MA532 Lecture

Timothy Kohl

Boston University

April 23, 2020

Timothy Kohl (Boston University) MA532 Lecture April 23, 2020 1 / 26 Cardinal Arithmetic

Recall that one may define addition and multiplication of ordinals

α = ot(A, A) β = ot(B, B )

α + β and α · β by constructing order relations on A ∪ B and B × A.

For cardinal numbers the foundations are somewhat similar, but also somewhat simpler since one need not refer to orderings.

Definition For sets A, B where |A| = α and |B| = β then α + β = |(A × {0}) ∪ (B × {1})|.

Timothy Kohl (Boston University) MA532 Lecture April 23, 2020 2 / 26 The curious part of the definition is the two sets A × {0} and B × {1} which can be viewed as subsets of the direct product

(A ∪ B) × {0, 1} which basically allows us to add |A| and |B|, in particular since, in the usual formula for the size of the union of two sets

|A ∪ B| = |A| + |B| − |A ∩ B| which in this case is bypassed since, by construction, (A × {0}) ∩ (B × {1})= ∅ regardless of the nature of A ∩ B.

Timothy Kohl (Boston University) MA532 Lecture April 23, 2020 3 / 26 Definition For sets A, B where |A| = α and |B| = β then α · β = |A × B|.

One immediate consequence of these definitions is the following. Proposition If m, n are finite ordinals, then as cardinals one has |m| + |n| = |m + n|, (where the addition on the right is ordinal addition in ω) meaning that ordinal addition and cardinal addition agree.

Proof. The simplest proof of this is to define a bijection f : (m × {0}) ∪ (n × {1}) → m + n by f (hr, 0i)= r for r ∈ m and f (hs, 1i)= m + s for s ∈ n.

Timothy Kohl (Boston University) MA532 Lecture April 23, 2020 4 / 26 And in a similar fashion, one has the relatively obvious fact: Proposition If m, n are finite ordinals, then as cardinals |m| · |n| = |m · n|, where the multiplication on the right is ordinal multiplication in ω.

Proof. Exercise.

Timothy Kohl (Boston University) MA532 Lecture April 23, 2020 5 / 26 The addition and multiplication of cardinals differs from that ordinals. Proposition For cardinal numbers α,β, and γ: (a) α + β = β + α (b) α +0= α (c) (α + β)+ γ = α + (β + γ) (d) α · 0 = 0 (e) α · 1= α (f) α · 2= α + α (and similarly for α · n) (g) (α · β) = (β · α) (h) (α · β) · γ = α · (β · γ) (i) α · (β + γ)= α · β + α · γ

Timothy Kohl (Boston University) MA532 Lecture April 23, 2020 6 / 26 Proof. (a) There is a fairly obvious bijection f : (A × {0}) ∪ (B × {1}) → (B × {0}) × (A × {1}), namely ha, 0i 7→ ha, 1i and hb, 1i 7→ hb, 0i, and so |A| + |B| =c |B| + |A|

(b) As ∅ × {1} = ∅ then (A × {0}) ∪ (∅ × {1}) = (A × {0}) ergo |A| + |∅| = |A|.

(c) [Exercise]

(d) For any set A we have A × ∅ = ∅ so |A| · |∅| = |∅|.

Timothy Kohl (Boston University) MA532 Lecture April 23, 2020 7 / 26 The remaining statements are proved similarly. Proof. (e) There is a bijection A × 1= A × {0}→ A given by ha, 0i 7→ a which implies |A| · |1| = |A|.

(f) As 2= {0, 1} there is yet another obvious bijection (A × {0}) ∪ (A × {1}) → A × 2= A × {0, 1} since, in fact,

(A × {0}) ∪ (A × {1})= A × {0, 1}

(g,h,i) Exercises.

Timothy Kohl (Boston University) MA532 Lecture April 23, 2020 8 / 26 For the definition of ordinal exponentiation, we recall that for sets A and B, the functions f : B → A is denoted B A.

Definition If α, β are cardinal numbers, then αβ denotes the unique cardinal γ such that, for sets A and B where |A| = α and |B| = β then γ = |B A|.

For finite sets, for example, 32, a function f : 3 → 2 is a mapping from 3= {0, 1, 2} to 2= {0, 1}, i.e. f (0) = 0 or 1,f (1) = 0 or 1,f (2) = 0 or 1, ergo |32| = 23 = 8, where, not coincidentally |32| =c |P(3)|.

And indeed, 2|A| = |P(A)| for any cardinality |A|.

And of course, for finite cardinals m, n in general, the set nm has mn elements, i.e. cardinal exponentiation and ordinal exponentiation (in ω) are the same thing.

Timothy Kohl (Boston University) MA532 Lecture April 23, 2020 9 / 26 We observe a number of facts about the exponentiation of cardinal numbers. Proposition For cardinals α, β, and γ: (a) α0 = 1 (b) α1 = α (c) α2 = α · α (d) αn = αn−1 · α for n a finite cardinal (e) if β 6= 0, 0β = 0 (f) 1β = 1 (g) αβ+γ = αβ · αγ (h) αβ·γ = (αβ)γ (i) (α · β)γ = αβ · βγ

Timothy Kohl (Boston University) MA532 Lecture April 23, 2020 10 / 26 Proof. (a) This is somewhat curious statement since α0 represents the cardinality of ∅α, the totality of functions from ∅ to α, but this consists of the empty set so it equals {∅} = {0} = 1.

(b) Each function in 1α has domain consisting of a single element, and so the number of functions is in bijective correspondence with the cardinality of α, ergo α1 = α.

(c,d) Exercises.

Timothy Kohl (Boston University) MA532 Lecture April 23, 2020 11 / 26 Here are the proofs of the remaining statements. Proof. (e) In contrast with α0, for 0β one is looking at the set of function B → ∅ for some set B with α = |B| then we must realize that no function with a non-empty domain can have range ∅.

(f) Exercise (g) For sets A, B, C, with cardinal numbers α, β and γ one compares functions (B × {0}) ∪ (C × {1}) → A with the direct product (B A) × (C A) and realize that since (B × {0}) ∩ (C × {1})= ∅ any f : B × {0}→ A and g : C × {1}→ A are ’independent’ and so determine a unique function hf , gi∈ (B A) × (C A).

(h,i) Exercises.

Timothy Kohl (Boston University) MA532 Lecture April 23, 2020 12 / 26 Cofinality

For the definition we give next, we need to give a definition of limit. Definition

If α> 0 is a limit ordinal, and hγζ | ζ < αi be a non-decreasing sequence of ordinals (i.e. ζ < η implies γζ ≤ γη) then the limit of the sequence

lim γζ = sup{γζ | ζ < α} ζ→α

and a sequence of ordinals hγα | α ∈ Oni is normal if it is increasing and continuous, i.e. for every limit ordinal α, γα = lim γζ ζ→α

Timothy Kohl (Boston University) MA532 Lecture April 23, 2020 13 / 26 Definition If α> 0 is a limit ordinal, we say that an increasing sequence hαζ | ζ < βi, β a limit ordinal, is cofinal in α if lim αζ = α. ζ→β Similarly, A ⊆ α is a cofinal in α if supA = α.

If α is an infinite limit ordinal, the confinality of α is

cf (α) = the least limit ordinal β such that there is an increasing

sequence hαζ | ζ < βi with lim αζ = α ζ→β

Timothy Kohl (Boston University) MA532 Lecture April 23, 2020 14 / 26 A slightly easier definition perhaps is this: Definition Let α, β be ordinals and let f : α → β be a function, f is cofinal (or α cofinally (in)to β) if for all γ < β, there exists δ < α such that γ ≤ f (δ).

Let β ∈ On then the confinality of β, denoted cf (β), is the minimal α such that there is a cofinal function from α to β.

So cf (β) ≤ β and is a cardinal for any β, and if β is a successor ordinal (i.e. not a limit ordinal) then cf (β) = 1 via the function f (0) = S β.

Timothy Kohl (Boston University) MA532 Lecture April 23, 2020 15 / 26 A more familiar way of viewing cofinality is from the theory of sequences from real analysis/calculus.

In particular, if {an} is a sequence, and if {kn} is defined as a sequence of indices such that lim kn = ∞ then {akn } is a subsequence of {an}. n→∞

One of the canonical properties of convergent sequences {an} is that if

L = lim an (finite) then for any subsequence {akn } one has lim akn = L n→∞ n→∞ as well.

Timothy Kohl (Boston University) MA532 Lecture April 23, 2020 16 / 26 1 1 For example, given the sequence, { 2n } we have that { 4n } is a subsequence, and both tend to zero as n →∞.

Although it’s not essential here, we observe that there exists sequences {an} with subsequences {bn} and {cn} such that lim bn = L1 and n→∞ lim cn = L2 (where L1 and L2 are finite) but where L1 6= L2. n→∞

In this situation we conclude that lim an does not exist, so the n→∞ convergence of the sequence implying the convergence of all subsequences is a pretty stringent condition.

Timothy Kohl (Boston University) MA532 Lecture April 23, 2020 17 / 26 And, in terms of sets, the cofinality can be be viewed as a kind of ’density’ 1 1 of one in the other, for example with the { 2n } and { 4n } sequences earlier, as sets 1 1 1 1 { } = { , , ,... } 4n 4 16 64 is cofinal in 1 1 1 1 1 1 1 1 { } = { , , , , , , ,... } 2n 2 4 8 16 32 64 128

1 1 since every other term of { 2n } is one of the elements of { 4n }.

Timothy Kohl (Boston University) MA532 Lecture April 23, 2020 18 / 26 So for a given ordinal α the cofinality is basically the order type of any index set of a sequence whose terms comprise a cofinal set.

1 e.g. A = { 2n } is indexed by the natural numbers 0, 1, 2,... and the 1 subsequence B = { 4n } consists of those terms of A indexed by the even natural numbers 0, 2, 4,...

So the ordinal α corresponding to A is ω and the ordinal indexing B is {0, 2, 4,... } has the same order type, namely ω

And it’s clear that any subsequence giving rise to a cofinal subset of A must have order type ω, so the cofinality of α is ω. i.e. cf (ω)= ω

Timothy Kohl (Boston University) MA532 Lecture April 23, 2020 19 / 26 Again, it’s clear that cf (α) is a limit ordinal, and cf (α) ≤ α.

The simpler way to view cofinality is the length of the shortest possible sequence leading up to it from below.

Examples: cf (ω + ω)= cf (ℵω)= ω because of the sequence ω,ω + 1,... .

Timothy Kohl (Boston University) MA532 Lecture April 23, 2020 20 / 26 Lemma cf (cf (α)) = cf (α)

Proof.

If hαζ | ζ < βi is cofinal in α and hζ(ν) | ν < γi is cofinal in β, then hαζ | ζ < γi is cofinal in α.

Lemma Let α> 0 be a limit ordinal.

(i) If A ⊆ α and supA = α, then ot(A) is at least cf (α).

(ii) If β0 ≤ β1 ≤ ...βζ ≤ . . . , ζ < γ is a non-decreasing sequence of ordinals in α and lim βζ = α then cf (γ)= cf (α). ζ→γ

Timothy Kohl (Boston University) MA532 Lecture April 23, 2020 21 / 26 Definition

An infinite cardinal ℵα is regular if cf (ωα)= ωα.

It is singular if cf (ωα) <ωα.

Lemma For every limit ordinal α, cf (α) is a regular cardinal.

Timothy Kohl (Boston University) MA532 Lecture April 23, 2020 22 / 26 Proof. If α is not a cardinal then one can construct a cofinal sequence in α of length ≤ |α| and so cf (α) < α.

Since cf (cf (α)) = cf (α) it follows that cf (α) is a cardinal and is regular.

Timothy Kohl (Boston University) MA532 Lecture April 23, 2020 23 / 26 Definition Let κ be a limit ordinal, a subset X ⊆ κ is bounded if supX < κ, and unbounded if supX = κ.

We quote the following without proof. Lemma (i) If X ⊆ κ and |X | < cf (κ) then X is bounded. (ii) If λ< cf (κ) and f : λ → κ then the range of f is bounded.

The implication of (i) is that every unbounded subset of a regular cardinal has cardinality κ.

Timothy Kohl (Boston University) MA532 Lecture April 23, 2020 24 / 26 The main point of all this technicality is that there are arbitrarily large singular cardinals.

For each α, ℵα+ω is a singular cardinal of cofinality ω

Timothy Kohl (Boston University) MA532 Lecture April 23, 2020 25 / 26 Theorem If κ is an infinite cardinal, then κ < κcf (κ).

Proof.

Let F be a collection of κ functions from cf (κ) to κ, F = {fα | α < κ}.

We show that there exists an f : cf (κ) → κ that is different from all the fα.

Let κ = lim then for ζ < cf (κ) let ζ→cf (κ)

f (ζ)= leastγ such that γ 6= fα(ζ) for all α < αζ and such a γ exists since

|{fα(ζ) | α < αζ } ≤ |αζ | < κ and obviously f 6= fα for all α < κ by construction. And in general, κ < κµ for any µ ≥ cf (κ). Timothy Kohl (Boston University) MA532 Lecture April 23, 2020 26 / 26