Page 210 - HOW TO PROVE IT: A Structured Approach, Second Edition

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196 Relations
3. Suppose R is a total order and b is a minimal element of B. Let x be an
arbitrary element of B.If x = b, then since R is reflexive, bRx. Now suppose
x = b. Since R is a total order, we know that either xRb or bRx. But xRb
can’t be true, since by combining xRb with our assumption that x = b we
could conclude that b is not minimal, thereby contradicting our assumption
that it is minimal. Thus, bRx must be true. Since x was arbitrary, we can
conclude that ∀x ∈ B(bRx), so b is the smallest element of B.

When comparing subsets of some set A, mathematicians often use the
partial order S ={(X, Y) ∈ P (A) × P (A) | X ⊆ Y}, although this is not al-
ways made explicit. Recall that if F ⊆ P (A) and X ∈ F, then according to
Definition 4.4.4, X is the S-smallest element of F iff ∀Y ∈ F(X ⊆ Y). In other
words, to say that an element of F is the smallest element means that it is a
subset of every element of F. Similarly, mathematicians sometimes talk of a
set being the smallest one with a certain property. Generally this means that
the set has the property in question, and furthermore it is a subset of every set
that has the property. For example, we might describe our conclusion in part
3 of Example 4.4.5 by saying that {2, 3} is the smallest set X ⊆ N with the
property that 2 ∈ X and 3 ∈ X. We will see more examples of this idea in the
next section and in later chapters.

Example 4.4.7.
1. Find the smallest set of real numbers X such that 5 ∈ X and for all real
numbers x and y,if x ∈ X and x < y then y ∈ X.
2. Find the smallest set of real numbers X such that X = ∅ and for all real
numbers x and y,if x ∈ X and x < y then y ∈ X.

Solutions

1. Another way to phrase the question would be to say that we are looking
for the smallest element of the family of sets F ={X ⊆ R | 5 ∈ X and
∀x∀y((x ∈ X ∧ x < y) → y ∈ X)}, where it is understood that smallest
means smallest with respect to the subset partial order. Now for any set X ∈
F we know that 5 ∈ X, and we know that ∀x∀y((x ∈ X ∧ x < y) → y ∈
X). In particular, since 5 ∈ X we can say that ∀y(5 < y → y ∈ X). Thus,
if we let A ={y ∈ R | 5 ≤ y}, then we can conclude that ∀X ∈ F(A ⊆ X).
But it is easy to see that A ∈ F,so A is the smallest element of F.
2. We must find the smallest element of the family of sets F ={X ⊆ R | X =
∅ and ∀x∀y((x ∈ X ∧ x < y) → y ∈ X)}. The set A ={y ∈ R | 5 ≤ y}
from part 1 is an element of F, but it is not the smallest element, or even a