Proving [imath]\mathbb{P}^2[/imath] is Hausdorff

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MathNugget

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I know that a space is Hausdorff if for every 2 different points, there's 2 disjoint open sets, each containing one of the points. Also, the [imath]\mathbb{P}^2[/imath] can be seen as [imath]\mathbb{R}^2[/imath] with a line at infinity, or as [imath]\mathbb{R}^3\backslash \{0\}/\sim[/imath], with [imath]P\sim Q \Leftrightarrow OP=OQ[/imath].

I have trouble defining the open sets of [imath]\mathbb{P}^2[/imath]. I am mostly looking at the first definition of the projective plane, as [imath]\mathbb{R}^2[/imath] is Hausdorff already. But how would I prove for 2 points at infinity, or 1 basic point and 1 at infinity?
 
I'd use the definition of [imath]\mathbb P^2[/imath] as a quotient space of [imath]\mathbb S^2[/imath].
 
blamocur said:
I'd use the definition of [imath]\mathbb P^2[/imath] as a quotient space of [imath]\mathbb S^2[/imath].
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I guess we talk here about about [imath]P\sim Q[/imath] if they're antipodal points? Meaning, [imath]P\sim -P[/imath]?
 
blamocur said:
I'd use the definition of [imath]\mathbb P^2[/imath] as a quotent space of [imath]\mathbb S^2[/imath].
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Forgot to mention that this relies on the fact that [imath]\mathbb S^2[/imath] is Hausdorff, but the latter is easy to prove using stereographic projection.
 
blamocur said:
Forgot to mention that this relies on the fact that [imath]\mathbb S^2[/imath] is Hausdorff, but the latter is easy to prove using stereographic projection.
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I guess [imath]S^2[/imath] is Hausdorff as a subset of [imath]\mathbb{R}^3[/imath]. But how would I prove after factoring through [imath]\sim[/imath] that I get a Hausdorff thing?
 
blamocur said:
Good point!

For two arbitrary points (in the definition of Hausdorff space) look at their "preimages" and the corresponding open neighborhoods.
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Seems I am using that this factorisation is a continuous function of sorts. But I know that for an open set [imath]U[/imath], [imath]f^{-1}(U)[/imath] would be open too, but I don't know about the opposite; would the image of an open set, be open?

I checked on the internet, it seems what I need is an 'open map', a function that maps open sets to open sets. Is it time to attempt to explicitly turn the factorisation into a function? It's a little troublesome to 'see' this factorisation, as I think it is physically impossible to deform a circle that way...
 
MathNugget said:
would the image of an open set, be open?
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Not in the general case.
MathNugget said:
It's a little troublesome to 'see' this factorisation, as I think it is physically impossible to deform a circle that way...
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The best visualization I know if is to take half sphere and "saw" together pairs of opposite edge points. Still, [imath]\mathbb P^2[/imath] cannot be embedded in [imath]\mathbb R^3[/imath] without self-intersections.

A hint: for any two points [imath]a,b\in \mathbb P^2[/imath] you can find proper open subset [imath]U\subset\mathbb P^2[/imath] which is homeomorphic to [imath]\mathbb R^2[/imath] and contains both [imath]a[/imath] and [imath]b[/imath].
 
blamocur said:
Not in the general case.

The best visualization I know if is to take half sphere and "saw" together pairs of opposite edge points. Still, [imath]\mathbb P^2[/imath] cannot be embedded in [imath]\mathbb R^3[/imath] without self-intersections.

A hint: for any two points [imath]a,b\in \mathbb P^2[/imath] you can find proper open subset [imath]U\subset\mathbb P^2[/imath] which is homeomorphic to [imath]\mathbb R^2[/imath] and contains both [imath]a[/imath] and [imath]b[/imath].
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That boils down to my other question... how do I know how the Open sets of the projective plane look like?

I am familiar with the formal definition of the family of open sets, but I think it is only good to check if something is open or not.
I assume discs would still be open sets on the [imath]\mathbb{R}^2[/imath] part of the projective plane (I suppose we can extend the distance function, so distance to and from a point at infinity somehow, and use the constructive definition of the projective plane, as adding the infinity line to real plane). How would an open set centered in an infinity point even look like? And how would I be able to guess what an open set looks like, if I didn't know the projective plane can be seen as the normal plane + something?
 
MathNugget said:
That boils down to my other question... how do I know how the Open sets of the projective plane look like?

I am familiar with the formal definition of the family of open sets, but I think it is only good to check if something is open or not.
I assume discs would still be open sets on the [imath]\mathbb{R}^2[/imath] part of the projective plane (I suppose we can extend the distance function, so distance to and from a point at infinity somehow, and use the constructive definition of the projective plane, as adding the infinity line to real plane). How would an open set centered in an infinity point even look like? And how would I be able to guess what an open set looks like, if I didn't know the projective plane can be seen as the normal plane + something?
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Visualizing topological space can be a challenge, especially once you move to higher dimensions. But just as in math in general, visualizing is not always necessary for developing intuition and skills. Here is one way to approach your problem:

By definition [imath]U\subset \mathbb P^2[/imath] is open if and only if [imath]\pi^{-1}(U)[/imath] is open, where [imath]\pi: \mathbb S^2 \rightarrow \mathbb P^2[/imath] is the defining projection.
One way to build an open subset in [imath]\mathbb P^2[/imath] is to pick an open subset [imath]V\subset \mathbb S^2[/imath] and prove that [imath]\pi(V)[/imath] is open. Hint: look at [imath]\pi^-1(\pi(V))[/imath].
 
well, [imath]\pi^{-1}(\pi(V))[/imath] would be [imath]V \cup -V[/imath], with [imath]-V=\{x \mid -x \in V\}[/imath].
blamocur said:
By definition [imath]U\subset \mathbb P^2[/imath] is open if and only if [imath]\pi^{-1}(U)[/imath] is open, where [imath]\pi: \mathbb S^2 \rightarrow \mathbb P^2[/imath] is the defining projection.
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Why is this so? I guess it's true for a homeomorphism; but while deforming the sfere into the projective plane is a continuous deformation, it doesn't seem bijective (meaning [imath]\pi[/imath] is not a homeomorphism.

By the way, this is one of the first exercises from Lee's "smooth manifolds" book. Forgot to mention.
 
MathNugget said:
You are right. I see now 😓 . I didn't check that.
So 2 open sets go to 1 open set (which is one of those 2). Since [imath]S^2[/imath] is Hausdorff, then so is this factor topology... And I guess same thing can be said about being second-countable space, since I can now see it as an inclusion.

Thank you!!
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You are welcome, although I am not sure if this will be considered a complete proof by your teacher.
 
blamocur said:
You are welcome, although I am not sure if this will be considered a complete proof by your teacher.
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It might not be, but being seemingly knowledgeable on the topic might be enough to pass, with enough luck. I'm still a long way from tricking anyone into thinking I know these things, though. :ROFLMAO:
 
MathNugget said:
It might not be, but being seemingly knowledgeable on the topic might be enough to pass, with enough luck. I'm still a long way from tricking anyone into thinking I know these things, though. :ROFLMAO:
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Good luck with your tricks :)
 
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