Addendum to the previous post

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:

\textbf{Lemma:} If X, Y are two homeomorphic topological spaces, with a homeomorphism given by f:X\to Y, and if S is any subspace of X, then S is homeomorphic to f(S) as a subspace of Y, with a homeomorphism given by f|_S

\textbf{Proof:} Let U be any open subset of f(S). By definition, this is of the form U' \cap f(S) for some U' which is open in Y. We have the following:

f|_S^{-1}(U) = f|_S^{-1}(U'\cap f(S)) = f^{-1}(U')\cap f^{-1}(f(S)) = f^{-1}(U')\cap S

The right hand side is open in S, so f|_S is continuous. Similarly, f|_S^{-1} 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 f be such a homeomorphism. For simplicity, the domain of f shall be [0,\infty). Consider S = (0,\infty) as a subset. We see that f(S) = (0,\infty) / f(0). However, in this case S is connected, but f(S) 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.

  1. Show that a figure 8 and a circle are not homeomorphic.
  2. Show that an uppercase X and an uppercase T are not homeomorphic.
  3. Show that S^1 and \mathbb{R} are not homeomorphic (see the previous post) without invoking their fundamental groups.

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