# Proof - Intersection S exists question from Jech's Set Theory [duplicate]

I have recently started studying Jech's Set Theory on my own and I am stuck on this question.

Prove that $\cap S$ exists for all $S \neq \phi$. Where is the assumption $S \neq \phi$ used in the proof.

I am new to this thing. I was thinking of saying something like

let there be two sets $A$ and $B$ so there exists a set $T$ such that $x \in A \cup B \mid P(x)$ where $P(x)$ is $x \in A \cap B$ by the axiom schema of comprehension. but this doesnt sound right

## 1 Answer

Notice that if $\bigcap \emptyset$ is a set, then $x\in \bigcap \emptyset$ for any $x$, i.e., $\bigcap\emptyset$ is the universe, which is not a set.

Now, if $S\ne\emptyset$, then there exists $A\in S$ and you may use the axiom of separation to define $\bigcap S=\{x\in A:P(x)\}$, for an appropriate formula $P(x)$.

• When you say appropriate formula do you mean the same formula that I have written above – arrhhh Mar 9 '17 at 5:51
• Not exactly: $P(x)$ is $\forall B(B\in S\rightarrow x\in B)$. – Renan Maneli Mezabarba Mar 9 '17 at 10:18
• Oh so this way x belongs to both A and B. Is this way more right because it conforms more closely to the axiom schema of comprehension while the way I did it originally starts off by already assigning x to the intersection which we must prove exists before we can use it. – arrhhh Mar 9 '17 at 10:33