How to show that the space $Y$ obtained by identifying points on the same straight line through origin in $R^{n+1}-\{0\}$ is Hausdorff?

Question

Background: $Y$ turns out to be projective space but I'm trying to prove this and have not proven it yet (i.e., $Y$ is homeomorphic to the space $X$ obtained by identifying antipodal points on the unit $n-$ sphere.)

One of the steps in proving the homeomorphism is to prove that $Y$ is Hausdorff. I am having difficulty in a step in showing the Hausdorffness of Y. I'm trying to use this solution https://math.stackexchange.com/a/241786/266435 in this case.

Let $p_2: \mathbb R^{n+1}-\{0\}\to Y: x\to [x]$ be the quotient map.

Suppose that $[x], [y]\in Y, [x]\ne [y], [l]$ denotes equivalence class of $l$. It follows that $p(x)=[x], p(y)=[y]$. Since $[x]\ne [y]$, it follows that $x\ne my$ for any $m\in \mathbb R-\{0\}$. (i.e., $x$ and $y$ cannot lies on the same line through the 'removed' origin.) $\mathbb R^{n+1}-\{0\}$ is Hausdorff so there exist $U, V$ disjoint open in $\mathbb R^{n+1}-\{0\}$ such that $x\in U, y\in V$.

Now, the following two things are to be shown:

1. $p(U), p(V)$ are open.
2. $p(U)\cap p(V)=\emptyset$

I have tried to show that 2) is true as follows: Suppose on the contrary that $p(U)\cap p(V)\ne\emptyset$. There exists $t\in p(U)\cap p(V)$. There exist $u\in U, v\in V$ such that $t=p(u)=p(v)\implies u=mv$ for some non zero real no. $m.\implies u,v$ lie on the same straight line through the origin. This contradicts the fact that $U, V$ are disjoint and that $U,V$ is 'centered' at different lines. But in view of red line in picture no. 2) in the picture below, this seems wrong.

About $(1)$, I tried to use the definition of quotient maps: $p_2(U)$ is open iff $p_2^{-1}(p_2(U))$ is open. But since $p_2^{-1}(p_2(U))\ne U$, I'm not sure how to do this either.

I tried to show the homeomorphism using the following picture:

Here, if I show that $Y$ is Hausdorff, then the induced map $p_2\circ i: S^n\to Y$ due to being onto continuous from compact $S^n$ to Hausdorff $Y$ will turn out to be a quotient map. This quotient map partitions $S^n$ into equivalence classes of antipodal points, so by the property of quotient maps, it follows that $X$ is homeomorphic to $Y$. But here the problem is in proving that $Y$ is Hausdorff, and this is what led me to ask this question.

Please advise. Thanks a lot.

Answer 1

$\mathbb R^{n+1}-\{0\}$ is Hausdorff so there exist $U, V$ disjoint open in $\mathbb R^{n+1}-\{0\}$ such that $x\in U, y\in V$.

This is correct, but it does not help you because you cannot expect that $p(U) \cap p(V) = \emptyset$. Your picture (2) gives a counterexample. By the way, one can show that $p$ is an open map, but this is somewhat tedious and above all unnecessary.

Anyway, you cannot take arbitrary open neighborhoods of $x$ and $y$, but have to make a careful choice.

Your observation that we can identify $Y$ with $S^n/{\sim}$, where $\sim$ identifies antipodal points, is a much better approach. Now look at Brian M. Scott's answer to How can I prove formally that the projective plane is a Hausdorff space? to see how to choose $U, V \subset S^n$.

Here are two questions which may be helpful:

• When is the restriction of a quotient map $p : X \to Y$ to a retract of $X$ again a quotient map?

• Real projective space is Hausdorff: is this proof correct?

Answer 2

Ok I try to sketch out a proof for why Y is hausdorff. This is not complete but I hope it will bring you forward. Notationwise, let $x = (x_1,...,x_{n+1}) \in \mathbb{R}^{n+1}$ and $[x]$ the equivalence class of $x$ in $Y$.

1. You can identify $Y$ with the upper Hemisphere $A$ of the n-Sphere $S^n$, where $x_{n+1} > 0$ be careful at the border $x_{n+1} = 0$, I'm not going to write the details out.
2. if $a \neq b \in A$ then there is a $n$-dimensional hyperplane $M$ defined by all points that are equally far away from $a$ and $b$. Note that $0$ is part of $M$ while $a$ and $b$ are not. Also it is possible to show, that if $x \in M$, then the straight line through $x$ and $0$ is also in $M$.
3. Since $M$ is closed, there is a point $m \in M$ which is closest to $a$ which is also closest to $b$. So let $\epsilon$ be the distance of $m$ and $a$ which is also the distance between $m$ and $b$. Now define the $\epsilon$-Ball $B_\epsilon(a)$ of radius $\epsilon$ around $a$ as well as $B_\epsilon(b)$. Since $$ B_\epsilon(a) \cap B_\epsilon(b) = \emptyset $$ the points $[a],[b] \in Y$ are also seperated by $[B_\epsilon(a)]$ and $[B_\epsilon(a)]$ where for a set $U$ $$[U]:= \{[x] \ | x \in U \}$$.

This is just a rough sketch but I hope it will help you out.