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


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 $P(x)$.

  • $\begingroup$ When you say appropriate formula do you mean the same formula that I have written above $\endgroup$ – arrhhh Mar 9 '17 at 5:51
  • $\begingroup$ Not exactly: $P(x)$ is $\forall B(B\in S\rightarrow x\in B)$. $\endgroup$ – Renan Maneli Mezabarba Mar 9 '17 at 10:18
  • 1
    $\begingroup$ 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. $\endgroup$ – arrhhh Mar 9 '17 at 10:33

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