Difference between revisions of "Exercises:Mond - Topology - 1/Question 9"

Revision as of 18:34, 10 October 2016 (view source) Alec (Talk | contribs) m (Altered formatting) ← Older edit		Revision as of 20:59, 10 October 2016 (view source) Alec (Talk | contribs) (Finishing proof) Newer edit →
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	So we must be careful to make sure our balls do not overlap at all!		So we must be careful to make sure our balls do not overlap at all!
		+
		+	Consider now {{M|1=\{x,-x\}=\pi^{-1}(a)}} and {{M|1=\{y,-y\}=\pi^{-1}(b)}}:
		+
		+	We notice also there is extra "structure" on {{M|\mathbb{R}^3}}, namely that it is a [[normed space]], {{M|(\mathbb{R}^3,\Vert\cdot\Vert)}}, and we consider the [[metric induced by the norm]] as the [[metric]], {{M|d}}, for a [[metric space]], {{M|(\mathbb{R}^3,d)}}, then we see:
		+	# {{M|1=d(x,y)=d(y,x)}} (by the ''symmetric'' property of a [[metric]]) and
		+	# {{M|1=d(-x,y)=\Vert -x-y\Vert = \Vert(-1)x+y\Vert=\Vert x+y\Vert=d(x,-y)}}
		+	# We don't need to consider {{M|d(x,-x)}} and {{M|d(y,-y)}}, also {{M|1=d(x,x)=d(y,y)=0}} is not very helpful
		+
		+	So take:
		+	* {{M|1=\epsilon\le\frac{1}{2}\text{min}(\{d(x,y),d(x,-y)\})}} (Note: {{Note|1=Smaller would work too, eg {{M|\frac{1}{4} }} rather than {{M|\frac{1}{2} }} - I hope I don't need to prove {{M|\frac{1}{2} }} is sufficient?}})
		+	** This should explain why {{M|d(x,-x)}} and {{M|d(y,-y)}} are of no use!
		+
		+	Then just place one of these [[open balls]] of radius {{M|\epsilon}} at each of the 4 points. Job done!
	====Solution====		====Solution====
	We wish to show that {{M|\mathbb{RP}^2}} is [[Hausdorff]].		We wish to show that {{M|\mathbb{RP}^2}} is [[Hausdorff]].
	* Let {{M|a,b\in\mathbb{RP}^2}} be given such that {{M|1=a\ne b}}, then		* Let {{M|a,b\in\mathbb{RP}^2}} be given such that {{M|1=a\ne b}}, then
	** there exist {{M|1=x,y\in\mathbb{S}^2}} such that {{M|1=\pi^{-1}(a)=\{x,-x\} }} and {{M|1=\pi^{-1}(b)=\{y,-y\} }}		** there exist {{M|1=x,y\in\mathbb{S}^2}} such that {{M|1=\pi^{-1}(a)=\{x,-x\} }} and {{M|1=\pi^{-1}(b)=\{y,-y\} }}
		+	*** Let {{M|1=\epsilon:=\frac{1}{2}\text{min}(\{d(x,y),d(x,-y)\})}} {{Note|Note that {{M|1=d(x,-y)=d(-x,y)}} - see above in the outline section}}
		+	**** Let {{M|1=V_a:=B_\epsilon(x)\cup B_\epsilon(-x)\subset\mathbb{R}^3}} and {{M|1=V_b:=B_\epsilon(y)\cup B_\epsilon(-y)\subset\mathbb{R}^3}}. These are open (in {{M|\mathbb{R}^3}}) as [[open balls]] are [[open sets]], and the union of open sets is open.
		+	***** Now define {{M|1=U_a:=V_a\cap\mathbb{S}^2}} and {{M|1=U_b:=V_b\cap\mathbb{S}^2}}, these are open in {{M|\mathbb{S}^2}} (considered with the [[subspace topology]] it inherits from {{M|\mathbb{R}^3}} - as mentioned in the outline)
		+	****** Recall {{M|U\in\mathcal{P}(\mathbb{RP}^2)}} is open {{iff}} {{M|\pi^{-1}(U)}} is open in {{M|\mathbb{S}^2}}
		+	****** Thus:
		+	******# {{M|1=\pi(U_a)}} is open {{iff}} {{M|\pi^{-1}(\pi(U_a))}} is open in {{M|\mathbb{S}^2}} and
		+	******# {{M|1=\pi(U_b)}} is open {{iff}} {{M|\pi^{-1}(\pi(U_b))}} is open in {{M|\mathbb{S}^2}}
		+	****** It should be clear that {{M|1=\pi^{-1}(\pi(U_a))=U_a}} and {{M|1=\pi^{-1}(\pi(U_b))=U_b}} (by their very construction)
		+	****** Thus:
		+	******# {{M|1=\pi(U_a)}} is open {{iff}} {{M|1=\pi^{-1}(\pi(U_a))=U_a}} is open in {{M|\mathbb{S}^2}} and
		+	******# {{M|1=\pi(U_b)}} is open {{iff}} {{M|1=\pi^{-1}(\pi(U_b))=U_b}} is open in {{M|\mathbb{S}^2}}
		+	****** As both right-hand-sides are true, we see {{M|\pi(U_a)}} and {{M|\pi(U_b)}} are both open in {{M|\mathbb{RP}^2}}
		+	****** We must now show {{M|U_a}} and {{M|U_b}} are [[disjoint]].
		+	******* Suppose there exists a {{M|1=p\in \pi(U_a)\cap\pi(U_b)}} (that is that they're not disjoint and {{M|x}} is in both of them), then:
		+	******** clearly {{M|1=\exists q\in \pi^{-1}(\pi(U_a))\cap\pi^{-1}(\pi(U_b))}} such that {{M|1=\pi(q)=p}}<ref group="Note">Note that by [[Properties of the pre-image of a map]] that {{M|1=\pi^{-1}\big(\pi(U_a)\cap\pi(U_b)\big)=\pi^{-1}(\pi(U_a))\cap\pi^{-1}(\pi(U_b))}}</ref>
		+	********* However {{M|1=\pi^{-1}(\pi(U_a))\cap\pi^{-1}(\pi(U_b))=U_a\cap U_b}} and {{M|1=U_a\cap U_b=\emptyset}} (by construction), so there does not exist such a {{M|q}}!
		+	********* If there is nothing in the pre-image of {{M|\pi(U_a)\cap\pi(U_b)}} that maps to {{M|p}} then we cannot have {{M|p\in \pi(U_a)\cap\pi(U_b)}} - a contradiction
		+	******* So there does not exist such a {{M|p}}, which means {{M|1=\pi(U_a)\cap\pi(U_b)=\emptyset}}, they're disjoint.
		+	This completes the proof.
	<noinclude>		<noinclude>
	==Notes==		==Notes==

---

Revision as of 20:59, 10 October 2016

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

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:

• $\epsilon\le\frac{1}{2}\text{min}(\{d(x,y),d(x,-y)\})$ (Note: Smaller would work too, eg $\frac{1}{4}$ rather than $\frac{1}{2}$ - I hope I don't need to prove $\frac{1}{2}$ is sufficient?)
  • 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 $\pi(q)=p$^{[Note 1]}
                • 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.

Notes

1. <cite_references_link_accessibility_label> ↑ Note that by Properties of the pre-image of a map that $\pi^{-1}\big(\pi(U_a)\cap\pi(U_b)\big)=\pi^{-1}(\pi(U_a))\cap\pi^{-1}(\pi(U_b))$

References