This one is my own doing. I will be showing the following:

is not homeomorphic to

First, we begin with a definition and a lemma. If is any topological space, we define the of as the topological space

We have the following lemma:

if are two homeomorphic topological spaces, then their one point compactifications are homeomorphic.

Because are homeomorphic, there is a homeomorphism between them. Denote the one point compactifications of as , respectively. Define such that , and such that for any non-infinite .

I claim that is a homeomorphism. Suppose that is an open subset of which does not contain . Then, by construction, we have that is an open subset of which does not contain . Suppose that is an open subset of which does contain . By construction, the complement of is closed and compact. So, is a compact and closed subset of by construction, so is the complement of a closed compact subset of and contains , so is open. Similarly, is continuous, so we are done.

With that out of the way, it is well known that is homeomorphic to , whose one point compactification is homeomorphic to , the unit circle. However, is homeomorhic to , whose one point compactification is homeomorphic to . I claim the following:

is not homeomorphic to

This can be seen by noting that while , and clearly these are not isomorphic. By the functoriality of , we have that is not homeomorphic to . By the contraposition of the lemma, we thus see that and are not homeomorphic.

### Like this:

Like Loading...

Test

LikeLiked by 1 person