What's $\bigcap \emptyset$? http://www.proofwiki.org/wiki/Intersection_of_Empty_Set
Qeustion1:
First of all, my set-theory is ZFC. So i cannot invoke the universe in the proof in the link.
By axiom of Union, $\bigcap \emptyset$ is a set.
Suppose $\exists a\in \bigcap \emptyset$.
Then $\exists A\in \emptyset (a\in A)$, which is false.
Thus, $\neg \exists a\in \bigcap \emptyset$, which implies that it is an emptyset itself.
Where did i go wrong?
 A: In $\sf{ZFC}$, $\bigcap A$ is a notational convenience for the unique class $\mathbf{X}$ (an object in the meta-language) defined by $$\{ x : ( \forall X \in A ) ( x \in X ) \},$$ which is to say that $x \in \bigcap A$ is an abbreviation for the statement $( \forall X \in A ) ( x \in X )$.  One can show (using the Axiom Schema of Separation) that if $\mathbf{A}$ is a nonempty class of sets (possibly proper), then $\bigcap \mathbf{A}$ is also a set.
But what about $\bigcap \varnothing$?  Well, given any $x$ the statement $( \forall X \in \varnothing ) ( x \in X )$ is vacuously true, and so $x$ satisfies the condition of being in $\bigcap \varnothing$.  Which is to say that $\bigcap \varnothing$ is the universe of all sets; it is the (proper) class usually denoted $\mathbf{V}$, defined to be $$\{ x : x = x \}.$$  In particular, $\bigcap \varnothing$ is not a set.  (Not a set, that is, unless $\sf{ZFC}$ is inconsistent.  But if $\sf{ZFC}$ is consistent, then essentially by Russell's Paradox we can show that $\mathbf{V}$ is not a set.)

One particular troubling point in your proof that $\bigcap \varnothing = \varnothing$ is the following line of reasoning:

Suppose $\exists a\in \bigcap \emptyset$.
Then $\exists A\in \emptyset (a\in A)$....

Note that this is not part of the definition of the $\bigcap$ notation.
A: Why "By axiom of union, $\bigcap \emptyset$" is a set?
You only have that $\bigcup\emptyset$ is a set!
A: $x\in\bigcap\varnothing $ iff $x\in A(\forall A\in\varnothing)$. So every set is in $\bigcap\varnothing$.
