Midterm 3 theorems

Theorem 26.11. Let A and B be sets, and let f : A → B be a function. We have (1) idB ◦f = f and (2) f ◦ idA =

(1) Note that idB ◦f : A → B and f : A → B have the same domain and codomain. Let a ∈ A. Notice that f(a) ∈ B. Then (idB ◦f)(a) = idB(f(a)) = f(a). Hence, we see that idB ◦f = f (2) We first show that f ◦ idA = f Both sides are functions from A to B. Therefore it suffices to show they have the same effect on any element of A. If a ∈ A then (f ◦ idA)(a) = f(idA(a)) = f(a). Thus f ◦ idA = f

Theorem 26.12. Let A, B, and C be sets, and let f : A → B and g : B → C be functions. (1) If g ◦ f is injective, then f is injective. (2) If g ◦ f is surjective, then g is surjective. (3) If f and g are injective, then g ◦ f is injective. (4) If f and g are surjective, then g ◦ f is surjective.

(1) Assume that g ◦ f : A → C is injective. To prove that f is injective, let a1, a2 ∈ A and assume f(a1) = f(a2); we will show a1 = a2. Applying g to both sides of the assumed equality yields g(f(a1)) = g(f(a2)), and so (g ◦ f)(a1) = (g ◦ f)(a2). (2) taking an arbitrary c∈C, concluding from surjectivity of g∘f that there is some a∈A such that (g∘f)(a)=c, and noting that f(a)∈B and g(f(a))=c. Since c was an arbitrary element of C then for all c∈C there exists b∈B such that g(b)=c (3)Let a,b∈A If (g∘f)(a)=(g∘f)(b), then f(g(a))=f(g(b)). Since g is one-to-one, we know that f(a)=f(b). And, since f is one-to-one is must be that a=b. Hence g∘f is one-to-one. (4) Assume that f and g are surjective. We wish to show that g ◦ f is surjective. Let c ∈ C. Since g : B → C is surjective, there is some b ∈ B such that g(b) = c. Since f : A → B is surjective, there is some a ∈ A such that f(a) = b. Then (g ◦ f)(a) = g(f(a)) = g(b) = c.

Theorem 27.2. Let A and B be sets, and let f : A → B be a function. If B is finite, then |im(f)| ≤ |B|. Moreover (still assuming B is finite) f is surjective if and only if |im(f)| = |B

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Corollary 29.7. The set Q is countably infinite.

Proof sketch. We have Q = Q>0 ∪ {0} ∪ Q<0. We know that Q>0 is countable by Theorem 29.5. Also, Q<0 is countable since there is an easy bijection Q>0 → Q<0. Also, {0} is finite, hence countable. By Theorem 29.1, the union of two countable sets is countable, so Q>0 ∪ {0} is countable. Applying Theorem 29.1 again, we find that Q = (Q>0 ∪ {0}) ∪ Q<0 is countable. It is also infinite, hence countably infinite.

Theorem 27.11 (The Pasting Together Theorem). Using the notation as given in Definition 27.9, each of the following holds: (1) If each fi is injective, then f is injective. (2) If each fi is surjective, then f is surjective. (3) If each fi is bijective, then f is bijective.

Proof. (1) Suppose that each fi is injective. We will show that f is injective, so let a1, a2 ∈ A and assume f(a1) = f(a2). We know that a1 ∈ Pi and a2 ∈ Pj for some i and j (since the P's partition A). Hence f(a1) = fi(a1) ∈ Qi, and f(a2) = fj (a2) ∈ Qj . Since f(a1) = f(a2) we have Qi ∩ Qj ̸= ∅. But the Q's partition B, hence Qi = Qj , or in other words i = j. Thus fi(a1) = f(a1) = f(a2) = fi(a2). Since fi is injective, a1 = a2. (2) Suppose that each fi is surjective. Let b ∈ B, so b ∈ Qi for some i. Since fi is surjective we can fix some a ∈ Pi with fi(a) = b. Hence, f(a) = fi(a) = b. Therefore, f is surjective. (3) This part follows from (1) and (2).

Theorem 27.1. Let A and B be sets, and let f : A → B be a function. If A is finite, then |im(f)| ≤ |A|. Moreover (still assuming A is finite) f is injective if and only if |im(f)| = |A|.

Proof. As a collection of ordered pairs, f has |A| elements. Hence, there are at most |A| second coordinates. Thus, there cannot be more than |A| elements in the image of f. W e now prove the last sentence. For the forward direction, assume that f is injective. Then all second coordinates of pairs in f are distinct, so the number of such second coordinates, which is |im(f)|, is equal to |f| = |A|. For the converse we work contrapositively. Assume that f is not injective. Then f has fewer than |A| distinct second coordinates (since at least two of the |A| second coordinates must be equal). Hence, |im(f)| < |A|.

Theorem 26.6. Let A, B, C, and D be sets, and let f : A → B, g : B → C, and h: C → D be three given functions. The two new functions h ◦ (g ◦ f): A → D and (h ◦ g) ◦ f : A → D are equal.

Proof. Both h ◦ (g ◦ f) and (h ◦ g) ◦ f have domain A and codomain D. Hence, to prove that they are equal, we need only check that for each a ∈ A both functions give the same image. Let a ∈ A. Then h ◦ (g ◦ f) (a) = h (g ◦ f)(a) = h(g(f(a)), and (h ◦ g) ◦ f (a) = (h ◦ g) f(a) = h(g(f(a))). Hence, we see that the two functions give the same image for a. Therefore, they are equal

Theorem 29.1. If A and B are countable sets, then A ∪ B is countable.

Proof. For the Finite set A we have a finite list of elements of a in A such that a1, a2,a3,...,an. Where n in N and n is the cardinality of the set A. For the countably infinite set B we have an infinite list of b in B. we can provide a natural number for each b in B. The elements of B create an infinite list such that b1, b2, b3 listing out both elements in a diagram we can cross A with B such that AXB = (a1,b1),(a2,b1),...,(an,b1),(a1,b2),(a2,b2)...,(an,b2)....(an,b2),(a1,b3) hence AXB is also countably infinite because can provide natural numbers for their cartesian product.

Theorem 28.14. Any infinite subset of a countably infinite set is countably infinite.

Proof. Let A be a countably infinite set. We can write the elements of A in an infinite list a1, a2, a3, . . .. Let B be an infinite subset of A. Let n1 be the smallest natural number with an1 ∈ B, which exists since B ̸= ∅ as B is infinite. Put b1 = an1 . Next let n2 be the smallest natural number with n2 > n1 and an2 ∈ B, which exists since B − {b1} ̸= ∅ as B is infinite. Put b2 = an2 . Repeating this process (by induction) we create an infinite list b1, b2, . . .. Clearly there are no repetitions in this list. This new list covers every element of B because we can also prove (by induction) that ni ≥ i for each i ∈ N; hence, we have worked all the way through the list of elements of A.

Theorem 31.5. If A is a set, then |A| < |P(A)|.

Proof. Let A be any set. First, we prove that |A| ≤ |P(A)|, so we need to find an injective function f : A → P(A). Define f by the rule f(a) = {a}. To prove that f is injective, let a1, a2 ∈ A and assume f(a1) = f(a2). Hence {a1} = {a2}. Therefore a1 = a2, since sets are equal exactly when they have the same elements. Next, let g : A → P(A) be an arbitrary function. We will show that g is never surjective, and hence there is no bijection between A and P(A). Define the (barber) set as B = {x ∈ A : x /∈ g(x)}. This is a subset of A, hence an element of P(A). We will show that B is not in the image of g. Let a ∈ A be arbitrary. There are two cases. Case 1: Assume a ∈ g(a). In this case a /∈ B, hence g(a) ̸= B. Case 2: Assume a /∈ g(a). Then a ∈ B, hence g(a) ̸= B. In every case B ̸= g(a). Since a ∈ A was arbitrary, this means B cannot be in the image of g (since it does not equal any element of the image of g).

Theorem 26.15. Let f : A → B be a function. Let g be the inverse relation to f. Then g is a function from B to A if and only if f is a bijection.

Proof. Let f : A → B be an arbitrary function, and let g = f −1 be the inverse relation. (⇐): Assume that f is a bijection. Then every element of B is the second coordinate of exactly one ordered pair in f. Hence, every element of B is the first coordinate of exactly one ordered pair in g. Therefore, g is a function. (⇒): Conversely, assume that g : B → A is a function. Hence, every element of B is the first coordinate of exactly one ordered pair in g. Thus, every element of B is the second coordinate of exactly one ordered pair in f. Therefore, f is a bijection.

Theorem 30.4. The set (0, 1) is uncountable.

Proof. Let f : N → (0, 1) be any function. We will show that f is not surjective. For each n ∈ N, write f(n) using a decimal expansion f(n) = 0.d1,nd2,nd3,n . . . (which doesn't end in repeating 9's). Fix x ∈ (0, 1) to be the number with decimal expansion x = 0.x1x2x3 . . . where xn = dig(dn,n). In other words, the nth digit of x is the digit change of the nth digit of f(n). Hence x ̸= f(n) for each n ∈ N. Therefore f is not surjective, as x is not in the image.

Theorem 29.5. The set Q>0 is countably infinite.

Proof. Put the elements of Q>0 into a diagram as below. (We put fractions which are not in lowest terms as light gray.) 1 1 1 2 1 3 1 4 2 1 2 2 2 3 2 4 3 1 3 2 3 3 3 4 4 1 4 2 4 3 4 4 . . . . . . . . . . . . . . . · · · · · · · · · · · · Now, we just list the elements as before, skipping over the elements in light gray, since they will be counted when they are in lowest terms. This counting procedure never repeats elements (since we skip those fractions not in lowest terms), and continues forever since Q>0 is infinite (since N is a subset; in other words, the top row of the diagram is infinite).

Theorem 27.18. Given a function f : A → B, the surjective reduction ˆf : A → im(f) is a surjective function.

Proof. Since ˆf consists of the same ordered pairs as f, it is a function. (Each element of the domain A is the first coordinate of exactly one ordered pair, and the second coordinate is an element of im(f).) For each b ∈ im(f), there is some a ∈ A such that f(a) = b (by the definition of the image). Then ˆf(a) = b, so ˆf is surjective.

Theorem 27.5. Let A and B be finite sets and assume |A| = |B|. A function f : A → B is injective if and only if it is surjective.

Proof. Suppose A and B are finite, |A| = |B|, and that f : A → B is a function. (⇒): Assume f is not surjective. Then |im(f)| < |B| = |A|, so f is not injective. (⇐): Assume f is not injective. Then |im(f)| < |A| = |B|, so f is not surjective.

Corollary 31.7. The set F(N) is uncountabl

Proof. The equality |P(N)| = |F(N)| follows from the previous theorem. We also know |N| < |P(N)|, hence F(N) is uncountable.

Theorem 25.30. Let A and B be sets, and f : A → B a function. Then f is surjective if and only if the range of f is equal to B.

Proof. The function f is surjective if and only if every b ∈ B is equal to f(a) for some a ∈ A. This happens if and only if B = {f(a) : a ∈ A} = im(f).

Theorem 26.20. Let f : A → B be a bijective function. The following each hold: (1) (f −1 ) −1 = f. (2) f −1 : B → A is a bijective function. (3) f −1 ◦ f : A → A is equal to idA. (4) f ◦ f −1 : B → B is equal to idB.

Proof. Throughout the proof we let f : A → B be a bijection. (1) This part holds for any relation f. If we reverse the ordered pairs in f, and then reverse again, we are back to the original relation. (2) By Theorem 26.15, we know f −1 is a function. Now, by part (1) above, its inverse relation is f, which is a function. Thus, by Theorem 26.15 again, we know f −1 is a bijection. (3) We note that f −1 ◦f and idA have the same domain and codomain. Now for any a ∈ A, we have (a, f(a)) ∈ f, so we see that (f(a), a) ∈ f −1 . This tells us that f −1 (f(a)) = a, proving that (f −1 ◦ f)(a) = a = idA(a). Hence, f −1 ◦ f = idA. (4) This is similar to part (3), by switching the roles of f and f −1 .

Proposition 30.6. The half open interval (0, 1] has the same cardinality as the open interval (0, 1), and hence it has continuum cardinality

Proof. We define a bijection f : (0, 1) → (0, 1], as follows. Fix S = 1 2 , 1 4 , 1 8 , . . . = 1 2 n : n ∈ N ⊊ (0, 1), and fix T = S ∪ {1} ⊊ (0, 1]. Now, define f as a piecewise function by the rule f(x) = ( x if x /∈ S 2x if x ∈ S. The first piece of this function is a bijection from (0, 1) − S to (0, 1] − T, and the second piece is a bijection from S to T. By pasting together, f is a bijection from (0, 1) to (0, 1]

Theorem 28.8. The relation of "having the same cardinality" as given in Definition 28.1 is an equivalence relation on the collection of sets.

Proof. We first prove this relation is reflexive. Let X be any set. The identity function idX : X → X is a bijection. Thus X is related to X. Next, we prove this relation is symmetric. Let X and Y be any sets, and assume X relates to Y . In other words, assume there is a bijection f : X → Y . Then f has an inverse function f −1 : Y → X which is also a bijection. Hence Y relates to X. Finally, we prove transitivity. Let X, Y , and Z be any sets, and assume there are bijections f : X → Y and g : Y → Z. The composite function g ◦ f : X → Z is a bijection, as needed.

Theorem 27.3. Let A and B both be finite sets, and let f : A → B be a function. (1) If f is injective, then |A| ≤ |B|. (2) If f is surjective, then |A| ≥ |B|. (3) If f is bijective, then |A| = |B|.

Proof. We make use of Theorems 27.1 and 27.2. (1) If f is injective, then |A| = |im(f)| ≤ |B| (since im(f) ⊆ B). (2) If f is surjective, then |A| ≥ |im(f)| = |B|. (3) This follows from (1) and (2).

Theorem 24.13. Let A and B be sets, and suppose that {P, Q} is a partition of A with two parts. If we are given a function g : P → B and a function h: Q → B, then f = g ∪ h defines a function f : A → B.

Proof. We may consider g as a subset of P × B, which is a subset of A × B, and h as a subset of Q × B, which is a subset of A × B. Hence, both g and h (considered as collections of ordered pairs) are subsets of A × B. In fact, g = {(x, g(x)) : x ∈ P} and h = {(x, h(x)) : x ∈ Q}. Let f = g ∪ h. Then f is a subset of A × B, so f is a relation from A to B. To see that f is a function, it suffices to check that each element of A is the first coordinate of exactly one ordered pair in f. To that end, let a ∈ A. Since A is partitioned into two parts, there are two cases; either a ∈ P or a ∈ Q. Without loss of generality, we may assume a ∈ P. We know that a is the first coordinate of exactly one ordered pair in g, namely (a, g(a)) (using the fact that g : P → B is a function). On the other hand, a is not the first coordinate of any ordered pairs in h (since every ordered pair in h has first coordinate in Q, and a /∈ Q). Hence, (a, g(a)) is the one and only ordered pair in f that has first coordinate a. Since a was an arbitrary element of A, we see that f is a function.

Theorem 31.6. If A is any set, then |P(A)| = |F(A)|.

Proof. We must construct a bijection between the two sets P(A) and F(A). The rule is this: send a subset S ⊆ A to its characteristic function, χS : A → {0, 1}. In other words, we define f : P(A) → F(A) by the rule f(S) = χS. We first show that f is injective. Let S, T ∈ P(A) and assume f(S) = f(T). We then have χS = χT . Plugging in an arbitrary element a ∈ A, we have χS(a) = χT (a). The left-hand side is 1 when a ∈ S and 0 otherwise, and similarly for the right-hand side. Thus, a ∈ S if and only if a ∈ T. In other words S = T. Finally, we show that f is surjective. Let φ: A → {0, 1} be any function. Put S = {a ∈ A : φ(a) = 1}. We then check directly that φ = χS. (They have the same domain and codomain, and the same rule.) Hence φ = f(S), so f is surjective.

Theorem 28.4. The cardinalities of N and Z are the same.

Proof. We need to construct a bijection f : N → Z; or, in other words, we need to "count" the elements of Z using the natural numbers. The idea is to first count 0, and then successively count the positive and negative integers, as in the following table n 1 2 3 4 5 6 7 8 9 10 f(n) 0 1 −1 2 −2 3 −3 4 −4 5 which yields the needed bijection between N and Z. Remark 28.5. The previous proof was very informal. First, the definition of the function f is sloppy. To be more precise we should define f as a piecewise function from N → Z by the rule f(n) = ( n/2 if n is even, −(n − 1)/2 if n is odd. This is exactly the same function as in Example 27.12. prove it is injective and surjective

Theorem 29.3. If A and B are countably infinite sets, then A × B is as well.

Proof. Without loss of generality, we just need to show that N × N is countably infinite. Consider the following diagram: (1, 1) (2, 1) (3, 1) (1, 2) (2, 2) (3, 2) (1, 3) (2, 3) (3, 3) . . . . . . . . . . . . · · · · · · · · · Travel along each arrow, starting at the smallest arrow, and passing to the next smallest arrow. This allows us to list the elements of N × N as (1, 1),(2, 1),(1, 2),(3, 1),(2, 2),(1, 3), . . . , according to when we pass through each ordered pair. We will hit each ordered pair exactly once.

Corollary 30.5. The set R is uncountable

The cardinality of R is called the continuum, and we write |R| = c. We have already seen that |(0, 1)| = |R|, so (0, 1) also has continuum cardinality. Here are some more examples of sets with continuum cardinality. (1) Any open interval (a, b) with a, b ∈ R. (We can also replace a with −∞, or b with ∞.) (2) Any half-open interval [a, b) with a, b ∈ R. (We can replace b with ∞.) (3) Any half-open interval (a, b] with a, b ∈ R. (We can replace a with −∞.) (4) Any closed interval [a, b] with a, b ∈ R. To give the idea behind how to prove these facts, we will show:

Theorem 30.7. Let A and B be sets, with A ⊆ B. If A is uncountable, then B is uncountable.

This is the contrapositive of Theorem 28.14, after noting that A and B must be infinite. Example 30.8