# 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.”

1. johnald42 says:

Test

Liked by 1 person