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Currently I'm reading through Lee's Introduction to Topological Manifolds. Although the full proof is not given until much later, he gives the following theorem:

Theorem 2.55 (Invariance of Dimension) If $m\neq n$, a nonempty topological space cannot be both an $m$-manifold and an $n$-manifold.

Again, the full proof is not given until much later, but he provides the following proof for the zero-dimensional case:

Proof of the zero-dimensional case. Suppose $M$ is a $0$-manifold. Then $M$ is a discrete space, so every singleton in $M$ is an open subset. $\color{red}{\textrm{But in an $n$-manifold for $n>0$, every nonempty open subset contains a coordinate ball}}$, which is uncountable, so no singleton can be open.

The text in $\color{red}{\textrm{red}}$ is what I don't quite understand. Here's my shot at it: Let $V$ be a nonempty open subset, then $V$ is itself an $n$-manifold. So for every $p\in V$, there is an open set $U$ in $V$ containing $p$ that is homeomorphic to a open ball in $\mathbb{R}^n$, that is $U\subseteq V$ is a cordinate ball.

Rough_Manifolds
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1 Answers1

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Your shot is more or less on target, but here's some additional details.

Assume that $M$ is an $n$-manifold for $n > 0$, and that $V \subset M$ in an open subset of $M$. Choose $x \in V$, and then choose any coordinate chart $\phi : U \to \mathbb R^n$ for $M$ such that $x \in U$, so $U \subset M$ is open, and $\phi(U) \subset \mathbb R^n$ is open, and $\phi$ is a diffeomorphism from $U$ onto $\phi(U)$.

The set $V \cap U$ is open in $U$, and so the set $\phi(V \cap U)$ is open in $\phi(U)$ and in $\mathbb R^n$.

Choose any $r>0$ so that the open ball $B(\phi(x),r)$ is a subset of $\phi(V \cap U)$. The set $\phi^{-1}(B(\phi(x),r))$ is what is called a "coordinate ball" in $M$, it is a subset of $V$, and it is uncountable. Therefore $V$ is uncountable.

Lee Mosher
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