Notes for Chapters 10 - 11. Properties of finite sets.
Eccles does this stuff in Chapter 10 but defers some proofs to 11. I think these notes below are in
a more logical order.
Definition and discussion. A set X is said to be finite if there is a bijection f : Nm → X for
some natural number m. In this case, the set X is said to have cardinality m (or cardinal number m)
and we write this as |X| = m. The empty set is also considered to be finite with cardinal number zero.
The issue that arises here is whether the cardinal number of the set X is well-defined. That is,
could X have two different cardinal numbers? As we will see later, Corollary IA says that this cannot
happen.
Proposition I. (This is Lemma 10.1.4. Pages 127 and 133 of Eccles). If f : Nm → Nn is
an injection then m ≤ n.
Proof: Suppose that m > n. First note that the natural numbers 1 through n belong to Nm and
that f (1), ..., f (n) are distinct elements of Nn why? . Therefore, these exhaust the elements of Nn .
Thus f (m) must be equal to one of these n function values. But since m > n, this contradicts the
assumption that f is an injection why? . QED
Proposition II. If f : Nm → Nn is a surjection then n ≤ m.
←
−
←
−
Proof: Since f is a surjection, f ({1}), ..., f ({n}) are nonempty why? pairwise disjoint1 why?
subsets of Nm . Now, define a function g : Nn → Nm as follows: For each j ∈ Nn let g(j) be an element
←
−
←
−
of f ({j}) (actually we can be specific and pick g(j) to be the smallest element of f ({j})2 . The
function g is an injection why? and therefore by proposition I, n ≤ m. QED
Corollary I. If f : Nm → Nn is a bijection then n = m.
Proof: This follows immediately from the two propositions above.
Corollary IA. A finite set has a unique cardinal number.
Proof: Suppose X has two cardinal numbers n and m. Then there’s a bijection f : Nm → X and
a bijection g : Nn → X. Since g is a bijection it has an inverse function g −1 that is also a bijection.
Thus, g −1 ◦ f : Nm → Nn is a bijection why? and therefore n = m why? . QED
Exercise 35 The proof of Proposition IA below is left as an exercise to be turned in. The proof uses
Proposition I along with the idea of the proof of Corollary IA. It’s an example of how a theorem that
has been proved for Nn can usually be restated and proved for any set with cardinality n.
Proposition IA: If X and Y are finite sets, and if f : X → Y is an injection then |X| ≤ |Y |.
Proposition III. If S is a nonempty subset of Nn then S is finite and |S| ≤ n.
Proof: We do this by induction on n.
1
2
A collection of sets is said to be pairwise disjoint if no two have an element in common.
See the box at the end of this section of the notes.
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The “base case” case n = 1 is trivial because in this case, the set S must be all of N1 = {1} and
therefore the identity function is a bijection from N1 onto S, so m = 1.
Now assume the proposition is true for some particular n. We want to prove it for n + 1. That
is, we want to suppose that S is a nonempty subset of Nn+1 and show that there is a natural number
m ≤ n + 1 and a bijection f : Nm → S.
So, let S be a nonempty subset of Nn+1 .
If S does not contain n + 1 then S is a subset of Nn and the conclusion follows from the induction
hypothesis supply the details .
Therefore, suppose that S does contain n + 1. We apply the induction hypothesis to the set S ∩ Nn
to deduce that there exists a natural number m1 ≤ n and a bijection g : Nm1 → S ∩Nn . Let m = m1 +1
Define f : Nm → S as follows:
f (i) = g(i) if 1 ≤ i ≤ m1 and f (m1 + 1) = n + 1.
Then f is a bijection from Nm onto S supply the proof . Finally, note that since m1 ≤ n, m1 + 1 ≤
n + 1, so m ≤ n + 1. QED.
All the theorems, propositions, etc. in Eccles Chapters 10 & 11 follow fairly easily
from the above results.
REMARK: If you look at the proof of Proposition II, you’ll see that what we did was
to construct a function by selecting an element from each set in a disjoint collection. The
argument went something like this: Suppose ∀ j ∈ J, Sj is a non-empty subset of a set
X. Then there’s a function f : J → X such that f (j) ∈ Sj ∀j. I define this function by
saying that for each j, Sj is nonempty, so I can pick any element I choose out of Sj and
define f (j) to be this element. Does this seem reasonable? The function f referred to here
is sometimes called a “choice function” or a “selection.”
Exercise 36 Prove that a “least element” of a set is unique. That is, suppose there are two such
elements and prove they must be equal.
Exercise 37 Suppose |X| = n. Prove that |Y | = n if and only if there is a bijection of Y onto X. Use
only the properties of bijections. Do not use any of the propositions as reasons.
Exercise 38 Answer the why and supply parts above.
Exercise 39 Number 10.2 p. 132 of Eccles. You may use Theorem 10.2.1.
Exercise 40 Problems 5, 6, 11 pp. 183 - 184 of Eccles.
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