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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?

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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.

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  • $\begingroup$ "inner formula" refers to ∃y∀z? Could you explain how that's read? $\endgroup$
    – LuxuryMode
    Dec 5, 2012 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. $\endgroup$
    – André Nicolas
    Dec 5, 2012 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, 2012 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. $\endgroup$
    – André Nicolas
    Dec 5, 2012 at 19:39
  • $\begingroup$ We are then y by identifying the elements of y? $\endgroup$
    – LuxuryMode
    Dec 5, 2012 at 19:48

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