Homeomorphisms on Topological Spaces Examples 2

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 more examples of homeomorphic topological spaces.

Example 1

Let $\mathbb{N}$ and $\mathbb{Z}$ be topological spaces with the subspace topology from $\mathbb{R}$ having the usual topology. Prove that $\mathbb{N}$ is homeomorphic to $\mathbb{Z}$.

We need to find a homeomorphism $f : \mathbb{N} \to \mathbb{Z}$. Consider the following map defined for all $n \in \mathbb{N}$ by:

(1)

\begin{align} \quad f(n) = \left\{\begin{matrix} -\frac{n+1}{2} & \mathrm{if} \: n \: \mathrm{if \: odd} \\ \frac{n}{2} & \mathrm{if} \: n \: \mathrm{is \: even} \end{matrix}\right. \end{align}

Then $f(0) = 0$, $f(1) = -1$, $f(2) = 1$, $f(3) = -2$, $f(4) = 2$, etc… We claim that $f$ is a homeomorphism. Clearly $f$ is both injective and surjective, i.e. bijective.

We now show that $f$ is continuous. Notice that the subspace topologies on $\mathbb{N}$ and $\mathbb{Z}$ from $\mathbb{R}$ give the discrete topology on these sets. So for any open set $U$ in $\mathbb{Z}$, $f^{-1}(U)$ will always be a subset of natural numbers which is open in $\mathbb{N}$. This shows that $f$ is continuous.

Similarly, for any open set $U$ in $\mathbb{N}$, $f(U)$ will be a subset of integers which will always be open in $\mathbb{Z}$. This shows that $f$ is an open map.

Hence $f$ is bijective, continuous, and open map, so $f$ is a homeomorphism from $\mathbb{N}$ to $\mathbb{Z}$, so $\mathbb{N}$ is homeomorphic to $\mathbb{Z}$.