What Is the Intersection of the Empty Family of Subsets of \(\mathbb{R}\)?

[doktordave]

I am beginning to study set theory and came across the following example:

Let $$\mathcal{A}$$ be the empty family of subsets of $$\mathbb{R}$$. Since $$\mathcal{A}$$ is empty, every member of $$\mathcal{A}$$ contains all real numbers. That is, $$((\forall A)(A\in\mathcal{A}\Rightarrow x\in A))$$ is true for all real numbers x. Thus $$\bigcap_{A\in\mathcal{A}} A = \mathbb{R}$$.

My problem is with the first sentence. Since a family is simply a set of sets, If we talk about an empty family wouldn't this simply be the empty set $$\emptyset$$? And since the empty set is defined not to contain anything, how could it contain any subsets of the set of real numbers?

 

[slider142]

It does not contain anything. The second sentence is vacuously true.

 

[Preno]

Actually, the intersection of the empty set is V, the class of all sets.

 

[HallsofIvy]

I am beginning to study set theory and came across the following example:

Let $$\mathcal{A}$$ be the empty family of subsets of $$\mathbb{R}$$. Since $$\mathcal{A}$$ is empty, every member of $$\mathcal{A}$$ contains all real numbers. That is, $$((\forall A)(A\in\mathcal{A}\Rightarrow x\in A))$$ is true for all real numbers x. Thus $$\bigcap_{A\in\mathcal{A}} A = \mathbb{R}$$.

My problem is with the first sentence. Since a family is simply a set of sets, If we talk about an empty family wouldn't this simply be the empty set $$\emptyset$$?

Yes, that's true. "every member of $$\mathcal{A}$$ contains all real numbers" is the same as "if U is a member of $$\mathcal{A}$$ then U contains all real numbers". The statement "if A then B" is true whenever A is false, irrespective of whether B is true or false (that is what slider142 means by "vacuously true"). Since "U is a member of $$\mathcal{a}$$ is always false, anything we say about U is true!

And since the empty set is defined not to contain anything, how could it contain any subsets of the set of real numbers?

It doesn't. That is not what the statement says!

 

[doktordave]

I think I understand now. Since the intersection over $$\mathcal{A}$$ is defined as $$\left\{x: (\forall A)(A\in \mathcal{A} \Rightarrow x\in A)\right\}$$ and the antecedent of the conditional is always false (there is nothing in $$\mathcal{A}$$), the conditional will always be true, because of the way the conditional operator is defined. So x can be anything in the universe. This seems a little backwards to my way of thinking, but I guess that's ok. I'll have to study that article on vacuous truth, it looks interesting. Thanks!

edit: Ah, thanks HallsofIvy. I was busy editing this post while you responded.