I have a question.
Suppose we have two arbitrary open interval on reals.
$A>x>B$
$C>x>D$
If we require $x$ to satisfy both intervals, is the following correct:
$x\in(B,A)\cap(D,C)$.
If so, is the following also an equivalent representation of $x$:
$min(A,C)>x>max(B,D).$
Yes to both questions: in the first instance the interval intersection you've used is just the rewriting of the inequalities you started off with. This is actually a nice point to see how good notation can be useful: the interval representation can be easily extended to more intervals, and can even be extended to countable or uncountable intersections (e.g. $\cap_{n=1}^{\infty}(A_n, B_n)$) with little extra writing, while the inequality version requires (essentially) an extra line of writing for each new inequality.
For the second question it's easy to see how it holds by considering what happens if it's not true (and proof by contradiction is commonplace in mathematics so it's worth doing): if $\min(A,C) \not> x$ then $x$ cannot lie in one of the intervals $(D,C)$ and $(B,A)$; likewise if $x \not> \max(D,B)$ then the same holds true. Thus $x\in(B,A)\cap(D,C)$ happens only if $\min(A,C) > x > \max(B,D)$.