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.