Maybe I'm just need to buff up on my logic notation, but I don't fully understand the following:

$$\exists y\forall z \left(\exists w(z\in w\wedge w\in x)\implies z\in y\right)$$

How should I unravel these statements generally? Starting with the innermost parens? As best I can tell the part starting with $\exists w$ means that there exists some $w$ such that $z$ is an element of $w$ and $w$ is an element $x$ which implies that $z$ is an element of $y$. But I dont understand how to parse $\exists y \forall z$ type statements (i.e. when they're up against each other like that). How do I even read that? "There's some element $y$ for all $z$'s"?

As you can tell, I'm generally confused. Can someone provide some guidance?


Read left to right. Remember that everything is a set. So $x$ is a set of (what else!) sets. For readability I would write $w\in x\land z\in w$ instead of the (equivalent) other way around.

The first thing to note is $\exists y$. It says there is a set with certain properties. In set theory, most of the time one describes the properties by specifying what the elements (of $y$ in this case) are. So next I see $z\in y$. This has something to do with describing the elements $z$ of $y$, though there will be a small complication.

Think of $x$ as a plastic bag full of plastic bags that contain stuff.

The part $\exists w(w\in x \land z\in w)$ says there is a plastic bag $w$ in $x$ such that $z$ is in that bag. Suppose magically the walls of the plastic bags in $x$ decay. Then all the $z$'s that were contained in any of these bags spill out. The union that the Axiom of Union produces is almost the combined contents of the plastic bags in $x$. But not quite.

The formula $(w\in x \land z\in w)\implies z\in y$actually only says that $y$ includes these contents. That's because by a separate axiom (usually called Separation, or Cut, or a consequence of Replacement) we can cut down $y$ to be precisely these contents.

It might have been better for clarity to say that $z\in y$ iff $\dots$. But it is considered unfashionable by some people to use a stronger-seeming axiom when a weaker one will do.

  • $\begingroup$ "inner formula" refers to ∃y∀z? Could you explain how that's read? $\endgroup$
    – LuxuryMode
    Dec 5 '12 at 19:08
  • $\begingroup$ Sorry, was talking to myself! By inner formula I meant the part without the quantifiers. For clarity, have replaced "inner formula" by the actual formula. $\endgroup$ Dec 5 '12 at 19:21
  • $\begingroup$ Thanks Andre. I'm almost there. You're saying that we're describing a set that contains other sets and showing that y includes the contents of all these sets. Is that correct? Also, the ∀z means what exactly? $\endgroup$
    – LuxuryMode
    Dec 5 '12 at 19:31
  • $\begingroup$ To make your life easier, pretend the $\implies$ is a biconditional. (It quite intentionally isn't, but trust me). We are then (remember pretend biconditional) $y$ by identifying the elements of $y$. So we are saying that for any $z$, $z\in y$ iff $z$ in an element of some element $w$ of $x$. Except for honesty, I should add that the actual formula only says that $y$ contains all such $z$, but might contain more stuff. Presumably less than a page later, or in an exercise, it is proved that there is a $y^\ast$ such that the binconditional holds. $\endgroup$ Dec 5 '12 at 19:39
  • $\begingroup$ We are then y by identifying the elements of y? $\endgroup$
    – LuxuryMode
    Dec 5 '12 at 19:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.