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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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    "inner formula" refers to ∃y∀z? Could you explain how that's read?2012-12-05
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    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.2012-12-05
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    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?2012-12-05
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    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.2012-12-05
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    We are then y by identifying the elements of y?2012-12-05
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    Yes, **if** you pretend it is a biconditional, which it isn't. To be precise, we are describing some $z$ that for sure are in $y$, without insisting that there are no other objects in $y$. The fact that there is a $y^\ast$ which has **exactly** these $z$ and no more is proved later. That's why I prefer the version of Union given Alex Karagila's answer.2012-12-05
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    Sorry, I was asking for clarification on the sentence. I think you were missing a word there. We are then WHAT y?2012-12-05
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    Oh, I see, a word or two got lost in editing. "We are then (prember pretend biconditional) *identifying* $y$ by $\dots$." Or describing $y$.2012-12-05
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    That's what I figured. Thanks so much!2012-12-05
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    So to me it's hard to read it straight from left to right. As I read it now, its "the set Y is the set for which any element z is an element of y if and only if there's some set w such that w is in x and z is in w"2012-12-05
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    Correct, if we assume $\implies$ is replaced by biconditional. What it actually says is that $y$ is **a** set such that (any $z$ for which there is some $w$ such that $z$ is in $w$) is in $y$.2012-12-05