Proposition

Show that a topological space is a $0$-manifold if and only if it is a countable discrete space.

Solution

Suppose that $X$ is $0$-manifold. Since it is a $0$-manifold, for each $x \in X$, there must exist a neighborhood $U$ that is homeomorphic to an open set in $\mathbb{R}^0$. The only open sets in $\mathbb{R}^0$ are $\emptyset$ and $\{ 0 \}$. Since a nonempty set cannot be mapped into the empty set, $U$ must be homeomorphic to $\{ 0 \}$. If $U$ contains more than $1$ element, then any map from $U$ into $\{ 0 \}$ is not injective. Thus $U$ must only contain $x$.

This shows that $\{ x \}$ is open for every $x \in X$, so $X$ has a discrete topology.

Let $\mathcal{B}$ be a countable basis of $X$. Let $x \in x$ be given. Then $\{ x \}$ is the union of elements of $\mathcal{B}$ by definition. This implies that $\{ x \} \in \mathcal{B}$.

Since $\mathcal{B}$ contains all singletons and $\mathcal{B}$ is countable, there are at most countably many singletons. In other words, $X$ must be countable.

Therefore, $X$ is a countable discrete space.

On the other hand, suppose that $X$ is a countable discrete space.

• $\{ \{ x \} \mid x \in X \}$ is a countable basis.
• $X$ is Hausdorff since any $x \ne y$ can be separated by $\{ x \}$ and $\{ y \}$.
• For every point $x \in X$, $\{ x \}$ is a neighborhood of $x$ that is locally homeomorphic to $\mathbb{R}^0$.

Thus $X$ is a $0$-manifold.