Page 207 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Ordering Relations 193
than 7, so ¬∃x ∈ B(x ≤ 7 ∧ x = 7). There are no other smallest or minimal
elements. Note that 7 is not a smallest or minimal element of C, since 7 /∈ C.
According to Definition 4.4.4, a smallest or minimal element of a set must
actually be an element of the set. In fact, C has no smallest or minimal
elements.
2. First of all, note that 6 and 9 are not minimal because both are divisible by
3, and 8 is not minimal because it is divisible by 4. All the other elements
of B are minimal elements, but none is a smallest element.
3. The set {2, 3} is a smallest element of F, since 2 and 3 are elements of every
set in F, and therefore ∀X ∈ F({2, 3}⊆ X). It is also a minimal element,
since no other element of F is a subset of it, and there are no other smallest
or minimal elements. The get G has two minimal elements, {2} and {3}.
Every other set in G must contain one of these two as a subset, so no other
set can be minimal. Neither set is smallest, since neither is a subset of the
other.

We are now ready to answer some of the questions we raised before
Definition 4.4.4.

Theorem 4.4.6. Suppose R is a partial order on a set A, and B ⊆ A.
1. If B has a smallest element, then this smallest element is unique. Thus, we
can speak of the smallest element of B rather than a smallest element.
2. Suppose b is the smallest element of B. Then b is also a minimal element
of B, and it is the only minimal element.
3. If R is a total order and b is a minimal element of B, then b is the smallest
element of B.

Scratch work

These proofs are somewhat harder than earlier ones in this chapter, so we do
some scratch work before the proofs.
1. Of course, we start by assuming that B has a smallest element, and because
this is an existential statement, we immediately introduce a name, say b, for
a smallest element of B. We must prove that b is the only smallest element.
As we saw in Section 3.6, this can be written ∀c(c is a smallest element of
B → b = c), so our next step should be to let c be arbitrary, assume it is also
a smallest element, and prove b = c.
At this point, we don’t know much about b and c. We know they’re both
elements of B, but we don’t even know what kinds of objects are in B – whether
they’re numbers, or sets, or some other type of object – so this doesn’t help us