# Proving set identities: empty set case.

I am currently refreshing my knowledge in naive set theory, and would like to prove that for all sets $$A,B,C$$ we have $$A\cap(B \cup C) = (A \cap B) \cup (A \cap C).$$

I understand that this can be done by proving both $$A\cap(B \cup C) \subset (A \cap B) \cup (A \cap C) \ \text{and} \ (A \cap B) \cup (A \cap C) \subset A\cap(B \cup C)$$ hold true.

We can do this by letting $$x$$ be an arbitrary element of $$A\cap(B \cup C)$$ and showing that it is an element of $$(A \cap B) \cup (A \cap C)$$, and vice versa.

But what about when $$A \cap (B \cup C)$$ is the empty set? Then I would think we can't let $$x$$ be an arbitrary element of $$A \cap (B \cup C)$$ since there are none. However, I am aware that $$A \cap (B \cup C) \subset (A \cap B) \cup (A \cap C)$$ is trivially true in this case.

In the discrete mathematics course I took at my university, I did not see such cases be brought to attention. Should they be mentioned in proofs of such identities? Why / why not?

If so, some suggestions as to how to incorporate them into proofs would be helpful :-).

• Because some of the two sides being the empty set is no problem. Your claim is: for any element here, I prove it also is an element there. If there is no element at all then that is not your problem (mathematicalwise, of course): the proof still holds! Anyway, if you still feel something else must be done (though it really mustn't), then you can make a separate case: if this side is empty then so and so. and also the other side is, and the other way around, too. But you really don't need that, – DonAntonio Dec 22 '18 at 18:35
• Words to live by: "vacuous truth". I am almost tempted to say this is a duplicate of many other questions that were asked before. For example, math.stackexchange.com/questions/734418/… math.stackexchange.com/questions/2723860/… math.stackexchange.com/questions/1953218/… and there are many others which are similar. – Asaf Karagila Dec 23 '18 at 10:24

In order to show that for two sets $$X$$ and $$Y$$ it holds that $$X\subseteq Y$$, you have to prove that

for every $$x$$, if $$x\in X$$, then $$x\in Y$$.

Note that a statement of the form “if $$\mathscr{A}$$ then $$\mathscr{B}$$” is true when

either $$\mathscr{A}$$ is false or both $$\mathscr{A}$$ and $$\mathscr{B}$$ are true

If $$X$$ is the empty set, then “$$x\in X$$” is false for every $$x$$; hence “if $$x\in X$$ then $$x\in Y$$” is true.

The phrase “take an arbitrary element $$x\in X$$” is possibly misleading, but its intended meaning is “suppose $$x\in X$$”.

• Ah, so if we write "suppose $x \in X$" and successfully deduce that "$x \in Y$" , we have shown that "if $x \in X$, then $x \in Y$" is true. But, if "$x \in X$" is always false, we have still shown "if $x \in X$ then $x \in Y$" is true, since as you wrote, the implication would be true for all $x$ in that case. I.e. we are allowed to suppose things are true, even when they may never be. Is my understanding correct? – user445909 Dec 22 '18 at 19:39
• @E-mu Basically so – egreg Dec 22 '18 at 20:27

If $$D$$ is the empty set then the following statement is always true:$$\text{If }x\in D\text{ then }\cdots$$no matter what the dots are standing for.

It is false that $$x\in D$$ and "ex falso sequitur quodlibet". A false statement implies whatever you want.

So the assumption will not bring you into troubles, but - on the contrary - will give you freedom to accept whatever you want.

Since the empty set is a subset of any set, there is no need of including that in a formal proof. However it is a good idea to be aware of the fact that the equality holds even in the case of empty sets .