For two points $x,y$ either $x,y \in U$ or $x \in U_{x}$ and $y\in U_{y}$ and $U_{x} \cap U_{y} = \varnothing$ implies Hausdorff Space

Question

Let $U_{i},U_{j},U_{k}$ be some chart areas in a locally euclidean space $(X,\tau)$, so that for two points $x,y$ it either follows $x,y \in U_{i}$ or $x \in U_{j}$ and $y \in U_{k}$ where $U_{j} \cap U_{k}=\varnothing$

1. Show that $X$ is Hausdorff.
2. Let $X = \bigcup\limits_{i \in I}U_{i}$ where $U_{i}$ are chart areas and $I$ is countable. Show that there exists a countable basis of the topology.

My ideas:

1. In the case where $x \in U_{j}$ and $y \in U_{k}$ where $U_{j} \cap U_{k}=\varnothing$ it is clear the Hausdorff property is satisfied. But when $x,y \in U_{i}$ I am not sure. By assumption, for all other chart areas $U_{j}$ and $U_{k}$ of $x$ and $y$ where $x \notin U_{k}$ and $y \notin U_{j}$ respectively, it would follow $U_{j} \cap U_{k}\neq\varnothing$. I do not see how I can construct disjoint neighborhoods of $x,y$ from this.

2. Let $X = \bigcup\limits_{i \in I}U_{i}$ where $U_{i}$ are chart areas, i.e. $M$ is a countable union of chart areas.
   I assume that $\tau_{B}:=\{U_{i}: i \in I\}$ be an appropriate basis for for $\tau$.
   Let $V \in \tau$ arbitrary, then for all $ p \in V$ there exists $i_{p} \in I$ so that $p \in U_{i_{p}}$ since $X = \bigcup \limits_{i \in I}U_{i}$

But how can I show that $V = \bigcup\limits_{j \in J}U_{j}$ where $ J \subseteq I$

Answer 1

The chart map is a homeomorphism between $U$ and a metric space (in fact a very specific one). So $U$ is Hausdorff (as a subspace, because all metric spaces are) and so the disjoint open subsets we get from that are still open in $X$ (open in open is open), and as required.

Every $U_i$ in the chart has itself a countable base $\mathcal{B}(U_i)$, say, (because it is a Euclidean subspace) and then the union $\mathcal{B}=\{\mathcal{B}(U_i): i \in I\}$ has size $|I| \cdot \aleph_0=|I|$, and $\mathcal{B}$ is then a base for $X$: all members are open in $X$ (open in open is open again) and each open set $O$ in $X$ can be written as

$$O=\bigcup_{i \in I} (O \cap U_i)$$

and each $O \cap U_i$ is a union of sets from $\mathcal{B}(U_i) \subseteq \mathcal{B}$ and hence so is $O$. So a countable chart gives a countable base this way.