The previous post is very inelegant because it invokes machinery such as the one point compactification (not too absurd for a problem of this nature I suppose) and the fundamental group (which is way overkill). The best way (in my opinion) to solve this problem is to use the following lemma:
If are two homeomorphic topological spaces, with a homeomorphism given by , and if is any subspace of , then is homeomorphic to as a subspace of , with a homeomorphism given by
Let be any open subset of . By definition, this is of the form for some which is open in . We have the following:
The right hand side is open in , so is continuous. Similarly, is continuous, so we are done.
Now, to use the lemma: suppose that the two spaces mentioned in the previous post were homeomorphic, and let be such a homeomorphism. For simplicity, the domain of shall be . Consider as a subset. We see that . However, in this case is connected, but isn’t, a contradiction.
This seemingly dumb trick can be used to prove that many different spaces are not homeomorphic. I leave the following as exercises to the reader using this.
- Show that a figure 8 and a circle are not homeomorphic.
- Show that an uppercase X and an uppercase T are not homeomorphic.
- Show that and are not homeomorphic (see the previous post) without invoking their fundamental groups.