How can we show that $S^0$ is a manifold?

Question

Recall $S^n = \{ (x^0, ..., x^n) \in \mathbb{R}^{n+1}: {x^0}^2 + ... + {x^n}^2 = 1 \}$

$S^0$ is a very cute set on $\mathbb{R}$ consisting of points $\{-1, 1\}$. How can we show that it satisfies the definition of a manifold?

Answer 1

That's rather trivial. Remember that $\mathbb R^0$ is a set consisting of exactly one element. It's topology is therefor $\tau_{\mathbb R^0} = \{\emptyset, \mathbb R^0\}$ and thus the only candidate for a codomain of a map is $\mathbb R^0$.
The topology on $S^0$ is $\tau_{S^0} = \{\emptyset, \{-1\}, \{1\}, S^0\}$. This leaves very little choice (exactly one) for an atlas of $S^0$ over $\mathbb R^0$. $$\mathcal A = ((\{-1\}, \phi_{-1}), (\{1\}, \phi_1))$$ Where $\phi_i$ is the constant map $\phi_i: \{i\} \to \mathbb R^0; ~~s\mapsto 0$ where $0\in\mathbb R^0$ is the only element in $\mathbb R^0$.
The $\phi_i$ are clearly diffeomorphic (with inverse $\phi_i^{-1}(0) = i$) since they have a one-element set as both domain and codomain. This proves that $S^0$ is a $0$-dimensional manifold.

Answer 2

Hint: is there some $n$ such that $S^0$ is locally homeomorphic to $\mathbb{R}^n$?