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

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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!
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 +
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)}}
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# 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\} }}  
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*** 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}}
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**** 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, [ilmath]\mathbb{RP}^2[/ilmath] is defined as the quotient of the sphere, [ilmath]\mathbb{S}^2[/ilmath], by the equivalence relation that defines (for [ilmath]x\in\mathbb{S}^2\subset\mathbb{R}^3[/ilmath]) [ilmath]x\sim -x[/ilmath], that is it identifies antipodal points.

Show that [ilmath]\mathbb{RP}^2[/ilmath] is Hausdorff

Definitions

Solution outline

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

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

  • [ilmath]a\in U_a[/ilmath], [ilmath]b\in U_b[/ilmath] and [ilmath]U_a\cap U_b=\emptyset[/ilmath]

Well:

  • [ilmath]U_a[/ilmath] is open in [ilmath]\mathbb{RP}^2[/ilmath]

if and only if

  • [ilmath]\pi^{-1}(U_a)[/ilmath] is open in [ilmath]\mathbb{S}^2[/ilmath]

if and only if

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

Of course, the open balls of [ilmath]\mathbb{R}^3[/ilmath] are a basis, so we can think of [ilmath]V_a[/ilmath] 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 [ilmath]a[/ilmath] and [ilmath]b[/ilmath], what size balls can we consider in [ilmath]\mathbb{R}^3[/ilmath] such that they're disjoint. There's a caveat here. This is what is shown in the diagram.

If [ilmath]a[/ilmath] and [ilmath]b[/ilmath] are "far apart" on [ilmath]\mathbb{RP}^2[/ilmath], it is entirely possible (in the pre-image under [ilmath]\pi[/ilmath]) 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 [ilmath]\{x,-x\}=\pi^{-1}(a)[/ilmath] and [ilmath]\{y,-y\}=\pi^{-1}(b)[/ilmath]:

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

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

So take:

  • [ilmath]\epsilon\le\frac{1}{2}\text{min}(\{d(x,y),d(x,-y)\})[/ilmath] (Note: Smaller would work too, eg [ilmath]\frac{1}{4} [/ilmath] rather than [ilmath]\frac{1}{2} [/ilmath] - I hope I don't need to prove [ilmath]\frac{1}{2} [/ilmath] is sufficient?)
    • This should explain why [ilmath]d(x,-x)[/ilmath] and [ilmath]d(y,-y)[/ilmath] are of no use!

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

Solution

We wish to show that [ilmath]\mathbb{RP}^2[/ilmath] is Hausdorff.

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

This completes the proof.

Notes

  1. ↑ Note that by Properties of the pre-image of a map that [ilmath]\pi^{-1}\big(\pi(U_a)\cap\pi(U_b)\big)=\pi^{-1}(\pi(U_a))\cap\pi^{-1}(\pi(U_b))[/ilmath]

References


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