Let $X$ be a set.
In Serge Lang's "Differential and Riemannian manifolds" the following definition of smooth $\mathbb{R}^{n}$-atlas on $X$ is given.
Let $I$ be some index set. An atlas of class $C^{\infty}$ on $X$ is a collection of pairs $\{(U_{i},\phi_{i})\}_{i \in I}$, satisfying the following conditions:
Each $U_{i}$ is a subset of $X$, and the $U_{i}$ cover $X$.
Each $\phi_{i}$ is a bijection of $U_{i}$ onto an open subset $\phi_{i}U_{i} \subseteq \mathbb{R}^{n}$. And for each $i,j \in I$, the set $\phi_{i}(U_{i} \cap U_{j})$ is open in $\mathbb{R}^{n}$.
For all $i,j \in I$, the map \begin{equation} \phi_{j}\phi_{i}^{-1}: \phi_{i}(U_{i} \cap U_{j}) \rightarrow \phi_{j}(U_{i} \cap U_{j}) \end{equation} is a diffeomorphism.
It then follows that the collection $\{ \phi_{i}^{-1}(V) \mid i \in I, \, V \text{ open in } \phi_{i}U_{i} \}$ is a base for a topology on $X$. The maps $\phi_{i}$ are continuous and open with respect to this topology, and hence homeomorphisms.
Now I have two questions.
Is $X$ automatically Hausdorff? If not, what condition needs to be put on the atlas to ensure that $X$ is Hausdorff?
It seems to me that because $\mathbb{R}^{n}$ is Hausdorff it follows that $X$ is as well, but I'm not quite seeing how to prove this.
Under what conditions is $X$ second-countable.
It seems to me that it is sufficient that the index set $I$ is countable is a sufficient condition. By taking the sets $V$ in the definition above to be members of a countable base for the topology of $\phi_{i} U_{i} \subseteq \mathbb{R}^{n}$ in the definition above one obtains a countable base for $X$, (it is a countable union of countable sets).
Can this condition be weakened in a sensible way, or can it be shown that this condition is "necessary"? (In the sense that if $X$ is second-countable, one can always find a countable subset $I' \subset I$ such that $\{(U_{i}, \phi_{i}) \}_{i \in I'}$ is an atlas that gives the same manifold structure to $X$.)