Page 133 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Proofs Involving Quantifiers 119
Because proofs involving quantifiers may require more practice than the
other proofs we have discussed so far, we end this section with two more
examples.

Example 3.3.5. Suppose B is a set and F is a family of sets. Prove that if
∪F ⊆ B then F ⊆ P (B).
Scratch Work
We assume ∪F ⊆ B and try to prove F ⊆ P (B). Because this goal means
∀x(x ∈ F → x ∈ P (B)), we let x be arbitrary, assume x ∈ F, and set x ∈
P (B) as our goal. Recall that F is a family of sets, so since x ∈ F, x is a set.
Thus, we now have the following givens and goal:

Givens Goal
∪F ⊆ B x ∈ P (B)
x ∈ F
To figure out how to prove this goal, we must use the definition of power set.
The statement x ∈ P (B) means x ⊆ B, or in other words ∀y(y ∈ x → y ∈ B).
We must therefore introduce another arbitrary object into the proof. We let y
be arbitrary, assume y ∈ x, and try to prove y ∈ B.
Givens Goal
∪F ⊆ B y ∈ B
x ∈ F
y ∈ x
The goal can be analyzed no further, so we must look more closely at the
givens. Our goal is y ∈ B, and the only given that even mentions B is the first.
In fact, the first given would enable us to reach this goal, if only we knew that
y ∈∪F. This suggests that we might try treating y ∈∪F as our goal. If we can
reach this goal, then we can just add one more step, applying the first given,
and the proof will be done.
Givens Goal
∪F ⊆ B y ∈∪F
x ∈ F
y ∈ x
Once again, we have a goal whose logical form can be analyzed, so we use
the form of the goal to guide our strategy. The goal means ∃A ∈ F(y ∈ A), so
to prove it we must find a set A such that A ∈ F and y ∈ A. Looking at the
givens, we see that x is such a set, so the proof is done.