Geometric interpretation of a given Alexandrov compactification

[prce]

What is the Alexandrov compactification of the following set and give the geometric interpretation of it:

[(x,y): x^2-y^2>=1, x>0] that is, the right part of the hyperbola along with the point in it.

This is a question from my todays exam in topology. I wrote that the given set is homeomorphic to the set [0,1)x[0,1) the alexandrov compactification of which is [0,1]x[0,1].

However I'm not sure at all. Is this correct?

 

[quasar987]

Well, you are right that the given set is homeomorphic to the set [0,1)x[0,1) and you are also right that the alexandrov compactification of it is [0,1]x[0,1].

But maybe you got the right answer for the wrong reason. For instance, what would you say is the Alexandroff compactification of the open square (0,1)x(0,1)?

It is not [0,1]x[0,1]; it is the 2-sphere S². The Alexandroff compactification is also called the one-point compactification because what it does typically is that it takes all the points "at infinity" and glue/collapse/identify them to a single point, which we usually write [itex]\infty[/itex]. This is the guiding picture to keep in mind while guessing what the Alexandroff compactification of a given space is. Once you've guessed what geometrical shape corresponds to the Alexandroff compactification, you need to produce an explicit homeomorphism between them in order to verify that your intuition was good.

For instance in the case of the compactification of (0,1)x(0,1), the homeomorphism f: S² --> (0,1)x(0,1) u {[itex]\infty[/itex]} is define by part by f=(stereographic projection on S²\{south pole}) and f(south pole)=[itex]\infty[/itex].

In your case, the "points at infinity" correspond to the points on the top side and right side of your "semi-open square" [0,1)x[0,1). Collapsing them onto a point gives something homeomorphic to [0,1]x[0,1].

 

[prce]

I kind of guessed and didnt put any reasons in the response. thinking now i should have procesed this way:
1. the given set i homeomorphic to the right semiplane. the compactification of alexandrov of which is clearly (by the stereographic projection) the semi-sphere (including the maximum circle).
2. The semi-sphere is homeomorphic to a closed disc which is homeomorphyc to [0,1]x[0,1].

But the question was: give the geometrical definition of the compactification. Whats "more right", semi-sphere, closed disc or [0,1]x[0,1]?

one other question asked for the intern of the set A in the subspace X (X subspace of E^2). A being the unit circle. B being [(x,y): x^2-y^2>=1, x>0] and C [(x,y): x^2-y^2>1, x<0]. X=AUBUC.
I put some crazy arguments and wrote that the intern of A is A itself.
But it is A\(1,0),(-1,0) is it?

On an exam, when proving homeomorphism is it necessary to give explicit formula of it or it is acceptable to only give a wordily, however logicaly consistent, description of the 1-1 mapping?

 

[quasar987]

I kind of guessed and didnt put any reasons in the response. thinking now i should have procesed this way:
1. the given set i homeomorphic to the right semiplane. the compactification of alexandrov of which is clearly (by the stereographic projection) the semi-sphere (including the maximum circle).
2. The semi-sphere is homeomorphic to a closed disc which is homeomorphyc to [0,1]x[0,1].

But the question was: give the geometrical definition of the compactification. Whats "more right", semi-sphere, closed disc or [0,1]x[0,1]?

I don't think one is more right that any other. I do like the closed disk best personally.

one other question asked for the intern of the set A in the subspace X (X subspace of E^2). A being the unit circle. B being [(x,y): x^2-y^2>=1, x>0] and C [(x,y): x^2-y^2>1, x<0]. X=AUBUC.
I put some crazy arguments and wrote that the intern of A is A itself.
But it is A\(1,0),(-1,0) is it?

I'm afraid so!

On an exam, when proving homeomorphism is it necessary to give explicit formula of it or it is acceptable to only give a wordily, however logicaly consistent, description of the 1-1 mapping?

It depends on your teacher. In the case of the compactification question, I would have given a picture to explain the map, and also to show why it is a homeomorphism.

 

[blue_raver22]

Yes, your answer is correct. The Alexandrov compactification of the given set is [0,1]x[0,1], which can be visualized as a square with its top and right edges identified. This compactification adds a single point at infinity, representing the "missing" points that complete the hyperbola.

Geometrically, this compactification can be interpreted as folding the hyperbola along the x-axis and "gluing" the two ends together to form a square. This allows us to visualize the entire hyperbola, including the point at infinity, in a compact and connected space.