Showing two sets are not homeomorphic in subspace topology.

[Kreizhn]

Homework Statement

The true problem is too complicated to present here, but hopefully somebody can give me a hand with this simplified version. Consider the set $H = \{ (x,y) \in \mathbb R^2 : y \geq 0 \}$. Denote by $\partial H = \{ (x,0) \}$. Let U and V be open sets (relative to H) such that $U \cap \partial H = \emptyset$ and $V \cap \partial H \neq \emptyset$. I want to show that U and V cannot be diffeormorphic. It should be possible to show that these sets are not homeomorphic so we content ourselves with that.

The Attempt at a Solution

It seems to me that this shouldn't be too hard, but I don't want to make any silly mistakes. First of all, these are both open sets in H viewed with the subspace topology inherited from $\mathbb R^2$. However, the fact that V intersects the "boundary" of H would imply that there is some topological property which we shouldn't be able to transfer between the two of them. Should I look for an intrinsic property in H? Or is it possible to view U and V as subsets of $\mathbb R^2$ and get a contradiction there?

 

[lippyka]

Any help would be appreciated. My thoughts:I think the easiest way to approach this problem is to show that U and V are not homeomorphic. To do this, we can look for a topological property which one of them has and the other does not. Since U does not intersect the boundary of H, it follows that it is path-connected. On the other hand, since V does intersect the boundary, it is not path-connected. Therefore, since path-connectivity is a topological invariant, U and V cannot be homeomorphic.