Are we only able and always able to refer to something that exists?

For instance, given two elements $$a$$ and $$b$$, the pairing axiom first states the existence of $$\{a,b,\cdots\}$$, and then the axiom of specification constructs the two-element set $$\{a,b\}$$. Similarly, given a set of the form $$\{\{a,b\},\{c,d\}\}$$, the union axiom first states the existence of $$\{a,b,c,d,\cdots\}$$, and then the axiom of specification constructs the union $$\{a,b,c,d\}$$.

However, I am confused when considering the intersection of a family of sets. Given a non-empty family of sets $$z$$, there indeed exists some $$u\in z$$ , and then we can use the axiom of specification to construct the intersection$$$$\{x\in u:\forall v\in z\,(x\in v)\} \tag{1}$$$$ But when $$z$$ is empty, there is no element belonging to $$z$$, so does this expression (1) completely lose its meaning — we cannot refer to $$u$$ in it? Nevertheless, this expression itself seems unrelated to whether $$z$$ is empty or not. So, when using the axiom of specification, how do we convey the information of the set being decomposed? Is it simply sufficient to refer to it? And can we only refer to things that exist?

Another question is, we allow the family of sets to contain infinitely many sets. Even if it is proven that some set indeed exists in a non-empty family of sets, can we truly find it among countless possibilities? Or in other words, is it correct that we can reference it once its existence has been proven?

• It's sometimes helpful to "shut up and calculate". When $z$ is empty your expression (1) doesn't make sense, but the formula $x\in\bigcap z$ still makes sense. It is an abbreviation for $\forall v\in z(x\in v)$, which is in turn an abbreviation for $\forall v(v\in z\rightarrow x\in v)$. By first order logic, this formula is always true when $z$ is empty, so every $x$ belongs to $\bigcap z$. In other words, the intersection of an empty family is the whole universe. Commented Jul 23 at 14:50
• If $z$ is nonempty, then it turns out the collection of all $x$ such that $x\in\bigcap z$ forms a set, as shown by picking an arbitrary $u\in z$ and then using specification. There is no real difference between this and how you define the group operation on the quotient group $G/H$: given two cosets pick any $g_1$ from the first coset and $g_2$ from the second coset, and consider the coset of $g_1g_2$. This is perfectly fine even if $H$ (and thus each coset) is infinite. Commented Jul 23 at 14:57
• The expression $\{x\in u:\forall v\in z\,(x\in v)\}$ makes sense for any $u$ and $z$. But this expression only defines $\bigcap z$ if we know that $u\in z$. If $z$ is empty, there is no choice of $u$ such that the expression $\{x\in u:\forall v\in z\,(x\in v)\}$ defines $\bigcap z$. Commented Jul 23 at 15:05
• Usually for an expression to successfully refer, you need both existence and uniqueness of a referent. Note in the case of an expression like "$\{a,b\}$", you also need the axiom of extensionality to guarantee that there is at most one set consisting of $a$ and $b$. Whether a referent is the thing you had in mind when you wrote an expression is a different matter. Commented Jul 23 at 15:32
• A two-headed floating cat is telling me that the answer is "no." Commented Jul 23 at 15:39

The approach to the definition of the generalised intersection $$\bigcap z$$ of a family $$z$$ of sets given in your formula (1) presupposes that $$z$$ is not empty, as you remark, because it requires you to choose an element $$u$$ of $$z$$. Your formula (1), doesn't lose its meaning if $$z$$ is empty, it just depends on a value $$u$$ that does not exist. So you have to accept that the generalised intersection either (a) does not always exist or (b) exists subject to assumptions about the context:
(a): if you're working in some standard presentation of set theory such as ZF, then the expression $$\bigcap \{\}$$ does not have a value: its defining property would be $$x \in \bigcap \{\} \Leftrightarrow (\forall y \in \{\}. x \in y)$$, which is vacuously true, so $$\bigcap \{\}$$ would be a set that contains every set, which is not possible in ZF.
(b): it is quite common to admit the notation $$\bigcap \{\}$$ when we are working in a more specific context, such as analysis, where, by convention, sets are usually sets of real (or maybe complex) numbers, so we can conveniently take $$\bigcap \{\}$$ to mean $$\Bbb{R}$$ or $$\Bbb{C}$$.
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