Exercises:Mond - Topology - 1/Question 9

Section B

Question 9

The real projective plane, $\mathbb{RP}^2$ is defined as the quotient of the sphere, $\mathbb{S}^2$, by the equivalence relation that defines (for $x\in\mathbb{S}^2\subset\mathbb{R}^3$) $x\sim -x$, that is it identifies antipodal points.

Show that $\mathbb{RP}^2$ is Hausdorff

Definitions

• We denote by $\pi:\mathbb{S}^2\rightarrow\frac{\mathbb{S}^2}{\sim}$ the canonical projection of the equivalence relation, $\sim$. Note that this is a quotient map when we consider $\frac{\mathbb{S}^2}{\sim}$ with the quotient topology.

Solution outline

Suppose we take the points $\alpha$ and $\beta$ on the sphere. We could have balls (in $\mathbb{R}^3$) that are centred at $\alpha$ and $\beta$ and they'd get quite large before touching (we want them to be disjoint!) However we must consider 4 open balls, one at each point $\alpha$, $\beta$, $\gamma$ and $\delta$, we see here even though $\alpha$ and $\beta$ are far apart that $\beta$ and $\gamma$ (which is antipodal to $\alpha$) are actually rather close!

So rather than $\epsilon\le\frac{1}{2}d(\alpha,\beta)$ (for $d$ being the Euclidean metric of $\mathbb{R}^3$) we must make sure that the ball at the antipodal point doesn't touch any others too!

It is clear that if $\epsilon\le\frac{1}{2} d(\alpha,\gamma)=1$ that the balls centred at $\alpha$ and $\gamma$ (or $\beta$ and $\delta$) wont touch, we must make sure that $\alpha$ and $\beta$ don't touch, so use $\epsilon\le d(\alpha,\beta)$.

But as the diagram shows, $\alpha$ and $\delta$ could be rather close! (or equivalently, $\gamma$ and $\beta$), so we need $\epsilon<\frac{1}{2} d(\alpha,\delta)$ too!

We can boil all these down into $\epsilon\le\frac{1}{2}\text{min}(\{d(\alpha,\beta),d(\alpha,\delta)\})$

We will deal with the open sets, $U$, in terms of $\pi^{-1}(U)$ (as by definition, $U\in\mathcal{P}(\frac{\mathbb{S}^2}{\sim})$ is open in $\frac{\mathbb{S}^2}{\sim}$ if and only if $\pi^{-1}(U)$ is open in $\mathbb{S}^2$, which we consider with the subspace topology inherited from $\mathbb{R}^3$ as usual) then we just have to find small enough open balls!

With this in mind, let $a,b\in\mathbb{RP}^2$ be given. We need to find two open sets (well... neighbourhoods will do... but open sets are neighbourhoods!), let's say $U_a$ and $U_b$ such that:

• $a\in U_a$, $b\in U_b$ and $U_a\cap U_b=\emptyset$

Well:

• $U_a$ is open in $\mathbb{RP}^2$

if and only if

• $\pi^{-1}(U_a)$ is open in $\mathbb{S}^2$

if and only if

• there exists an open set, $V_a$ in $\mathbb{R}^3$ such that $V_a\cap\mathbb{S}^2=\pi^{-1}(U_a)$

Of course, the open balls of $\mathbb{R}^3$ are a basis, so we can think of $V_a$ as a union of open balls, or possibly just an open ball (as basis sets themselves are open, and also as an open ball can be expressed as a union of open balls).

This changes the question into, in terms of $a$ and $b$, what size balls can we consider in $\mathbb{R}^3$ such that they're disjoint. There's a caveat here. This is what is shown in the diagram.

If $a$ and $b$ are "far apart" on $\mathbb{RP}^2$, it is entirely possible (in the pre-image under $\pi$) that the antipodal point of one is near the other!

So we must be careful to make sure our balls do not overlap at all!

Consider now $\{x,-x\}=\pi^{-1}(a)$ and $\{y,-y\}=\pi^{-1}(b)$:

We notice also there is extra "structure" on $\mathbb{R}^3$, namely that it is a normed space, $(\mathbb{R}^3,\Vert\cdot\Vert)$, and we consider the metric induced by the norm as the metric, $d$, for a metric space, $(\mathbb{R}^3,d)$, then we see:

1. $d(x,y)=d(y,x)$ (by the symmetric property of a metric) and
2. $d(-x,y)=\Vert -x-y\Vert = \Vert(-1)x+y\Vert=\Vert x+y\Vert=d(x,-y)$
3. We don't need to consider $d(x,-x)$ and $d(y,-y)$, also $d(x,x)=d(y,y)=0$ is not very helpful

So take:

• $0<\epsilon\le\frac{1}{2}\text{min}(\{d(x,y),d(x,-y)\})$ (Note: there are "safer" choices for the upper bound to put on $\epsilon$, eg $\frac{1}{4} \text{min}(\{d(x,y),d(x,-y)\})$ rather than $\frac{1}{2} \text{min}(\{d(x,y),d(x,-y)\})$ - I hope I don't need to prove $\frac{1}{2}$ is sufficient? However for a discussion see the caption of the picture on the right.)
  • This should explain why $d(x,-x)$ and $d(y,-y)$ are of no use!

Then just place one of these open balls of radius $\epsilon$ at each of the 4 points. Job done!

Solution

We wish to show that $\mathbb{RP}^2$ is Hausdorff.

• Let $a,b\in\mathbb{RP}^2$ be given such that $a\ne b$, then
  • there exist $x,y\in\mathbb{S}^2$ such that $\pi^{-1}(a)=\{x,-x\}$ and $\pi^{-1}(b)=\{y,-y\}$
    • Let $\epsilon:=\frac{1}{2}\text{min}(\{d(x,y),d(x,-y)\})$ Note that $d(x,-y)=d(-x,y)$ - see above in the outline section
      • Let $V_a:=B_\epsilon(x)\cup B_\epsilon(-x)\subset\mathbb{R}^3$ and $V_b:=B_\epsilon(y)\cup B_\epsilon(-y)\subset\mathbb{R}^3$. These are open (in $\mathbb{R}^3$) as open balls are open sets, and the union of open sets is open.
        • Now define $U_a:=V_a\cap\mathbb{S}^2$ and $U_b:=V_b\cap\mathbb{S}^2$, these are open in $\mathbb{S}^2$ (considered with the subspace topology it inherits from $\mathbb{R}^3$ - as mentioned in the outline)
          • Recall $U\in\mathcal{P}(\mathbb{RP}^2)$ is open if and only if $\pi^{-1}(U)$ is open in $\mathbb{S}^2$
          • Thus:
            1. $\pi(U_a)$ is open if and only if $\pi^{-1}(\pi(U_a))$ is open in $\mathbb{S}^2$ and
            2. $\pi(U_b)$ is open if and only if $\pi^{-1}(\pi(U_b))$ is open in $\mathbb{S}^2$
          • It should be clear that $\pi^{-1}(\pi(U_a))=U_a$ and $\pi^{-1}(\pi(U_b))=U_b$ (by their very construction)
          • Thus:
            1. $\pi(U_a)$ is open if and only if $\pi^{-1}(\pi(U_a))=U_a$ is open in $\mathbb{S}^2$ and
            2. $\pi(U_b)$ is open if and only if $\pi^{-1}(\pi(U_b))=U_b$ is open in $\mathbb{S}^2$
          • As both right-hand-sides are true, we see $\pi(U_a)$ and $\pi(U_b)$ are both open in $\mathbb{RP}^2$
          • We must now show $U_a$ and $U_b$ are disjoint.
            • Suppose there exists a $p\in \pi(U_a)\cap\pi(U_b)$ (that is that they're not disjoint and $x$ is in both of them), then:
              • clearly $\exists q\in \pi^{-1}(\pi(U_a))\cap\pi^{-1}(\pi(U_b))$ such that $\pi(q)=p$
                • However $\pi^{-1}(\pi(U_a))\cap\pi^{-1}(\pi(U_b))=U_a\cap U_b$ and $U_a\cap U_b=\emptyset$ (by construction), so there does not exist such a $q$!
                • If there is nothing in the pre-image of $\pi(U_a)\cap\pi(U_b)$ that maps to $p$ then we cannot have $p\in \pi(U_a)\cap\pi(U_b)$ - a contradiction
            • So there does not exist such a $p$, which means $\pi(U_a)\cap\pi(U_b)=\emptyset$, they're disjoint.

This completes the proof.