R^2 and R^2 - {(0,0)} are not homeomorphic

[poet_3000]

Let R denote the real line and R^2 the real plane.

How do I show that R^2 and R^2 - {(0,0)} are not homeomorphic? I don't even know where to start since I cannot think of a topological property that is in one but not in the other. Can anyone give me a clue on what that property is?

A similar problem is this:
Consider S = the union of the x and y- axes together with the subspace topology inherited from R^2. Show that S and R are not homeomorphic.

Thanks in advance!

 

[micromass]

Let R denote the real line and R^2 the real plane.

How do I show that R^2 and R^2 - {(0,0)} are not homeomorphic? I don't even know where to start since I cannot think of a topological property that is in one but not in the other. Can anyone give me a clue on what that property is?

Simply connectedness. Or equivalent: calculate the fundamental group of the two spaces. Of R^2 the group is trivial, but of R^2\{(0,0)} the group is Z.

A similar problem is this:
Consider S = the union of the x and y- axes together with the subspace topology inherited from R^2. Show that S and R are not homeomorphic.

Thanks in advance!

In S, there exists a point such that if you remove the point, then there are 4 connected components. But there is no such point in R: any point that you remove in R will yield 2 connected components.

 

[lavinia]

Let R denote the real line and R^2 the real plane.

How do I show that R^2 and R^2 - {(0,0)} are not homeomorphic? I don't even know where to start since I cannot think of a topological property that is in one but not in the other. Can anyone give me a clue on what that property is?

A similar problem is this:
Consider S = the union of the x and y- axes together with the subspace topology inherited from R^2. Show that S and R are not homeomorphic.

Thanks in advance!

for the first one there are classic algebraic topology proofs. Do you want one of these or would you like something more elementary?

 

[poet_3000]

Oh. Thank you for your replies! It really helped me. Thank you micromass and lavinia.

lavinia, may I know your other suggestion?

 

[lavinia]

Oh. Thank you for your replies! It really helped me. Thank you micromass and lavinia.

lavinia, may I know your other suggestion?

I never tried a simpler proof but let's give it a try.

If there were a homeomorphism from R^-0 to R^2, a Cauchy sequence that converges to the origin must be mapped to infinity. This means that the punctured disc around the origin must be mapped to a neighborhood of infinity that is bounded by a simple closed curve.

But then the neighborhood of infinity must be mapped into the inside this simple closed curve. A sequence that diverges to infinity would be mapped to a bounded sequence which must have a limit point. that is impossible.This generalizes to higher dimensions.

It might be fun to think up some others.

 

[lavinia]

For the second problem, here is a more complicated proof.

If there is a homeomorphism from the union of the x and y axes then consider the image of the x axis. This is homeomorphic to R so the image of the origin is surrounded by an open interval. That does it because now the image of the y-axis can not penetrate this interval (which it must by continuity) because the map is 1-1.

 

[Deveno]

the way i said it for my senior thesis (i am not joking) is:

ok we have da loop, and we have da group. we loop de loop and group da loop, to da loop group.

but onoz! we have hole. so sorry, no homeomorphism for j00...

(and then i popped a balloon with a pin, and said: see?)

 

[poet_3000]

the way i said it for my senior thesis (i am not joking) is:

ok we have da loop, and we have da group. we loop de loop and group da loop, to da loop group.

but onoz! we have hole. so sorry, no homeomorphism for j00...

(and then i popped a balloon with a pin, and said: see?)

that is entertaining deveno! LOL! :p