptyset $. Then ${\mathbb{P}}^n$ is Hausdorff
But I don't think my proof is correct since for it to work we need $f$ to have the property stated at (1) but the function $f$ that they propose later has that property only when taking one of the points fixed, say $a$ as $a \sim w$ as done in (*), and not for any two points as needed for the definition of Hausdorff space. Besides if w is the center of a circle (in the 2-dimensional case for instance there are two lines that have the same distance to w, that is the tangent lines, so (1) does not hold for any two different $[a] $and $[b]$)
2)why $f \circ \pi $ continuous implies $f$ is continuous? don't think I can compose with a continuous $\pi^{-1}$, since that function is not well defined since $\pi$ is not injective
Your proof of $(1)$ is correct. In fact, you can generalize to this lemma:
Note that $Y$ and $f$ are allowed to depend on $x,x'$. Also the converse is true (simply take $Y = X$ and $f = id$ for all $x,x'$).
Concerning $(2)$: The construction of $f$ does not make sense in the present form. Writing $f([\nu])$ means that you want to define it for the elements of $\mathbb P^n$. For each $\omega \in \mathbb R^{n+1} \setminus \{0\}$ define
$$ f_\omega :\mathbb R^{n+1} \setminus \{0\} \to \mathbb R, f_\omega(\nu) = \lvert \omega \rvert^2 \left(1- \frac{(\omega \cdot \nu)^2}{\lvert \omega \rvert^2 \lvert \nu \rvert^2}\right) .$$ It is obvious that if $\nu \sim \nu'$, then $f_\omega(\nu) = f_\omega(\nu')$. Thus $f$ induces a unique function $F_\omega : \mathbb P^n \to \mathbb R$ such that $f_\omega = F_\omega \circ \pi$. By the universal property of the quotient topology $F_\omega$ is continuous. Note that $F_\omega([\omega]) = 0$.
Now consider any two distinct points $[\nu],[\omega]$ in $\mathbb P^n$. This means that $\nu$ does not lie on the line $\mathbb R \omega$, thus $(\omega \cdot \nu)^2 < \lvert \omega \rvert^2 \lvert \nu \rvert^2$ (Cauchy-Schwarz). Therefore $F_\omega ([\nu]) \ne 0 = F_\omega([\omega])$. The function $F_\omega$ depends on $[\omega]$, but this is allowed.
Remark:
We could also work with $f_\omega(\nu) =\frac{(\omega \cdot \nu)^2}{\lvert \omega \rvert^2 \lvert \nu \rvert^2}$. This is the square of the cosine of the angle between $\omega$ and $\nu$. Then $F_\omega ([\omega]) = 1$ and $F_\omega ([\nu]) < 1$.