Showing two easy spaces are not homeomorphic.

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

(0,\infty) is not homeomorphic to [0, \infty)

First, we begin with a definition and a lemma. If X = (A, \tau) is any topological space, we define the \textbf{one point compactifiation} of X as the topological space

(A \cup \{ \infty_X \}, \tau \cup \{B \cup \{ \infty_X \} \mid A/B \text{ is a compact closed subset of X} \})

We have the following lemma:

\textbf{Lemma:} if X, Y are two homeomorphic topological spaces, then their one point compactifications are homeomorphic.

\textbf{Proof:} Because X, Y are homeomorphic, there is a homeomorphism f:X\to Y between them. Denote the one point compactifications of X, Y as \hat{X}, \hat{Y}, respectively. Define \hat{f}:\hat{X}\to\hat{Y} such that f(\infty_X)=\infty_Y, and such that \hat{f}(x) = f(x) for any non-infinite x\in X.

I claim that \hat{f} is a homeomorphism. Suppose that U\subset \hat{Y} is an open subset of \hat{Y} which does not contain \infty_Y. Then, by construction, we have that \hat{f}^{-1}(U) is an open subset of \hat{X} which does not contain \infty_X. Suppose that U\subset \hat{Y} is an open subset of \hat{Y} which does contain \infty_Y. By construction, the complement of U is closed and compact. So, f^{-1}(\hat{Y}/U) is a compact and closed subset of \hat{X} by construction, so \hat{f}^{-1}(U) is the complement of a closed compact subset of X and contains \infty_X, so is open. Similarly, \hat{f}^{-1} is continuous, so we are done.

With that out of the way, it is well known that (0, \infty) is homeomorphic to \mathbb{R}, whose one point compactification is homeomorphic to S^1, the unit circle. However, [0, \infty) is homeomorhic to [0, 1), whose one point compactification is homeomorphic to [0, 1]. I claim the following:

[0, 1] is not homeomorphic to S^1

This can be seen by noting that \pi_1([0, 1], 1)\cong\{0\} while \pi_1(S^1, 1)\cong\mathbb{Z}, and clearly these are not isomorphic. By the functoriality of \pi_1, we have that [0, 1] is not homeomorphic to S^1. By the contraposition of the lemma, we thus see that (0, \infty) and [0, \infty) are not homeomorphic.

One thought on “Showing two easy spaces are not homeomorphic.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

Create your website with WordPress.com
Get started
%d bloggers like this: