Prove that the topology induced by the following definition of an atlas is Hausdorff and second-countable

Question

So we start off with an arbitrary set $X$, and define a smooth atlas $\mathcal{A}$ as a set of pairs of the form $(U, \varphi)$, where $U$ is a subset of $X$ and $\varphi$ is a function bijective to a disk $D^{\circ}_{\varphi}$ of $\mathbb{R}^n$. For any choice of two charts $(U, \varphi)$ and $(U^{'}, \varphi^{'})$, $\varphi(U \cap U')$ and $\varphi'(U \cap U')$ are open (with the usual topology on $\mathbb{R}^n$) subsets of the disks $D^{\circ}_{\varphi}$ and $D^{\circ}_{\varphi'}$ respectively. Then we require that the transition maps (i.e. the functions $\varphi \circ \varphi'^{-1} : \varphi'( U \cap U') \to \varphi (U \cap U')$ and $\varphi' \circ \varphi^{-1}: \varphi (U \cap U') \to \varphi'( U \cap U')$) be smooth maps. Finally, the patches (i.e. the "$U$" sets) cover all of $X$. That is, for any $x \in X$, there exists a patch $U_{i}$ (where $i$ is an element of an indexing set $I$) such that $x \in U_i$.

Now, we define the open sets of $X$ as the subsets $V$ such that for all charts $(U, \varphi)$, $\varphi(V \cap U)$ maps to an open subset of $D^{\circ}_{\varphi}$ in $\mathbb{R}^n$.

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My attempt for Hausdorff space: Here, a Hausdorff space is defined as a topological space $(X, \tau)$ such that if $x_1, x_2 \in X$ and $x_1 \neq x_2$, then there exist open sets $\mathcal{O}_1$ and $\mathcal{O}_2 \in \tau$ such that $x_1 \in \mathcal{O_1}$, $x_2 \in \mathcal{O_2}$, and $\mathcal{O}_1 \cap \mathcal{O}_2 = \emptyset$.

First, assume that for such $x_1, x_2$, there exist patches $U_1$ and $U_2$ such that $x_1 \in U_1$, $x_2 \in U_2$, and $U_1 \cap U_2 = \emptyset$. Note that all patches are open sets, since by definition $\varphi (U \cap U')$ and $\varphi'(U \cap U')$ are open sets for any two $(U, \varphi)$, $(U', \varphi')$. So if such patches exist, then we are done.

If not, then let $U_1$ be any patch containing $x_1$ and $U_2$ be any patch containing $x_2$. Then $\varphi_1(U_1 \cap U_2)$ is a nonempty open set. Since the open disk in $\mathbb{R}^n$ is Hausdorff, there exist sets open in $D^{\circ}_{\varphi_1}$ $V$ and $W$ such that $\varphi_1(x_1) \in V$, $\varphi_1(x_2) \in W$, and $V \cap W = \emptyset$. Wishful thinking (and the fact that the exercise I'm referencing eventually wants us to prove that the functions are homeomorphisms) would lead us to believe that $\varphi_1^{-1} (V)$ and $\varphi_1^{-1} (W)$ are open sets. And that's where I've been stuck. I try to note that for them to be open then it must be the case that if $(U', \varphi')$ is an arbitrary chart, then $\varphi'(U \cap \varphi_1^{-1} (V))$ must be open. The fact that $\varphi'\circ \varphi^{-1}: \varphi( U_1 \cap U') \to \varphi'( U \cap U_1)$ is a smooth map and so is its inverse must be involved somehow, but I don't exactly see how to put the pieces together.

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My attempt for second-countability: We are also asked to prove that if the atlas $\mathcal{A}$ is countable, then the topology is second-countable. For the record, "second-countable" here is defined as having a countable basis, where a basis $\mathcal{B}$ is defined a subset of the topology such that for all $x \in X$, if $\mathcal{O}$ is an open set containing $x$, then there exists $B \in \mathcal{O}$ such that $x \in B \subseteq \mathcal{O}$. Here I ultimately run into a very similar problem.

We let $\mathcal{B}'$ be a countable basis for $\mathbb{R}^n$. Then, knowing that the charts are indexed under $\mathbb{N}$ (it works out very similarly if it's just a finite subset), we define $\mathcal{B}$ as $\{\varphi_i^{-1}(B): B \in \mathcal{B}', i \in \mathbb{N}\}$. This is a countable set. It remains to be proven that it's a basis. Note that $\varphi_i(\mathcal{O} \cap U_i) = \cup_{i \in I \subseteq \mathbb{N}} B_i$ Then, $\mathcal{O} \cap U_i = \varphi_i^{-1}(\cup_{i \in I \subseteq \mathbb{N}} B_i) = \bigcup_{i \in I \subseteq \mathbb{N}} \varphi_i^{-1}(B_i)$. Finally, since the $U_i$ cover $x$, we can take the union of all $O \cap U_i$ to get that $O$ is equal to a union of $\varphi_i^{-1}(B_i)$. The only thing that remains to be proven is that $\varphi_i^{-1}(B_i)$ is an open set, which is equivalent to proving that for any $j \in \mathbb{N}$, $\varphi_j(U_j \cap \varphi_i^{-1} (B_i))$ is an open set. And again, here I am stuck.

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Any help filling in the blanks (or pointing out errors) would be much appreciated. Thanks!

Answer 1

You cannot prove that $X$ is Hausdorff or second countable because it is not true under your assumptions.

1. Hausdorff: See Is a differentiable manifold Hausdorff?.

2. Second countable: Take an uncountable set $A$ and define $X = \mathbb R \times A$. As an atlas in your sense take all $p^J_a : J \times \{a\} \to J, p^J_a(x,a) = x$, where $J$ is any open interval and $a \in A$. The resulting topology on $X$ is the product topology on $\mathbb R \times A$ where $A$ carries the discrete topology. This is not second countable.