why $\bigcap\mathscr A= A ?$ Taken from General Topology by stephen willard
My confusion given in red box

My attempt : here its given $\bigcap\mathscr A= A$
I think it should be  $\bigcap\mathscr A = \emptyset$
Note : Im not able to write  that letter in mathsjax
 A: The assertion $\bigcap \emptyset=\emptyset$ is unwarranted because it is quite clear that the condition $[\forall y\in\emptyset,\ x\in y]$ is satisfied for all objects $x$. Now, in these terms, the intersection of (all the elements of) the empty set should be the class of all sets. However, there is a valid argument for assuming implicitly that, when an ambient set $U$ is obvious from context, the notation $\bigcap \mathscr A$, for some $\mathscr A\subseteq \mathcal P(U)$, stands for $\{x\in U\,:\, \forall y\in\mathscr A, x\in y\}$. In that sense, $\bigcap \emptyset=U$.
A: For subsets of $A$ and $\mathcal{A}$ a collection of subsets of $A$ we define
$$\bigcap \mathcal{A}=\{x \in A\mid \forall C \in \mathcal{A}: x \in C\}$$
And when $\mathcal{A}=\emptyset$ the $\forall$ statement is always true (void truth) so the statement after the $\mid$ holds for any $x \in A$. So $\bigcap \emptyset = A$.
Note that this is an ugly "trick": if we did not specify the "universe" to be $A$ (but $B \supseteq A$ e.g.) this same intersection would have been $B$ instead. Or in naive set theory e would get "the set of all sets" which cannot exist (or co-exist with comprehension, rather) by Russell's paradox. It's a handy convention in topology, as it allows any collection of subsets to be a subbase for a topology (later in the book).
For now, I'd accept this reasoning and use it as a handy convention.
A: The following conversation concerns any $x\in A$.
"Sorry, sir. I have some $x\in A$ here and he wants to know whether he is an element of $\bigcap\mathscr A$."
"For that it must be checked whether there is some $U\in\mathscr A$ such that $x\notin U$. Such $U$ would prevent it."
"But there is no $U$ that satisfies $U\in\mathscr A$."
"Fine, then there is no veto, hence he is indeed a member of $\bigcap\mathscr A$."
