Why Read This Book? - Index of

168 Chapter 5 The Real Numbers
Although ɛ is not generally thought to be of any particular size, it is usually present
in theorems and proofs because smaller values of ɛ represent the primary obstacle
to overcome in concocting the proof. Theorem 5.1.5 is a natural way to visualize
the LUB property in terms of how intervals to the left and right of L intersect the
set A. Every interval of the form (L, L + ɛ), no matter how large ɛ is, must not
contain any elements of A. Also, every interval of the form (L − ɛ, L], no matter
how small ɛ is, must contain some element of A.
Here is a suggestion about how to tackle the proof of Theorem 5.1.5. Given a
set A and a real number L, the theorem says (L1 ∧ L2) ↔ (M1 ∧ M2). For the →
direction, Exercises 1.2.18(f) and (i) imply that
(L1 ∧ L2) → (M1 ∧ M2) ⇔[(L1 ∧ L2) → M1]∧[(L1 ∧ L2) → M2]
⇔[(L2 ∧¬M1) →¬L1]∧[(L1 ∧¬M2) →¬L2]
(5.5)
The statement ¬M1 →¬L1 is at least as strong as (L2 ∧¬M1) →¬L1. So to
prove the → direction of Theorem 5.1.5, you will want to show L1 → M1 by
contrapositive and then show (L1 ∧ L2) → M2 by showing (L1 ∧¬M2) →¬L2.
Then you can prove ← by showing ¬L1 →¬M1 and ¬L2 →¬M2.
EXERCISE 5.1.6 Prove Theorem 5.1.5.
Example 5.1.7 Show that the set {1} has LUB 1 by appealing to properties
M1–M2.
Solution Pick ɛ>0. Because (1, 1 + ɛ) ∩{1} is empty, property M1 holds. Also,
because 1 ∈ (1 − ɛ, 1]∩{1}, property M2 holds. Thus 1 is the LUB of {1}. �
EXERCISE 5.1.8 Show that b is the LUB of the interval (a, b) by appealing to
properties M1–M2. (An identical argument will work for [a, b].)
5.1.2 Greatest Lower Bounds
Now let’s turn the LUB property upside down and discuss bounds from below.
We do not have to make any new assumptions concerning the existence of the
greatest lower bound of a set, for we can derive results from the LUB property
by flipping a set upside down, so to speak.
Definition 5.1.9 A real number G is said to be a greatest lower bound (GLB)
for a set A if G has the following properties:
(G1) For every a ∈ A, we have that a ≥ G, and
(G2) If N is any lower bound for A, it must be that N ≤ G.