Page 131 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Proofs Involving Quantifiers 117
particular, we should keep in mind that if we ever come across an element of F
while trying to figure out the proof, we can plug it in for A in the second given
and conclude that it contains x as an element. The first given, however, starts
with ∃A, so we should use it immediately. It says that there is some object that
is an element of F ∩ G. By existential instantiation, we can introduce a name,
say A 0 , for this object. Thus, we can treat A 0 ∈ F ∩ G as a given from now on.
Because we now have a name, A 0 , for a particular element of F ∩ G, it would
be redundant to continue to discuss the given statement ∃A(A ∈ F ∩ G), so we
will drop it from our list of givens. Since our new given A 0 ∈ F ∩ G means
A 0 ∈ F and A 0 ∈ G, we now have the following situation:

Givens Goal
A 0 ∈ F ∃A ∈ G(x ∈ A)
A 0 ∈ G
∀A ∈ F(x ∈ A)

If you’ve been paying close attention, you should know what the next step
should be. We decided before to keep our eyes open for any elements of F that
might come up during the proof, because we might want to plug them in for A
in the last given. An element of F has come up: A 0 ! Plugging A 0 in for A in
the last given, we can conclude that x ∈ A 0 . Any conclusions can be treated in
the future as givens, so you can add this statement to the givens column if you
like.
Remember that we decided to look at the givens because we didn’t know
what value to assign to A in the goal. What we need is a value for A that is in G
and that will make the statement x ∈ A come out true. Has this consideration
of the givens suggested a value to use for A?Yes! Use A = A 0 .
Although we translated the given statements x ∈∩F, x ∈∪G, and F ∩
G = ∅ into logical symbols in order to figure out how to use them in the
proof, these translations are not usually written out when the proof is written
up in final form. In the final proof we just write these statements in their original
form and leave it to the reader of the proof to work out their logical forms in
order to follow our reasoning.

Solution
Theorem. Suppose F and G are families of sets, and F ∩ G =∅. Then
∩F ⊆∪G.
Proof. Suppose x ∈∩F. Since F ∩ G =∅, we can let A 0 be an element of
F ∩ G. Thus, A 0 ∈ F and A 0 ∈ G. Since x ∈∩F and A 0 ∈ F, it follows that
x ∈ A 0 . But we also know that A 0 ∈ G, so we can conclude that x ∈∪G.