Set theory: how to interpret multiple quantifiers How does one interpret this ZFC Union axiom? I can't quite understand what is meant after "There exists some elements y for all elements z"? I'm also wondering if the x is a typo.


  $\exists y \forall z(\exists w(z\in w \space \land w \in x)\implies z\in y)$
The $x$ is not a typo, but there should be a $\forall x$ in front of the whole thing:

$$\forall x\,\exists y\,\forall z\,\big(\exists w\,(z\in w\land w\in x)\to z\in y\big)\;.\tag{1}$$

Given a set $x$, the expression asserts the existence of a set $y$ with the following property: for any object $z$, if $z\in w$ for some member $w$ of $x$, then $z\in y$. That is, $y$ contains as an element each object that is an element of some member of $x$.

We want $\bigcup x$ to be $\bigcup\{w:w\in x\}$, meaning that $z\in\bigcup x$ if and only if there is some $w\in x$ with $z\in w$. $(1)$ says that there is a set $y$ that contains everything that we want to have in $\bigcup x$. This $y$ may, however, contain other things as well, since $(1)$ has only a one-way implication ($\to$, not $\leftrightarrow$). However, once we have this $y$, we can use the axiom schema of comprehension (or separation) to cut it down to exactly $\bigcup x$:

$$\bigcup x=\{z\in y:\exists w\in x\,(z\in w)\}\;.$$
The reading is "There exists a $y$ such that, for all $z$, the following statement is true: 'If there is a $w$ such that $z$ is in $w$ and $w$ is in $x$, then $z$ is in $y$'".

I think that there is a missing $\forall x$ before the entire statement. If you include that, then you're saying that for any set $x$, there is a set $y$ that is the union of the sets that are elements of $x$. The breakdown is that you consider all possible $z$, and then you check - is it true that there's a $w$ such that $z$ is in $w$ and $w$ is in $x$? In other words, is it true that $z$ is an element of at least one set that is an element of $x$? If so, then $z$ is in the set $y$.