intersection of the empty set and vacuous truth

Question

Let $\mathbb S = \varnothing$.

Then from the definition: $ \bigcap \mathbb S = \left\{{x: \forall X \in \mathbb S: x \in X}\right\}$

Consider any $x \in \mathbb U$.

Then as $\mathbb S = \varnothing$, it follows that: $\forall X \in \mathbb S: x \in X$ from the definition of vacuous truth.

It follows directly that: $\bigcap \mathbb S = \left\{{x: x \in \mathbb U}\right\}$

That is: $\bigcap \mathbb S = \mathbb U$.

Proofwiki uses the above "proof" to "prove" that intersection of the empty set is the whole universe.

My question is, is the use of vacuous truth really allowed in axiomatic set theory, like ZFC? I don't see how the use of vacuous truth is justified.

The next problem I can think of is that we cannot really "define" the elements of empty set (to my knowledge, there is no element in empty set) so how can we then prove as the above proof did? This seems to contradict the use of vacuous truth.

And of course, there is issue of using the whole universe as a set, and I don't think this is allowed.... (Maybe proof above is using a different axiomatic set theory, as I am using ZF-minded thoughts...)

Answer 1

There’s nothing wrong with the ‘vacuous truth’ part of the argument. It’s perfectly correct that if $X$ is any set, then $\left\{x\in X:x\in\bigcap\varnothing\right\}=X$. To see this, note that if $x\in X$, then $x\notin\bigcap\varnothing$ if and only if there is an $A\in\varnothing$ such that $x\notin A$, and since there is no $A\in\varnothing$ at all, this is not the case.

The problem with the argument is that nothing in $\mathsf{ZF}$ permits the formation of $\left\{x:x\in\bigcap\varnothing\right\}$: this an example of unrestricted comprehension, which is not permitted in $\mathsf{ZF}$. $\mathsf{ZF}$ permits only restricted comprehension, using a formula to pick elements from an already existing set, not from the universe at large.

Answer 2

Your confusion about how the intersection over a set can result in a proper class is justified.

In some places the definition of the intersection is bounded, so the result is always a set, i.e. $$\bigcap\mathcal A=\left\{x\in\bigcup\mathcal A\mid\forall A\in\mathcal A.x\in A\right\}$$

The philosophical justification is that the intersection over a set should result in a set, so we take only elements from $\bigcup\cal A$, which by the axiom of union is a set. The result is only different for the empty set, that is if $\cal A\neq\varnothing$ then we can easily forget about this bound, but when $\cal A=\varnothing$ we need to decide whether or not we do that.

This is the set theoretical equivalent of $0^0$ being indeterminate in analysis.

And as a side remark, vacuous argument reside in the logic, not in the axiomatic systems.

Answer 3

Let me suggest following point of view: suppose $(X_{\tau})_{\tau \in I}$ is family of sets and $I$ is its index set. Intersection of family is $\bigcap\limits_{\tau \in I}X_{\tau}=\left\lbrace x:\forall \tau(\tau \in I \Rightarrow x \in X_{\tau})\right\rbrace$. Usually this definition needs $I \ne \emptyset.$ But what if nevertheless we take $I = \emptyset$. Then implication in definition becomes true for any $x$. Really this is not set, but can make understandable meaning of "universal set".