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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: $\bigcap \mathbb S = \mathbb U$. (http://www.proofwiki.org/wiki/Intersection_of_Empty_Set)

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...)

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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.

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  • $\begingroup$ That's what I thought. so proofwiki is wrong here, or they were thinking of other axiomatic systems. $\endgroup$ – Zeus Apr 2 '13 at 4:04
  • $\begingroup$ @Zeus: My guess is that it’s simply wrong. $\endgroup$ – Brian M. Scott Apr 2 '13 at 4:06
  • $\begingroup$ @BrianM.Scott I think the proofwiki is just saying it is equal to the universe, which is a contradiction in ZF and thus $\bigcap \varnothing$ is undefined. $\endgroup$ – gowrath Jan 13 '17 at 5:03
  • $\begingroup$ @gowrath: That is clearly not what it is saying. $\endgroup$ – Brian M. Scott Jan 13 '17 at 9:34
  • $\begingroup$ @BrianM.Scott It may not say the implication outright but it clearly says this is a paradoxical result. Unrestricted comprehension was axiomatically forbidden for this reason, and this is a sound way to demonstrate why that axiom is needed. I don't think the proof is wrong because it is disregarding the axiom it is meant to demonstrate the importance of. $\endgroup$ – gowrath Jan 13 '17 at 16:30
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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.

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  • $\begingroup$ Defining $\bigcap \emptyset = \emptyset$ creates other problems. You would no longer have the property $A \subseteq B \implies \bigcap B \subseteq \bigcap A$, nor would you have $\bigcap (A \cup B) = \bigcap A \cap \bigcap B$. $\endgroup$ – Mathemanic Nov 26 '16 at 8:10
  • $\begingroup$ Of course it all still works, only under the assumption that the sets are not empty. You can't have your cake and eat it too. $\endgroup$ – Asaf Karagila Nov 26 '16 at 15:26
  • $\begingroup$ @AsafKaragila The only place where I have seen $$\bigcap\mathcal A=\left\{x\in\bigcup\mathcal A\mid\forall A\in\mathcal A.x\in A\right\}$$ is in Isabelle where also "define" x/0 = 0 do you also find it acceptable? $\endgroup$ – Carlos Freites Jun 5 '18 at 8:26
  • $\begingroup$ @Carlos: I didn't mean to offend. Sorry. $\endgroup$ – Asaf Karagila Jun 5 '18 at 11:03
  • $\begingroup$ @AsafKaragila Don't worry. Thank you! :-) $\endgroup$ – Carlos Freites Jun 5 '18 at 11:35
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Consider a fixed universal set $E$. Consider an arbitrary collection of sets $C \subset P(E)$, where $P(E)$ is the power set of $E$. Then given the following definition:

$$\bigcap_{X \in C} X = \{x \in E : \forall_{X \in C} x \in X \} $$ then:

$$\bigcap_{X \in \emptyset } X = E$$.

Proof:

just substitute $C = \emptyset$ in the definition:

$$\bigcap_{X \in \emptyset} X = \{x \in E : \forall_{X \in \emptyset} x \in X \} = \{x \in E \} = E $$

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  • $\begingroup$ I'm not sure how rewriting the proof that OP posted contributes anything. $\endgroup$ – Alberto Takase Sep 2 at 12:24

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