How can I prove formally that the projective plane is a Hausdorff space? I want to prove the Hausdorff property of the projective space with this definition: the sphere $S^n$ with the antipodal points identified. It's seems easy, but I can't prove formally with this definition. I take neighborhoods of the points $p$, $q$ in $S^n/\sim $, with $p\neq q$. In $S^n$ we will have four points $p$, $q$, $-p$, $-q$, with 4 disjoints neighborhoods by the Hausdorff property of $\mathbb R^{n+1}$. I don't know how to prove with rigour why when we pass to the quotient we will have two disjoints neighborhoods of $p$ and $q$.
Thanks
 A: You did ask explicitly for a proof that uses the construction of projective space as a quotient of $S^n$, but consider this for fun. If you are willing to consider the projective space as the set of lines through $0$ in $\mathbb R^{n+1}$, then there is an intuative geometric way of seeing the Hausdorff property, that one should be able to convert into a rigorous proof:
Let $x$ and $y$ be distinct points in $\mathbb P^n$. Let $l_x$ and $l_y$ be the corresponding lines in $\mathbb R^{n+1}$. The Hausdorff property of projective space follows from the fact that we can fit the lines into two cones in $\mathbb R^{n+1}$ that only have $0$ in common, whose projections to projective space then give disjoint open sets that contain $x$ and $y$. It's helpful to draw a picture of the 2-dimensional situation to get a feel for this; to construct the cones you can fix an inner product and take the set of lines whose angle with $l_x$ (resp. $l_y$) is less than some $\epsilon > 0$. (By doing this we're basically making $\mathbb P^n$ into a metric space, whence the Hausdorff property follows.)
A: Recall that an equivalence relation $\sim$ on a topological space X is said to be $\textit{open}$ if for every open subset $A$ of $X$, the set $$[A] := \{x \in X | x \sim a \text{ for some a} \in A\}$$ is also open. It is easy to see that group actions on topological spaces give open equivalence relations. It is not hard to show the following
$\textbf{Lemma. }$ Let $\sim$ be an open equivalence relation on a space $X$. Then $X /\sim$ is Hausdorff iff the set $\{(x,y)| x \sim y\}$ is closed in $X \times X$.
Let $X = \mathbb{R}^{n+1} - 0$ and $\sim$ be the quotient by the action of $\mathbb{R}^{+}$ by scaling. Define a real valued function $f$ on $X \times X$ as 
$$f(x_1, ..., x_{n+1}, y_1,...,y_{n+1}) = \sum_{i \neq j}(x_i y_j - y_i x_j)^2 . $$Then $f(x,y)$ is clearly continuous and vanishes iff $y = \lambda x$ for some $\lambda \in \mathbb{R}^{+}$, that is, iff $x \sim y.$ Thus we have that $$f^{-1}(0) = \{(x,y)| x \sim y\} $$ is closed in $X \times X$ and by the lemma, $\mathbb{RP}^n$ is Hausdorff.
A: Perhaps the nicest approach to prove that $\Bbb{R}P^n$ is Hausdorff is to notice that it is the orbit space $S^n/(\Bbb{Z}/2\Bbb{Z})$ where the action of $\Bbb{Z}/2\Bbb{Z}$ is given by the antipodal map. The following problem (which you should try to do) will now complete the proof that real projective space is Hausdorff:


Let $G$ be a finite group that acts on a Hausdorff space $X$. Then the orbit space $X/G$ is Hausdorff.


A: Another way to see this is to note that $\mathbb{R}P^n$ is a CW-complex with one cell in each dimension $i\leq n$ i.e 
$$\mathbb{R}P^n=\bigcup_{i=0} ^n e^n.$$
Since CW-complexes are Hausdorff, $\mathbb{R}P^n$ is. 
A: Let $q:S^n\to P^n$ be the quotient map, and let $u,v\in P^n$ with $u\ne v$; there are $x,y\in S^n$ such that $q^{-1}[\{u\}]=\{x,-x\}$ and $q^{-1}[\{v\}]=\{y,-y\}$. Let $\epsilon=\frac13\min\{\|x-y\|,\|x+y\|\}$, and set
$$U=B(x,\epsilon)\cap S^n\quad\text{and}\quad V=B(y,\epsilon)\cap S^n\;,$$
where the open balls are taken in $\Bbb R^n$. Then $U,V,-U$, and $-V$ are pairwise disjoint open nbhds of $x,y,-x$, and $-y$, respectively, in $S^n$. Moreover, $q^{-1}\big[q[U]\big]=-U\cup U$ and $q^{-1}\big[q[V]\big]=-V\cup V$. Show that $q[U]$ and $q[V]$ are disjoint open nbhds of $u$ and $v$ in $P^n$.
A: Let $G$ be a compact Hausdorff group.
Let $X$ be a locally compact Hausdorff space.
Suppose $G$ acts continuously on $X$.
We will prove that the orbit space $X/G$ is Hausdorff.
The proof is due to Bourbaki.
As an immediate corollary, projective spaces over $\mathbb{R}$ and $\mathbb{C}$ are Hausdorff.
Definition 1
Let $X, Y$ be locally compact Haudsorff spaces.
Let $f\colon X \rightarrow Y$ be a continuous map.
If $f^{-1}(K)$ is compact for every compact subset $K$ of $Y$, $f$ is called a proper map.
Lemma 1
Let $X$ be a topological space.
Let $(A_i)_{i\in I}$ be a family of subsets of $X$.
Suppose $X = \bigcup_i int(A_i)$, where $int(A_i)$ is the interior of $A_i$.
Let $B$ be a subset of $X$.
Suppose $B \cap A_i$ is closed in $A_i$ for every $i \in I$.
Then $B$ is closed in $X$.
Proof:
Let $C = X - B$.
Since $C \cap A_i = A_i - (B \cap A_i)$, $C \cap A_i$ is open in $A_i$.
Hence there exists an open subset $V_i$ of $X$ such that $C \cap A_i = V_i \cap A_i$.
Then $C \cap int(A_i) = (C \cap A_i) \cap int(A_i) = (V_i \cap A_i) \cap int(A_i)
= V_i \cap int(A_i)$.
Hence $C \cap int(A_i)$ is open in $X$.
Since $X = \bigcup int(A_i)$, $C = \bigcup (C \cap int(A_i))$.
Hence $C$ is open in $X$.
Hence $B$ is closed in $X$.
QED
Lemma 2
Let $X, Y$ be topological spaces.
Let $f\colon X \rightarrow Y$ be a continuous map.
Let $(A_i)_{i\in I}$ be a family of subsets of $Y$.
Suppose $Y = \bigcup_i int(A_i)$.
Let $f_i\colon f^{-1}(A_i) \rightarrow A_i$ be the restriction of $f$ for each $i \in I$.
Suppose $f_i$ is a closed map for every $i \in I$.
Then $f$ is a closed map.
Proof:
Let $B$ be a closed subset of $X$.
Since $f(B) \cap A_i = f_i(B \cap f^{-1}(A_i))$, $f(B) \cap A_i$ is closed in $A_i$ for every $i \in I$.
Hence $f$ is a closed map by Lemma 1.
QED
Lemma 3
Let $X, Y$ be locally compact Hausdorff spaces.
Let $f\colon X \rightarrow Y$ be a proper map.
Then $f$ is a closed map.
Proof:
There exists a family $(A_i)_{i\in I}$ of compact subsets of $Y$ such that $Y = \bigcup_i int(A_i)$.Let $f_i\colon f^{-1}(A_i) \rightarrow A_i$ be the restriction of $f$ for each $i \in I$.
Since $f$ is proper, $f^{-1}(A_i)$ is compact.
Hence $f_i$ is a closed map.
Hence $f$ is a closed map by Lemma 2.
QED
Definition 2
Let $G$ be a locally compact Hausdorff group.
Let $X$ a locally compact Hausdorff space.
Suppose $G$ acts on $X$ continuously.
Suppose the map $f\colon G\times X \rightarrow X\times X$ defined by $f(s, x) = (x, sx)$ is proper.
Then we say $G$ acts properly on $X$.
Definition 3
Suppose a group $G$ acts on a set $X$.
Let $K, L$ be subsets of $X$.
We denote $P(K, L) = \{s \in G| sK \cap L \neq \emptyset\}$.
Lemma 4
Let $X, Y$ be Hausdorff spaces.
Let $C$ be a compact subset of $X\times Y$.
Then there exists a compact subset $K$(resp. $L$) of $X$(resp. $L$) such that $C \subset K\times L$.
Proof:
Let $K$(resp. $L$) be the image of $C$ by the projection map on the first(resp. the second) factor of $X\times X$.
Then $K$ and $L$ are compact and $C \subset K\times L$.
QED
Lemma 5
Let $f\colon X_1 \rightarrow Y_1$(resp. $g\colon X_2 \rightarrow Y_2$) be a proper map.
Then $f\times g\colon X_1\times X_2 \rightarrow Y_1 \times Y_2$ is proper.
Proof:
Let $C$ be a compact subset of $Y_1\times Y_2$.
By Lemma 4, there exists a compact subset $K$(resp. $L$) of $Y_1$(resp. $Y_2$) such that $C \subset K\times L$.
Since $(f\times g)^{-1}(K\times L) = f^{-1}(K)\times g^{-1}(L)$ is compact, $(f\times g)^{-1}(C)$ is compact.
QED
Lemma 6
Let $X$ be a compact Haudsorff space.
Let $Y$ be a locally compact Hausdorff space.
Then the projection map $\pi\colon X\times Y \rightarrow Y$ is proper.
Proof:
Let $p$ be a one point space.
Then $f\colon X \rightarrow p$ is proper.
Hence, by Lemma 5, $\pi = f\times id_Y\colon X \times Y \rightarrow p\times Y = Y$ is proper.
QED
Lemma 7
Let $G$ be a locally compact Hausdorff group.
Let $X$ a locally compact Hausdorff space.
Suppose $G$ acts on $X$ continuously.
Then $P(K, L)$ is closed for any compact subsets $K, L$ of $X$.
Proof:
Let $f\colon G\times X \rightarrow X$ be the map defined by $f(s, x) = sx$.
Since $f$ is continuous, $f^{-1}(L)$ is closed.
Let $\pi\colon G\times K \rightarrow G$ be the projection.
By Lemma 6, $\pi$ is proper.
Hence, by Lemma 3, $\pi$ is a closed map.
Hence $P(K, L) = \pi(f^{-1}(L))$ is closed.
QED
Lemma 8
Let $G$ be a locally compact Hausdorff group.
Let $X$ a locally compact Hausdorff space.
Suppose $G$ acts on $X$ continuously.
Suppose $P(K, L)$ is compact for any compact subsets $K, L$ of $X$.
Then $G$ acts properly on $X$.
Proof:
Let $h\colon G\times X \rightarrow X\times X$ be the map defined by $h(s, x) = (x, sx)$.
Let $C$ be a compact subset of $X\times X$.
By Lemma 4, there exist compact subsets $K, L$ of $X$ such that $C \subset K\times L$.
Then $h^{-1}(K\times L) \subset P(K, L)\times K$.
Since $P(K, L)\times K$ is compact and $h^{-1}(K\times L)$ is closed, $h^{-1}(K\times L)$ is compact.
Since $C \subset K\times L$, $h^{-1}(C) \subset h^{-1}(K\times L)$.
Since $h^{-1}(C)$ is closed, it is compact.
Hence $h$ is proper.
Hence $G$ acts proplerly on $X$.
QED
Lemma 9
Let $X$ be a topological space.
Let $R$ be an equivalence relation on $X$.
Let $X/R$ be the quotient space.
Let $\pi\colon X \rightarrow X/R$ be the canonical map.
Let $\Gamma = \{(x, y)\in X\times X| x \equiv y$ (mod $R)\}$.
Suppose $\pi$ is an open map and $\Gamma$ is closed.
Then $X/R$ is Hausdorff.
Proof:
Let $\rho := \pi\times \pi\colon X\times X \rightarrow (X/R)\times (X/R)$.
Since $\pi$ is an open map, so is $\rho$.
Since $\rho(\Gamma)$ is the diagonal subset of $(X/R)\times (X/R)$, it suffices to prove that $\rho(\Gamma)$ is closed.
Since $\Gamma = \rho^{-1}(\rho(\Gamma)$, $X\times X - \Gamma = \rho^{-1}((X/R)\times (X/R) - \rho(\Gamma))$.
Since $\rho$ is surjective, $\rho(X\times X - \Gamma) = (X/R)\times (X/R) - \rho(\Gamma)$.
Since $\rho$ is an open map and $X\times X - \Gamma$ is open, $(X/R)\times (X/R) - \rho(\Gamma)$ is open.
Hence $\rho(\Gamma)$ is closed.
QED
Lemma 10
Let $G$ be a locally compact Hausdorff group.
Let $X$ a locally compact Hausdorff space.
Suppose $G$ acts properly on $X$.
Then the orbit space $X/G$ is Hausdorff.
Proof:
Let $\pi\colon X \rightarrow X/G$ be the canonical map.
Let $U$ be an open subset of $X$.
Then $\pi^{-1}(\pi(U)) = GU$.
Since $GU$ is open, $\pi$ is an open map.
By Lemma 9, it suffices to prove that $\Gamma = \{(x, y) \in X\times X| x \equiv y$ (mod $G)\}$ is closed.
Let $h\colon G\times X \rightarrow X\times X$ be the map defined by $h(s, x) = (x, sx)$.
Since $G$ acts properly on $X$, $h$ is proper.
Hence $h$ is closed by Lemma 3.
Hence $\Gamma = h(G\times X)$ is closed.
QED
Proposition
Let $G$ be a compact Hausdorff group.
Let $X$ be a locally compact Hausdorff space.
Suppose $G$ acts continuously on $X$.
Then $G$ acts properly on $X$.
Hence the orbit space $X/G$ is Hausdorff.
Proof:
By Lemma 7, $P(K, L)$ is closed for any compact subsets $K, L$ of $X$.
Since $G$ is compact, $P(K, L)$ is compact.
Hence, by Lemma 8, $G$ acts proplerly on $X$.
Hence, by Lemma 10, $X/G$ is Hausdorff.
QED
Corollary
Let $K$ be the field of real numbers or the field of complex numbers.
Then the projective space $P^n(K)$ over $K$ is Hausdorff.
Proof:
Let $x = (x_1, \dots, x_n) \in K^n$.
We denote $||x|| = (\sum_i x_i \bar x_i)^{\frac{1}{2}}$.
Let $G = \{x \in K|\ |x| = 1\}$.
Then $G$ is a subgroup of the multiplicative group of $K$.
Let $X = \{x \in K^n|\ ||x|| = 1\}$.
Clearly $G$ acts on $X$ continuously.
Since $G$ and $X$ are compact Hausdorff spaces, $X/G$ is Hausdorff by the proposition.
Since $P^n(K) = X/G$, the assertion follows.
QED
Remark 1
The above proposition holds without assuming $X$ is locally Hausdorff, but only assuming it is Hausdorff as the answer to this question shows.
Remark 2
If $G$ and $X$ are both compact Hausdorff, it is easy to prove that $X/G$ is Hausdorff.
Let $\Gamma = \{(x, y) \in X\times X| x \equiv y$ (mod $G)\}$.
Let $h\colon G\times X \rightarrow X\times X$ be the map defined by $h(s, x) = (x, sx)$.
Since $h$ is continuous, $\Gamma = h(G\times X)$ is compact, hence closed.
By Lemma 9, $X/G$ is Hausdorff.
A: Let $K$ be the field of real numbers or the field of complex numbers.
Let $G = K^*$ be the multiplicative group of $K$.
Let $X = K^{n+1} - {0}$.
Then $G$ acts on $X$.
We regard $X$ as a topological subspace of $K^{n+1}$.
Then the projective space $P^n$ over $K$ is, by defintion, the orbit space $X/G$ with the quotient topology.
Let $\pi\colon X \rightarrow P^n$ be the canonical map.
Let $x = (x_0,x_1,\dots,x_n) \in X$.
We denote $\pi(x)$ by $[x_0,x_1,\dots,x_n]$.
Let $\Delta = \{(x,x) \in P^n\times P^n| x \in P^n\}$.
To prove that $P^n$ is Hausdorff, it suffices to prove that $\Delta$ is closed in $P^n\times P^n$.
Let $([x_0,x_1,\dots,x_n], [y_0,y_1,\dots,y_n]) \in \Delta$.
Suppose $x_i \neq 0$.
Then $y_i \neq 0$ and $x_j/x_i = y_j/y_i$ for $i \neq j$.
Hence $x_iy_j = x_jy_i$ for $i \neq j, 0\le i, j\le n$.
Conversely suppose $([x_0,x_1,\dots,x_n], [y_0,y_1,\dots,y_n]) \in P^n\times P^n$.
Suppose $x_iy_j = x_jy_i$ for $i \neq j, 0\le i, j\le n$.
Suppose $x_i \neq 0$. If $y_i = 0$, then $y_j = 0$ for all $j$.
This is a contradiction.
Hence $y_i \neq 0$.
Hence $x_j/x_i = y_j/y_i$ for all $j \neq i$.
Hence $([x_0,x_1,\dots,x_n], [y_0,y_1,\dots,y_n]) \in \Delta$.
Therefore $\Delta = \{([x_0,x_1,\dots,x_n], [y_0,y_1,\dots,y_n]) \in P^n\times P^n| x_iy_j = x_jy_i, i\neq j, 0\le i,j\le n\}$.
Let $U_i = \{[x_0,x_1,\dots,x_n] \in P^n|\ x_i \neq 0\}$.
Then $U_i$ is an open subset of $P^n$ and it is is canonically isomorphic to $K^n$(see here).
Clearly $(U_i)_{i \in I}$ is an open cover of $P^n$, where $I = \{0,1,\dots,n\}$.
Hence $(U_i\times U_j)_{(i, j) \in I\times I}$ is an open cover of $P^n\times P^n$.
Since $\Delta$ is the common zeros of bihomogeneous polynomials $x_iy_j - x_jy_i, i\neq j$,
$(U_i\times U_j) \cap \Delta$ is canonically isomorphic to an algebraic subset of $K^n\times K^n$(e.g. see Mychael Joyce's answer to this question).
Hence it is closed in $U_i\times U_j$.
Hence $\Delta$ is closed.
A: The proof I'll give bellow works for the fields $\mathbb K = \mathbb R$ and $\mathbb C$. It also works for $\mathbb  K = \mathbb H$. Let $S$ be the unit sphere in $\mathbb K^{n+1}$ and $S_{\mathbb K}$ be the unit sphere in $\mathbb K$, i.e.,
$$ S_{\mathbb R} = \{-1, 1\}\quad \text{and}\quad \quad S_{\mathbb C} = \mathbb S^1.$$
We have the surjective map $\pi:S \to \mathbb P_{\mathbb K}^n$ given by $\pi(x) = [x]$. The topology considered in the projective space is the quotient topology. Notice that if $U \subset S$ is open, then 
$$\pi^{-1}(\pi(U)) = \cup_{\gamma\in S_{\mathbb K}} \gamma U$$
is open
and therefore $\pi$ is an open map.
Now, take $[x],[y] \in \mathbb P_{\mathbb K}^n$, $[x] \neq [y]$. Notice $$\pi^{-1}[x] = \{\gamma x \in S:\gamma \in S_{\mathbb K}\} = S_{\mathbb K} x.$$
The map $S_{\mathbb K} \to \pi^{-1}[x]$ given by $\gamma \mapsto \gamma x$ is homeomorphism and therefore $\pi^{-1}[x]$ is compact. So, we have $\pi^{-1}[x]$ and $\pi^{-1}[y]$ are closed subsets of $S$. Clearly they are also disjoint.
By the Urysohn's lemma, there is a continuous function $g:S \to \mathbb R$ such that $g|_{\pi^{-1}[x]} = 1$ and $g|_{\pi^{-1}[y]} = 0$. 
But this function is not necessarly invariant under action of $S_{\mathbb K}$, so we define $h: S \to \mathbb R$ by the formula
$$h(x) = \int_{S_{\mathbb K}} g(\gamma x) d\gamma,$$
where $d\gamma$ is the Haar measure of $S_{\mathbb K}$. More explicitly,
$$h(x) = \frac{1}{2}(g(x)+g(-x))\quad \text{for} \quad \mathbb K = \mathbb R$$
and
$$h(x) = \frac{1}{2\pi}\int_{0}^{2\pi} g(e^{i\theta}x)d\theta\quad \text{for} \quad \mathbb K = \mathbb C.$$
Now, since the sets $\pi^{-1}[x]$ and $\pi^{-1}[y]$ are invariant under action of $S_{\mathbb K}$, we have $h|_{\pi^{-1}[x]} = 1$ and $h|_{\pi^{-1}[y]} = 0$.
This function $h$ is a function on the projective space. More precisely, define $\widehat h: \mathbb P_{\mathbb K}^n \to \mathbb R$ by the formula $\widehat h([z]) = h(z)$. Since, $h(\gamma z) = h(z)$ for $\gamma \in S_{\mathbb K}$, the function $\widehat h$ is well defined. Besides that, since $h = \widehat h \circ \pi$ and $\pi$ is open map, we obtain $\widehat h$ is continuous.
Taking the open sets $U = \widehat h^{-1}(1/2,\infty)$ and $V = \widehat h^{-1}(-\infty,1/2)$ we have $[x] \in U$, $[y] \in V$ and $U \cap V = \varnothing$, proving $\mathbb P_{\mathbb K}^n$ is Hausdorff.
A: I think in this case, and in may others, it is way easier to use the following equaivalent definition of Hasudorff:
Definition: A topological space X is Hausdorff iff the Diagonal $\,\Delta_X:=\{(x,x)\in X^2\}\,$ is closed in the product space $\,X\times X\,$
