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.
Test
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