# Definition of topological manifolds with dimension zero/locally euclidean of dimension zero

A topological $$n$$-manifold $$M$$ is locally Euclidean of dimension $$n$$ (each point of $$M$$ has a neighborhood that is homeomorphic to an open subset of $$\mathbb{R}^n)$$.

But what does locally Euclidean of dimension 0 mean?

• Discrete${}{}$? – Angina Seng Jun 3 '20 at 13:58
• @AnginaSeng What does that mean? – Filippo Jun 3 '20 at 14:04
• It means that "locally Euclidean of dimension zero" is equivalent to "discrete". – Lee Mosher Jun 3 '20 at 14:45
• @LeeMosher A topological space $M$ is discrete if (and only if) every subset of $M$ is open, right? – Filippo Jun 3 '20 at 15:18
• @Filippo right, all subsets are open. However manifolds are requested to be second countable also, hence uncountable discrete spaces are not allowed. – InsideOut Jun 3 '20 at 15:28

## 1 Answer

$$\Bbb R^0$$ is just a point, isn't it? Hence locally Euclidean of dimension zero means that locally homeomorphic to $$\Bbb R^0=\{pt\}$$. A $$0-$$topological manifold is then a countable set endowed with the discrete topology.

• Only countable family if you have having a countable base as a part of the definition of a manifold. The OP might well not assume that. – Henno Brandsma Jun 3 '20 at 15:31
• Maybe we are using different definitions of manifolds. I have always assumed manifolds to be second-countable and then the existence of a countable basis. However, if we drop the second-countability condition also uncountable union of points endowed with the discrete topology are allowed. – InsideOut Jun 3 '20 at 15:35
• I am currently reading John Lee's book about smooth manifolds and i just found the equation/definition $\mathbb{R}^0=\{0\}$ on page 25. Since $\{\{0\},\emptyset\}$ is the only existing topology on $\{0\}$, we can even make sense of the definition in my question for the case $n=0$ and we can prove what InsideOut said. – Filippo Jun 3 '20 at 19:52