Homeomorphisms on Topological Spaces Examples 1

Homeomorphisms on Topological Spaces Examples 1

Recall from the Homeomorphisms on Topological Spaces page that if $X$ and $Y$ are topological spaces then a bijective map $f : X \to Y$ is said to be a homeomorphism of these two spaces if $f$ is also open and continuous, or equivalently, both $f$ and $f^{-1}$ are continuous.

We will now look at some examples of homeomorphic topological spaces.

Example 1

Show that the topological spaces $(0, 1)$ and $(0, \infty)$ (with their topologies being the unions of open balls resulting from the usual Euclidean metric on these subsets of $\mathbb{R}$) are homeomorphic.

To show that these two topological spaces are homeomorphic we must find a continuous bijection $f : X \to Y$ such that $f^{-1}$ is also continuous.

Consider the following function $f : (0, 1) \to (1, \infty)$ given by:

(1)
\begin{align} \quad f(x) = \frac{1}{x} \end{align}

We first show that $f$ is bijection. Let $x, y \in (0, 1)$ and suppose that $f(x) = f(y)$. Then:

(2)
\begin{align} \quad \frac{1}{x} = \frac{1}{y} \\ \end{align}

Cross multiplying gives us that then $x = y$, so $f$ is injective.

Now let $b \in (1, \infty)$. Since $b > 1$ we have that $0 < \frac{1}{b} < 1$, and so let $a = \frac{1}{b}$. Then:

(3)
\begin{align} \quad f(a) = \frac{1}{a} = \frac{1}{\frac{1}{b}} = b \end{align}

So for all $b \in (1, \infty)$ there exists an $a \in (0, 1)$ such that $f(a) = b$, so $f$ is surjective.

It's not hard to see that $f$ is a continuous map. Furthermore, $f^{-1} : (1, \infty) \to (0, 1)$ is also given by $f^{-1}(x) = \frac{1}{x}$ (which is continuous), and so $f$ is a homeomorphism between $(0, 1)$ and $(1, \infty)$, so these spaces are homeomorphic.

Example 2

Show that the spaces $(-r, r)$, $r > 0$ and $\mathbb{R}$ with the topologies obtained by the unions of open balls with respect to the usual Euclidean metric are homeomorphic.

Consider the following function $f : (- r, r) \to \mathbb{R}$ given by:

(4)
\begin{align} \quad f(x) = \tan \left ( \frac{\pi x}{2r} \right ) \end{align}

Then $f$ is clearly continuous as $f$ will always have the following form: Further it should be clear that $f^{-1}$ will always always be continuous: Therefore $f$ is a homeomorphism between $(-r, r)$ and $\mathbb{R}$ so these spaces are homeomorphic.