Cardinality and Ordinal Numbers

The cardinality |A| of a finite set A is simply the number of elements in it. When it comes to infinite sets, we no longer can speak of the number of elements in such a set. We can, however, try to match up the elements of two infinite sets A and B one by one. If this is possible, i.e. if there is a bijective function h : A → B, we say that A and B are of the same cardinality and denote this fact by |A| = |B|. If two (finite or infinite) sets A and B are not of the same cardinality, we can try to compare which one of the two sets has at least as many elements as the other. There are basically two ways of doing that: if we can pair up every element a of the set A with one and only one element b of the set B so that no two elements of B are paired with the same element of A (i.e. if there is an injective function f : A → B), then B must have at least as many elements as A. Alternatively, one could detect this by exhibiting a surjective function g : B → A, because that would mean that there are enough elements in B to cover all elements in A. It is a consequence of the following two theorems that these two notions of “has at least as many elements as” agree.

Theorem. For a function f : A → B, the following four statements are equivalent:

(a) f : A → B is injective.

(b) f(S ∩ T ) = f(S) ∩ f(T ) for all S,T ⊆ A.

(d) There is a function g : B → A such that g ◦ f = idA.

Remark. Recall that the identity function idA is deﬁned by idA(a) = a for all a ∈ A.

PROOF. “(a)⇒(b)” Exercise. “(b)⇒(c)” Suppose f(S ∩ T ) = f(S) ∩ f(T ) for all S,T ⊆ A. Let S ⊆ A. We wish to sow that f −1(f(S)) = S. First we show that S ⊆ f −1(f(S)): let s ∈ S, then f(s) ∈ f(S). So, by definition of preimage, s ∈ f −1(f(S)). Conversely, let a ∈ f −1(f(S)). This means that f(a) ∈ f(S). Hence, f(a) ∈ f(S) ∩ f({a}) = f(S ∩ {a}), where we use T = {a}. In particular, S ∩ {a}= 6 ∅. Consequently, a ∈ S. “(c)⇒(d)” Suppose f −1(f(S)) = S for all S ⊆ A. Assuming, as we may, that A is not empty, fix any a0 ∈ A. We now define a function g : B → A as a subset of B × A as follows

g = {(b, a) ∈ B × A | (b 6∈ f(A) ∧ a = a0) ∨ (b ∈ f(A) ∧ f(a) = b)}.

Let us verify that g is indeed a function, i.e., that it passes the vertical line test: Suppose (b, a1), (b, a2) ∈ g. If b 6∈ f(A), then a1 = a2 = a0 and all is well. If −1 b ∈ f(A), then f(a1) = f(a2) = b, so that a1 ∈ f (f({a2})) = {a2}, where we use our assumption with S = {a2}. So, again, a1 = a2. Therefore, g is a function. Finally, in order to check if g ◦ f = idA, we let a ∈ A and put b = f(a). Then (g ◦ f)(a) = g(f(a)) = g(b) = a by deﬁnition of g. “(d)⇒(a)” Exercise.

Theorem. For a function g : B → A, the following three statements are equivalent:

(a) g : B → A is surjective.

(b) g(g−1(S)) = S for all S ⊆ A.

Remark. If you expected a forth statement here, recall that g(S ∪ T ) = g(S) ∪ g(T ) is always true for all S,T ⊆ B, whether g is surjective or not.

PROOF. “(a)⇒(b)” Suppose that g : B → A is surjective. Let S ⊆ A be any subset. We wish to show that g(g−1(S)) = S. We start with proving that g(g−1(S)) ⊆ S: let a ∈ g(g−1(S)), then a = g(b) for some b ∈ g−1(S). So, a = g(b) ∈ S. (Notice that this is always true.) Conversely, let s ∈ S. Since g : B → A is surjective and s ∈ A, then s = g(b) for some b ∈ B. Hence, b ∈ g−1(S), and consequently s = g(b) ∈ g(g−1(S)). This proves that S ⊆ g(g−1(S)). “(b)⇒(c)” Let us now assume that statement (b) holds. We construct the desired function f as follows: for every a ∈ A we can form the set

Ga = {(a, b) ∈ A × B | a = g(b)}

of all pairs whose second coordinate b maps to its ﬁrst coordinate a under g. Since, by assumption, g(g−1({a})) = {a} for all a ∈ A, the set g−1({a}) is not empty, guaranteeing the existence of at least one element b ∈ B with g(b) = a. Consequently,

none of the sets Ga is empty. Moreover, if a1 6= a2 then Ga1 ∩ Ga2 = ∅, as can be seen by looking at the ﬁrst coordinates. We are now in a position to apply the axiom

of choice: let f be a set which contains exactly one element from each of the sets Ga with a ∈ A. Then f ⊆ A × B. In fact, since by the very choice of f no two elements of f have the same ﬁrst coordinate, f : A → B is a function. Also, g(f(a)) = a for

all a ∈ A by deﬁnition of the sets Ga.

“(c)⇒(a)” Suppose there is a function f : A → B such that g ◦ f = idA. This allows us to show that g : B → A is surjective. For if a ∈ A, we can deﬁne b = f(a) ∈ B. Then g(b) = g(f(a)) = a, proving that g is surjective. Corollary. For any two sets A and B, the following two statements are equivalent:

(i) There is an injective function f : A → B.

(ii) There is a surjective function g : B → A.

PROOF. The above theorems imply that being injective is equivalent with having a “left inverse” and being surjective is equivalent with having a “right inverse”. To prove the corollary one only has to observe that a function with a “right inverse” is the “left inverse” of that function and vice versa.

This, ﬁnally, allows us to make the following

Deﬁnition. Let A and B be two sets. If there is an injective function f : A → B, or, equivalently, if there is a surjective function g : B → A, then we write |A| 6 |B|. We will use the notation |A| < |B| to mean (|A| 6 |B| ∧ |A|= 6 |B|). If |A| < |N|, then we call A ﬁnite. If |A| 6 |N|, then we call A countable. A set, which is not countable is called uncountable.

Example. |A| < |P(A)| for every set A.

PROOF. Since f : A → P(A) given by f(a) = {a} is clearly an injective function, we have |A| 6 |P(A)|. So, we only have to show that |A| 6= |P(A)|. We do this by way of contradiction. Suppose that there is a surjective function g : A → P(A). Then g(a) is a subset of A for every a ∈ A. Consider the set B = {a ∈ A | a 6∈ g(a)}. Since B ⊆ A, i.e. B ∈ P(A), and g is surjective, we have B = g(b) for some b ∈ A. Now, either b ∈ B or b 6∈ B. We will show that neither is possible, thus establishing the desired contradiction. If b ∈ B, then (by the very deﬁnition of the set B) b 6∈ g(b) = B; but this is impossible! On the other hand, if b 6∈ B, then (again by the deﬁnition of B) b ∈ g(b) = B; another impossibility.

Remark (Continuum hypothesis). The following statement is known as the continuum hypothesis: “there is a set B such that |N| < |B| < |R|.” Recall that |R| = |P(N)|. Accordingly, there is a generalized continuum hypothesis: “given an inﬁnite set A, there is a set B such that |A| < |B| < |P(A)|.” It has been shown that neither of these statements can be proved or disproved using the standard axioms of set theory!

Theorem (Schröder-Bernstein). If |A| 6 |B| and |B| 6 |A|, then |A| = |B|. PROOF. Suppose there are two injective functions f : A → B and g : B → A. Then their composition h : A → A given by h = g ◦ f is also injective. (Verify that!) Consider the set C = A \ g(B). We claim that none of the sets h(C), h(h(C)), h(h(h(C))), ··· (which are all subsets of A) shares any elements with the set C. To see this, notice that each one of the sets h(C), h(h(C)), h(h(h(C))), ··· is a subset of h(A) = g(f(A)) which, in turn, is a subset of g(B) = A \ C. Put D = C ∪h(C)∪h(h(C))∪· · ·. Then h(D) = h(C)∪h(h(C))∪h(h(h(C)))∪· · · . Therefore, due to the above claim, h(D) = D \ C . Since h is injective, the function k : D → h(D) defined by k(x) = h(x) is bijective. Hence, defining k(x) = x for x ∈ A\D, extends k to a bijective function k : A → A\C. (Recall that C ⊆ D ⊆ A.) Consequently, |A| = |A\C|. We also have |B| = |A\C|, because g : B → g(B) = A\C is a bijection. Thus, |A| = |B|.

Deﬁnition. A relation 4 on a set A is called a partial order if for all a, b, c ∈ A the following are true

(i) a 4 a (Reﬂexivity);

(ii) (a 4 b ∧ b 4 c) → a 4 c (Transitivity);

(iii) (a 4 b ∧ b 4 a) → a = b (Antisymmetry).

We will often write a ≺ x instead of (a 4 x ∧ a 6= x).

Deﬁnition. A partial order 4 on A is called a linear order if for every two elements a, b ∈ A either a 4 b or b 4 a (Comparability).

Deﬁnition. Two linearly ordered sets A and B, with order relations 4A and 4B, respectively, are said to have the same order type, if there is a bijective function h : A → B such that x 4A y if and only if h(x) 4B h(y). (Note that this is equivalent to requiring that h : A → B be surjective and that h(x) ≺B h(y) whenever x ≺A y.)

Remark. Diﬀerent linear orders on the same set may result in diﬀerent order types!

Deﬁnition. A linearly ordered set A is called well-ordered if every nonempty subset S of A has a smallest element s0, that is, if there is an s0 ∈ S such that s0 4 s for all s ∈ S. The order type of a well-ordered set is called its ordinal number.

One advantage of having a well-order on a set is that it allows us to perform an induction:

Theorem (Transﬁnite induction). Let A be a well-ordered set and let P (x) be some predicate. Suppose that P (a) is true whenever P (x) is true for all x ≺ a. Then P (a) is true for all a ∈ A.

PROOF. Exercise!

Deﬁnition. For an element x of a well-ordered set A we deﬁne the set Sx(A) by

Sx(A) = {a ∈ A | a 4 x and a 6= x} and call it the section of A by x. We state the following theorem without proof.

Theorem (Trichotomy of ordinal numbers). Suppose A and B are two well-ordered sets. Then exactly one of the following three statements is true:

(i) A and B have the same order type; or

(ii) A has the order type of a section of B; or

(iii) B has the order type of a section of A.

Remark. It is customary to denote the statements of the above theorem as (i) ||A|| = ||B||; (ii) ||A|| < ||B||; (iii) ||B|| < ||A||. Caution: it is possible to have ||A|| < ||B|| while |A| = |B|. Can you think of an example?

Corollary (Comparibility of cardinalilties). For any two sets A and B either |A| 6 |B| or |B| 6 |A|.

PROOF. By the well-ordering theorem, there are linear orders 4A and 4B on A and B, respectively, which make A and B well-ordered. If A and B have the same order type, then there is an order preserving bijective function h : A → B, so that

A and B must have the same cardinality. If A has the order type of a section Sb(B) of B, then there is an order preserving bijection f : A → Sb(B). Since this function f : A → B is injective, we see that |A| 6 |B|. The corollary now follows from the preceding theorem.