Exercise I2.5 13 from Apostol

Question

Just came up with a funny proof of this identity from Apostol calculus 1:

Exercise: prove that $A \cap \emptyset = \emptyset$

So I did it by contrapositive: suppose that there is some elements $x$ and $y$. Then,

$\big(x \not\in \emptyset \Rightarrow x \not\in A \cap \emptyset\big) \Leftrightarrow \big(y \in A \cap \emptyset \Rightarrow y \in \emptyset\big)$

This is equal to the proof that $A \cap \emptyset \subseteq \emptyset$. Then we prove that $\emptyset \subseteq A \cap \emptyset$, and conclude the equality proof.

But I have a feeling there is something wrong with this proof (or first part of it), since it somehow acknowledges that there might be an element in the empty set. Any ideas if it is valid/invalid and why?

Answer 1

I think it's okay, although your notation is very hard to read with no words. Personally I find reasoning about vacuous things kind of fun, so here's a more direct proof to give you a different flavor:

Since any element $x$ in $A\cap\varnothing$ must be in $\varnothing$, we see that $A\cap\varnothing \subset\varnothing$. But on the other hand, the empty set is a subset of any set, so $\varnothing \subset A\cap\varnothing$, so the sets must be equal.

Answer 2

Let's admit that

(1) the complement of the empty set is U ( the universal set)

(2) De Morgan's Law : Complement of ( X n Y) = Complement of X u Complement of Y ( and reciprocally)

(3) The union of any set X and of the universal set U is equal to the universal set , in symbols : X u U = U

(4) if two sets are equal, their complements are equal, and reciprocally

By (4) the proposition to be proved : A n ∅ = ∅ , is equivalent to :

complement of ( A n ∅) = Complement of ∅

Which is equivalent ( using (1) and (2)) to :

Complement-of-A u U = U

And this last proposition is an instance of the law (3).

Being equivalent to a law of the algebra of sets ( namely to law (3)) , the initial proposition is necessarily true.