Suppose That Every Member of a Has Cardinality at Most Κ, Then Card S a ≤ (Carda) · Κ

MATH 135: SOLUTIONS TO HOMEWORK SET #9

1 (page 161, # 26) Suppose that every member of A has cardinality at most κ, then card S A ≤ (cardA) · κ. Proof. Define Y := {ha, xi : x ∈ a ∈ A}. Then by definition of the union, the function f : Y → S A defined by ha, xi 7→ x is surjective. Hence, A Y . It suffices to show that Y A × κ. Let R := {ha, fi ∈ A × (S A × κ): f is an injective function with domain a and range contained in κ}. By hypothe- sis, every a ∈ A has cardinality at most κ so that there is some injective f : a ,→ κ. That is, domR = A. By the axiom of choice there is a function F ⊆ R with domF = domR = A. Define G : Y → A × κ by ha, xi 7→ ha, F (a)(x)i. If y = ha, xi and y0 = ha0, x0i are two elements of Y with G(y) = G(y0), then a = a0 and F (a)(x) = F (a0)(x0) = F (a)(x0). As F (a) is injective, we have x = x0 as well, so that y = y0. Thus, G is injective showing that Y A × κ. S Therefore, card A ≤ (cardA) · κ.

2 (page 161, # 27) (a) Let A be a collection of circular disks in the plane, no two of which intersect. Show that A is countable.

Proof. As Q2 is dense in R2, if D is any disk in the plane, then we must 2 S 2 have D ∩ Q 6= ∅. Let R := {D ∩ Q |D ∈ A}. For x ∈ R, by definition, there is some D ∈ A with x ∈ D. As the disks in A are disjoint, there can be at most one such D. Define G : R → A by G := {hx, Di ∈ Q2 × A|x ∈ D}. The above remark shows that R is a surjective function. Hence, we may 2 find some injective f : A,→ R with G ◦ f = idA. As R ⊆ Q , we have |R| ≤ ℵ0. Hence, |A| ≤ |ℵ0|. (b) Let B be a collection of circles in the plane, no two of which intersect. Need B be countable? 2 2 2 No. For r a positive real number, let Sr := {(x, y) ∈ R : x + y = r}. Then Sr is a circle and for r 6= s we have Sr ∩ Ss = ∅. Set C := {Sr|r ∈ R, r > 0}. (c) Let C be a collection of figure eights in the plane, no two of which intersect. Need C be countable? Yes. Recall that we defined a figure eight to be a set of the form X ∪ Y where X and Y are circles whose bounded disks meet at exactly one point. For each pair of positive rational numbers p and q with p ≤ q, we define Cp,q to be the set of figure eights in C whose smaller circle has radius r with p ≤ r and whose large circle has radius R with q ≤ R < q + p. Note that S C = p≤q Cp,q is a countable union of the sets Cp,q. If we show that each Cp,q is countable, then we know that C itself is. Note that if F,E ∈ Cp,q are two distinct figure eights with the specified conditions on their radii, then not only are F and E disjoint, but the disks that they bound are also 1 2 MATH 135: SOLUTIONS TO HOMEWORK SET #9

disjoint as if F were a subset of the union of the disks bounded by F , we would need the sum of the radii of the circles of F to be less than the radius of the larger circle of E, but this sum is at least p + q. Thus, exactly as in 2 part (a) we may deﬁne an injective function f : Cp,q → Q showing that Cp,q is countable.

3 (page 161, # 28) Find a set A of open intervals in R such that every rational S number belongs√ to one√ of those intervals but A 6= R. Let A = {( 2 + n, 2 + n + 1) : n ∈ Z}. 4 (page 161, # 29) Let A be a set of positive real numbers. Assume that there is a bound b such that the sum of any ﬁnite subset of A is less than b. Show that A is countable.

1 Proof. For each positive integer n define An := A ∩ ( n , ∞). I claim that for each n, the set An is finite. Suppose this fails for some n. Let k ∈ ω be a natural number for which k > nb. If An were infinite, then we could find a1, . . . , ak ∈ A distinct. Pk Pk 1 k nb We have b ≥ i=1 ai > i=1 n = n > n = b.With this contradiction, we see that S∞ An is finite. Thus, A = n=1 An being a countable union of at most countable sets is itself at most countable.

5 (page 161, # 30) Assume that A is a set with at least two elements. Show that Sq(A) ωA.

S n Proof. By definition of Sq(A), we have Sq(A) = n∈ω A. We break the argument into two cases at this point. If A is finite, then each of the sets nA is also finite and, thus, at most countable. Therefore, Sq(A) is also at most countable. However, as 2 A, we see that Sq(A) ω ≺ ω2 ωA. If A is infinite, then A ≈ A ∪ {∗} =: A0 where ∗ ∈/ A. Let g : A0 → A be a bijection and let G : ωA0 → ωA be the induced bijection given by σ 7→ g ◦ σ. Define a function F : Sq(A) → ωA0 by ( σ(n) if n ∈ domσ F (σ)(n) = ∗ otherwise

Then F is injective as if F (σ) = F (τ) we see that dom(σ) = F (σ)−1[[A]] = −1 F (τ) [[A]]dom(τ) and that we have σ = F (σ) dom(σ) = F (τ) dom(σ) = F (τ) dom(τ) = τ. ω Composing F with G, we see that Sq(A) A.

6 (page 165, # 32) Let FA be the collection of all ﬁnite subsets of A. Show that if A is inﬁnite, then A ≈ FA.

Proof. We claim that for n ∈ ω and n > 0 we have A ≈ nA. We prove this by induction on n. For n = 1, there is an obvious bijection between A and 1A given MATH 135: SOLUTIONS TO HOMEWORK SET #9 3 by a 7→ [0 7→ a]. Assuming the result for n, we compute

+ n A = n∪{n}A ≈ nA × {n}A by Thm 6I(4) ≈ nA × A by the base case ≈ A × A by IH ≈ A by Lemma 6R

Thus, by problem 26 of page 161, we have that |Sq(A)| ≤ ℵ0 ·|A| = |A|. However, as 1A ⊆ Sq(A), we certainly have |A| ≤ |Sq(A)|. Therefore, A ≈ Sq(A). Now, define a function Sq(A) → FA by σ 7→ rngσ. By definition of a finite set, this function is surjective. Thus, |FA| ≤ |Sq(A)| = |A|. Again, as every singleton in A is a finite set, we have |A| ≤ |FA|. Therefore, A ≈ FA.

7 (page 165, # 35) Find a collection A of 2ℵ0 sets of natural numbers such that any two distinct members of A have ﬁnite intersection.

Proof. We prove a claim. Claim: Let K be any inﬁnite set with cardK = κ. Let IK := {X ∈ PK | |X| ≥ κ ℵ0}. Then |IK| = 2 . Proof of claim: We may write PK as the disjoint union of IK and FK. By the result of the previous problem, we have cardFA = κ. Letting λ := cardIK, We compute 2κ = cardPK = card(FK ∪ IK) = card(FK) + card(IK) = κ + λ by Problem 6 = max{κ, λ} by the absorption law = λ as κ < 2κ

z By Euclid’s theorem, the set P of prime numbers is a countably infinite set. Hence, the set IP of infinite sets of primes has cardinality 2ℵ0 . Q For each X ∈ IP define AX := { p≤n,p∈X p | n ∈ ω} ⊆ ω}. This association defined a function F : IP → Pω. Let A := rngF . Note that for any X ∈ IP the set AX is infinite as X is infinite so that there are elements of AX with arbitrarily many prime divisors. If X 6= Y are two distinct elements of IP , then there is some prime which belongs to one set but not the other. Without loss of generality, we may assume p ∈ X \ Y . 0 Q 00 0 Let AX := { `≤n,`∈X ` | n < p} and AX := AX \ AX . Note that if y ∈ AY , then p 00 does not divide y while p divides every element of AX . Thus, AX ∩AY is contained 0 0 in AX ∩ AY ⊆ AX ⊆ {m ∈ ω : m < p!} which is a finite set. Hence, AX ∩ AY is finite. As AX is infinite, this implies in particular that AX 6= AY . Thus, the map ℵ F : IP → A is a bijection so that A has cardinality 2 0 and is almost disjoint.

8 Show that for any inﬁnite cardinal κ, we have κ! = 2κ. 4 MATH 135: SOLUTIONS TO HOMEWORK SET #9

Proof. Let K be any set with cardK = κ. Let Sym(K) := {σ : K → K|σ a bijection }. Then by definition, we have κ! = cardSym(K). As every permutation is a function from K to K (and in particular a relation with field K), we have Sym(K) ⊆ P(K × K) so that κ! ≤ 2κ·κ = 2κ. For the other inequality, define a function F : Sym(K) → PK by σ 7→ {x ∈ K|σ(x) = x}. Let A := rngF . To compute the cardinality of A we need a claim. Claim: Let X be any set with cardX ≥ 2. Then there is a permutation π : X → X having no fixed points. Proof of claim: If X is finite, then X ≈ n for some n ∈ ω with n > 2. Let f : n → X be a bijection. Define σ : n → n by σ(i) = i+ if i < n − 1 and σ(n − 1) = 0. Define π : X → X by π = f ◦ π ◦ f −1. If X is infinite of cardinality λ, then as λ = λ × λ, we can find a bijection g : X → X ×2. Define σ : X ×2 → X ×2 by hx, ii → hx, 1−ii. Let π := g−1 ◦σ ◦g. z From the claim it follows that B = {X ∈ PK | card(K \ X) 6= 1}. For if X = K, then X = F (idK ) and if K \ X =: Y has more than one element, then we can find (by the claim) a permutation π : Y → Y having no fixed points so that if σ = π ∪ idX we have F (σ) = X. Finally, if K \ X = {a} is a singleton, then X cannot be in the image of F as any permutation which fixes X must fix its complement setwise, and in the case of a singleton, being fixed setwise is the same as being fixed pointwise. There is an obvious bijection between K and the set of subsets of K whose complements are singletons given by a 7→ K \{a}. Hence, |PK \ B| = κ. As in the proof the claim in problem 7, we conclude that |B| = 2κ. As F : Sym(K) → B is surjective, we conclude that κ! ≥ 2κ. κ Combining this with the other inequality, we conclude that κ! = 2 .