Mathematics for Computer Science

“mcs” — 2017/3/3 — 11:21 — page 34 — #42
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Chapter 2
The Well Ordering Principle
Definition 2.4.2. A lower bound (respectively, upper bound) for a set S of real
numbers is a number b such that b s (respectively, b s) for every s 2 S.
Note that a lower or upper bound of set S is not required to be in the set.
Corollary 2.4.3. Any set of integers with a lower bound is well ordered.
Proof. A set of integers with a lower bound b 2 R will also have the integer n D
bbc as a lower bound, where bbc, called the floor of b, is gotten by rounding down
b to the nearest integer. So Theorem 2.4.1 implies the set is well ordered.
Corollary 2.4.4. Any nonempty set of integers with an upper bound has a maximum
element.
Proof. Suppose a set S of integers has an upper bound b 2 R. Now multiply each
element of S by -1; let’s call this new set of elements S. Now, of course, b is a
lower bound of S. So S has a minimum element m by Corollary 2.4.3. But
then it’s easy to see that m is the maximum element of S.
2.4.1 A Different Well Ordered Set (Optional)
Another example of a well ordered set of numbers is the set F of fractions that can
be expressed in the form n=.n C 1/:
0
1 ; 1 2 ; 2 3 ; 3 4 ; : : : ; n
n C 1 ; : : : :
The minimum element of any nonempty subset of F is simply the one with the
minimum numerator when expressed in the form n=.n C 1/.
Now we can define a very different well ordered set by adding nonnegative integers
to numbers in F. That is, we take all the numbers of the form n C f where n is
a nonnegative integer and f is a number in F. Let’s call this set of numbers—you
guessed it—N C F. There is a simple recipe for finding the minimum number in
any nonempty subset of N C F, which explains why this set is well ordered:
Lemma 2.4.5. N C F is well ordered.
Proof. Given any nonempty subset S of N C F, look at all the nonnegative integers
n such that n C f is in S for some f 2 F. This is a nonempty set nonnegative
integers, so by the WOP, there is a minimum one; call it n s .
By definition of n s , there is some f 2 F such that n S C f is in the set S. So
the set all fractions f such that n S C f 2 S is a nonempty subset of F, and since
F is well ordered, this nonempty set contains a minimum element; call it f S . Now
it easy to verify that n S C f S is the minimum element of S (Problem 2.19).