Confusion on the proof for finite intersections of open sets being open I have looked at the following source here (part (b)) for this proof.  But I have a hard time seeing that it is completely correct.
If we take $r = \min(r_1, r_2)$ as in the reference how can we be sure that $B_r(x) \subset A_1 \cap A_2$  What if the portion of $A_1$ or $A_2$ that we needed for this to be true was not in the intersection?  My confusion is that since the intersection is smaller than either of the two sets alone, the intersection doesn't necessarily inherit the necessary properties so easily.
 A: If $y$ is in $B_r(x)$, it is in $B_{r_1}(x)$ as $r\leq r_1$ and is therefore in $A_1$ by hypothesis, and it is also in $B_{r_2}(x)$ as $r\leq r_2$ and is therefore in $A_2$ by hypothesis. Being in $A_1$ and in $A_2$ it is also in $A_1 \cap A_2$ by definition of $A_1 \cap A_2$.
A: The point of the argument is that since $r \le r_1$, then
$$B_r(x) \subseteq B_{r_1}(x) \subseteq A_1,$$
and similarly for $r_2$ and $A_2$.
A: Since $r = \min(r_1,r_2)$ this means that $r \le r_1$ and $r \le r_2$. Thus $B_r(x) \subset B_{r_1}(x)$ and $B_r(x) \subset B_{r_2}(x)$. Thus $B_r(x) \subset B_{r_1}(x) \cap B_{r_2}(x) \subset A_1 \cap A_2$.
In other words, if $y \in B_r(x)$, then $y \in B_{r_1}(x)$ and so $y \in A_1$ by construction. But also if $y \in B_r(x)$ we have $y \in B_{r_2}(x)$ and so $y \in A_2$. Therefore, any $y \in B_r(x)$ is actually in the intersection of $A_1 \cap A_2$.
We have made the radius small enough that the ball actually lives inside of the intersection of $A_1$ and $A_2$.
A: To show that $A_1\cap A_2$ is open you want to show that for each $x\in A_1\cap A_2$ there exists $r>0$ (which may depend on $x$) such that $B_r(x)\subseteq A_1\cap A_2$. Let $x\in A_1\cap A_2$ be arbitrary. Then there exists $r_1>0$ such that $B_{r_1}(x)\subseteq A_1$ 8because $A_1$ is open and $x\in A_1$). Likewise, there exixts $r_2>0$ such that $B_{r_2}(x)\subseteq A_2$. With $r=\min\{r_1,r_2\}$ we achieve that $r>0$ and $B_r(x)\subseteq B_{r_1}(x)\subseteq A_1$ and $B_r(x)\subseteq B_{r_2}(x)\subseteq A_2$, i.e., $B_r(x)\subseteq A_1\cap A_2$ as desired.
(The argument in the pdf may be confusing as - reading its formulation carefully - literally only shows  that $A_1\cap A_2$ is either empty or a neighbourhood for at least one of its points $x$, and this confusion may even be amplified by the superfluous or even confusing special treatment of the disjoint case; compare with part (a) where the author correctly argues  that "$\forall x\in A\exists r>0\ldots$", whereas in (b) the argument is "$A=\emptyset\lor \exists x\in A\exists r>0\ldots$" and that is of course not a formally correct argument for openness).
