Revisions for Sets and Classes, 2007.03.02 ed.

David Pierce

April 2, 2007

Some corrections and changes.

1 General

• A list of symbols should be provided.

• The relation symbolized by ∈ and called containment on p. 34 would be better called membership (as on p. 19).

2 Significant changes

• p. 10, ¶ 1.1.2. The latter part of this paragraph needs to be rethought. Set and class are not the most ‘generally applicable’ collective nouns; they are the most abstract. For us, set will be the name of something whose members are other sets.

• P. 11: expand ¶ 1.2.2: say more about this ‘correspondence’ between ∪ and ∨.

• P. 21, ¶ 2.2.4, after the list: the comment ‘depending on the axioms’ makes the truth-value of a ∈ a sound arbitrary. The Foundation Axiom will say that the sentence is false; but ¶ 8.3.5 shows that this axiom can be understood merely as a definition of the sets that we choose to study.

• ¶ 2.2.5: The sets that we study can be called pure sets (Moschovakis) or hereditary sets (Kunen).

• P. 22, ¶ 2.2.9: The alternative formulation of V should also use the ‘official’ language: so make it {x: x ∈ x ⇒ x ∈ x ⇒ x ∈ x}.

• P. 25: Not all of our theorems will have formal deductions even in principle: In ¶ 8.3.5, it will be noted that WF is a model of ZF. This conclusion can be formulated as an infinite list of sentences in the official language, each with a formal proof. We conclude that ZF is consistent: this can be formulated (as Gödel showed) as a single sentence of the official language; but it has no formal proof.

• Exercises might be added to Ch. 2.

• P. 32, bottom: the definition of field is not really needed.

• P. 34, ¶ 3.5.6 (ii): the reference should be to ¶ 3.5.5.

• P. 35, ¶ 3.5.7: a least element is also a minimum element.

• P. 36, ¶ 3.6.2: Instead of ‘virtual class’ here, we might speak of a family of classes. Then a class C and a binary relation R determine the families {xR: x ∈ C} and {Rx: x ∈ C}. In particular, if E ⊆ C × C and is an equivalence-relation on C, then C/E is the family {xE: x ∈ C}.

• P. 37, ¶ 3.6.4: item (v) should be item (ii); also, in (vi), the notation π_{0} and π_{1} that will be used in ¶ 5.4.2 (Recursion with Parameter) can be introduced.

• P. 46: replace ¶ 4.2.4 with:

1. Suppose (C, F , i) is a recursive structure. Then C can be denoted suggestively by

{i, F (i), F (F (i)), … }.

Some possibilities are depicted in Fig. 4.1. Note well that possibly F is not injective, and possibly i ∈ F [ C]. (However, these possibilities seem to be mutually exclusive.)

{par:rec}

2. Suppose again (C, F , i) is a recursive structure, and (D, G, j) is another iterative structure (not necessarily recursive). There may be a function H from C to D such that

(i) H(i) = j,

(ii) a ∈ C ⇒ H(F (a)) = G(H(a)), that is, H ∘ F = G ∘ H on C.

The first rule says what H(i) is; the second says how to obtain H(F (a)) from H(a). By induction, H is uniquely determined by these rules: see Corollary 5 below. In this case, we say that H is recursively defined by the given rules.

{par:rec2}

3. Note first another possible kind of recursive definition: (C, F , i) is recursive, E ⊆ D, and G : D → D, then perhaps there is a sub-class R of C × D such that (in the notation of ¶ 3.6.2)

(i) iR = E,

(ii) a ∈ C ⇒ F (a)R = G[ aR].

Then R too is uniquely determined by these rules, so it too is recursively defined:

{thm:rec-uni}

4 Theorem. Suppose (C, F , i) is recursive, G : D → D, and E ⊆ D. Then there is at most one relation R as in ¶ 3.

Proof. Suppose R_{0} and R_{1} are two such relations. Let

C_{1} = {x: x ∈ C & xR_{0} = xR_{1}}.

Since iR_{0} = E = iR_{1}, we have i ∈ C_{1}. Suppose a ∈ C_{1}, so aR_{0} = aR_{1}. Then

F (a)R_{0} = G[ aR_{0} ] = G[ aR_{1} ] = F (a)R_{1},

so F (a) ∈ C_{1}. By induction (and Lemma 4.2.3), C_{1} = C. Since dom(R_{0}) ⊆ C and dom(R_{1}) ⊆ C, we conclude that R_{0} = R_{1}.

{cor:rec-uni}

5 Corollary. Suppose (C, F , i) is recursive, and (D, G, j) is iterative. Then there is at most one function H as in ¶ 3.

Proof. The function H (if it exists) is a relation, namely a sub-class R of C ×D. Let E = {H(i)}. Then

(i) iR = {H(i)} = E;

(ii) a ∈ C ⇒ F (a)R = {H(F (a))} = {G(H(a))} = G[ {H(a)}] = G[ aR].

By the theorem, R is unique, so H is unique.

• P. 48, ¶ 4.3.5: The Recursion Theorem should be given more generally:

{thm:rec}

6 Theorem (Recursion). Suppose (C, F , i) is an arithmetic structure, G: D → D, and E ⊆ D. Then there is (uniquely, by Theorem 4) a sub-class R of C ×D such that

(i) iR = E,

(ii) a ∈ C ⇒ F (a)R = G[ aR].

Proof. Let B be the sub-class

{x: ∀y (y ∈ x ⇒ ∃z (z ∈ E & y = (i, z)) ∨

∨ ∃u ∃v ((u, v) ∈ C × D ∩ x & y = (F (u), G(v))))}

of P(C × D), and let R = ⋃ B, so R ⊆ C × D. If a ∈ E, then {(i, a)} is a set (by the Pairing Axiom), so it belongs to B, and hence i R a. Suppose b R c. Then (b, c) ∈ d for some d in B, so d ∪ {(F (b), G(c))} is a set (by the Weak Union Axiom), and this set belongs to B, so F (b) R G(c). We now have the following characterization of R:

a R b ⇔ ((a = i & b ∈ E) ∨ ∃u ∃v (u R v & a = F (u) & b = G(v))).

Since i ∉ F [ C], we have i R a ⇔ a ∈ E, so iR = E. Since F is injective, if a ∈ C, we have F (a) R b ⇔ b ∈ G[ aR], so F (a)R = G[ aR].

{cor:rec}

7 Corollary. Suppose (C, F , i) is an arithmetic structure and (D, G, j) is an iterative structure. Then there is (uniquely, by Corollary 5) a function H from C to D such that (i) H(i) = j, (ii) a ∈ C ⇒ H(F (a)) = G(H(a)), that is, H ∘ F = G ∘ H. Proof. Exercise.

• P. 51, end of § 4.4: ‘We seem to have this if i = ∅, and F and G are both x ↦ x ∪ {x}.’

• P. 52, proof of 4.5.4, part (i): ‘Then b ≠ α, since α is well-ordered by containment, and such orderings are by definition strict’.

• P. 53, ¶ 4.5.8 should begin: ‘The structure (ON, x ↦ x′, ∅) satisfies the hypothesis of Theorem 4.4.3 with respect to the ordering ∈; hence there is a class {∅, ∅′, ∅′′} … ’.

• ¶¶ 5.2.1–2 can be replaced with the following:

8. We shall define the binary operation of addition on ℕ so that (i) m + 0 = m, (ii) m + n^{+} = (m + n)^{+}. These rules tell how to add 0, and they tell how to add n^{+}, provided one can add n. The rules will determine a unique operation, by a variant of the Recursion Theorem (¶ 6). Moreover, suppose (A, S, i) is a recursive structure. (This will be so throughout this section.) Then we shall be able to define addition on A by the rules

(i) a + i = a, (ii) a + S(b) = S(a + b), even though the Recursion Theorem does not apply generally to all recursive structures.

{lem:+}

9 Lemma. Suppose F : B → C and G: C → C. Then there is a unique function H from B × ℕ to C such that

(i) a H 0 = F (a), (ii) a H n^{+} = G(a H n). Proof. By the Recursion Theorem, there is a unique sub-class R of ℕ×(B ×C) such that

(i) 0R = F, (ii) n ∈ ℕ ⇒ n^{+}R = ((x, y) ↦ (x, G(y))[ nR] = {(x, G(y)): n R (x, y)}. By induction, if n ∈ ℕ, then nR is a function from B to C. Indeed, 0R is such a function (namely F), and if nR is such a function, then n^{+}R is its composition with G. Let the function nR be denoted by K_{n}; then K_{n^{+}} = G ∘ K_{n}. We can now define the binary function H as (x, y) ↦ K_{y}(x) on A × ℕ. Then (i) a H 0 = K_{0}(a) = F (a),

(ii) a H n^{+} = K_{n^{+}}(a) = G(K_{n}(a)) = G(a H n).

So H is as desired. To see that H is unique, note that R determines H, and conversely. Indeed,

H = {((x, y), z): y R (x, z)},
R = {(y, (x, z)): x H y = z}.

Since R uniquely satisfies the given conditions, so does H.

{thm:+}

10 Theorem and Definition. Suppose (A, S, i) is recursive. Then there is a unique binary operation of addition on A given by

(i) a + i = a,

(ii) a + S(b) = S(a + b).

Proof. By the lemma, there is a unique function H from A × ℕ to A such that

(i) a H 0 = a,

(ii) a H n^{+} = S(a H n).

So H is recursively defined in its second argument. We shall show that it is also recursively definable in its first argument. First, let F be the function x ↦ i H x from ℕ into A. Then

\begin{aligned}F (0) & = i, \\ F (n^{+}) & = S(F (n)). \end{aligned}  (1) \{eqn:+i\}

(So F is the unique homomorphism from (ℕ,^{+}, 0) into (A, S, i) guaranteed by Corollary 7.) By induction, rng(F) = A; indeed, i ∈ rng(F), and a ∈ rng(F) ⇒ S(a) ∈ rng(F). The equation

S(a) H n = S(a H n)  (2) {eqn:+n}

holds when n = 0, since S(a) H 0 = S(a) = S(a H 0). Suppose (2) holds for some n in ℕ. Then

S(a) H n^{+} = S(S(a) H n)  [by definition of H]
= S(S(a H n))  [by inductive hypothesis]
= S(a H n^{+}).  [by definition of H]

So (2) holds for all n in ℕ. Therefore each of the operations x ↦ x H n is the operation G_{n} recursively defined by

(i) G_{n}(i) = F (n),

(ii) G_{n}(S(a)) = S(G_{n}(a)).

In particular,

G_{m} = G_{n} ⇔ F (m) = F (n).

Now we can define addition on A by

a + b = c ⇔ ∃x (F (x) = b & G_{x}(a) = c).

Then a + i = G_{0}(a) = a H 0 = a. Also, if b = F (n), so that S(b) = F (n^{+}), then

a + S(b) = G_{n^{+}}(a) = a H n^{+} = S(a H n) = S(G_{n}(a)) = S(a + b).

Thus + is as desired; it is unique by Theorem 4.

• ¶ 5.2.3 should have a reference to Landau. ¶ 5.2.4 can be slightly rewritten:

11. Suppose (a, s, i) is a recursive set, so that all operations on a are sets. Then we can establish addition on a as follows. By Corollary 5, for each b in a, there is at most one singulary operation f_{b} on a such that

(i) f_{b}(i) = b, (ii) f_{b} ∘ s = s ∘ f_{b}. Let a_{0} be the subset of a comprising those b such that f_{b} does exist; note well how this definition of a_{0} requires f_{b} to be a set. Then f_{i} exists and is id_{a}, so i ∈ a_{0}. If b ∈ a_{0}, then f_{s(b)} exists and is s ∘ f_{b}. By induction, a_{0} = a. Now we can define b + c = f_{b}(c).

• At the beginning of § 5.2 should be inserted the following:

{lem:.}

12 Lemma. Suppose F : B → C and G: C × B → C. Then there is a unique function H from B × ℕ to C such that

(i) a H 0 = F (a), (ii) a H n^{+} = (a H n) G a. Proof. By the Recursion Theorem, there is a unique sub-class R of ℕ×(B ×C) such that

(i) 0R = F, (ii) n ∈ ℕ ⇒ n^{+}R = ((x, y) ↦ (x, y G x))[ nR] = {(x, y G x)): n R (x, y)}. By induction, if n ∈ ℕ, then nR is a function K_{n} from B to C. Indeed, 0R is such a function, namely F; this then is K_{0}. If nR is such a function, as K_{n}, then n^{+}R is x ↦ H_{n}(x) G x; this then is K_{n^{+}}. We can now define the binary function H as (x, y) ↦ K_{y}(x) on A × ℕ. Then

(i) a H 0 = K_{0}(a) = F (a), (ii) a H n^{+} = K_{n^{+}}(a) = K_{n}(a) G a = (a H n) G a. So H is as desired; its uniqueness is as in Lemma 9.

• The proof of Theorem 5.3.2 can be supplied as follows:

Proof. We follow the pattern of the proof of Theorem 10. By the lemma, there is a unique function H from A × ℕ into A such that

(i) a H 0 = i,

(ii) a H n^{+} = a H n + a.

By induction, i H n = i for all n in ℕ; indeed, this is given when n = 0, and if it holds when n = m, then i H m^{+} = i H m + i = i H m = i. Let F be the unique homomorphism from (ℕ,^{+}, 0) into (A, S, i). The equation

S(a) H n = a H n + F (n)  (3) {eqn:.n}

holds when n = 0, since S(a) H 0 = i = i+i = a H 0+F (0). Suppose (3) holds for some n in ℕ. Then

S(a) H n^{+} = S(a) H n + S(a)  [by definition of H]
= (a H n + F (n)) + S(a) [by inductive hypothesis]
= a H n + (F (n) + S(a)) [by associativity of +]
= a H n + S(F (n) + a)  [by definition of +]
= a H n + (S(F (n)) + a) [by Lemma 5.2.5]
= a H n + (F (n^{+}) + a)  [because F is a homomorphism]
= a H n + (a + F (n^{+}))  [by commutativity of +]
= (a H n + a) + F (n^{+})  [by associativity of +]
= a H n^{+} + F (n^{+}).  [by definition of H]

So (3) holds for all n in ℕ. Therefore each of the operations x ↦ x H n is the operation G_{n} recursively defined by

(i) G_{n}(i) = i,

(ii) G_{n}(S(a)) = G_{n}(a) + F (n).

In particular,

G_{m} = G_{n} ⇔ F (m) = F (n).

Now we can define multiplication on A by

a · b = c ⇔ ∃x (F (x) = b & G_{x}(a) = c).

Then a · i = G_{0}(a) = a H 0 = i. Also, if b = F (n), so that S(b) = F (n^{+}), then

a · S(b) = G_{n^{+}}(a) = a H n^{+} = a H n + a = G_{n}(a) + a = a · b + a.

Thus · is as desired; it is unique by Theorem 4.

• The proof of Theorem 5.3.5 can be replaced with a reference to Lemma 12.

Trivial changes

• ¶ 4.4.3: it can be noted that the last part of the proof is by contradiction.

• P. 4, item (i), after ‘V in these notes’: insert reference to ¶ 2.2.7.

• P. 8: include Table 2.1 on p. 18 (best done by changing the Table to a Figure).

• P. 9: capitalize the letters after the hyphens in ‘Replacement-scheme’ and ‘Power-set’.

• P. 10: transpose ¶ 1.1.1 to read:

A set is a thing that contains other things. Those other things are called members or elements of the set. The set cannot be separated from its elements the way a box can be emptied of its contents: the set comprises its members, and the members compose the set. A set is its elements, considered as one thing. It is a multitude that is also a unity.

• P. 15, caption to Table 1.1: Replace sentence ‘A terminal ω… ’ with The vowels α, η, and ω may have an iota subscript (�, �, �).

• P. 16, after the first list of 3 items: Delete repeated ‘recursively’ (and add to index).

• After the second list of 3 items: change ‘is’ to ‘of’; don’t capitalize ‘Parts’.

• P. 20, n. 4: ‘The latter sequence that gives… ’: delete ‘that’.

• P. 21, ¶ 2.2.4, item (iii): ‘Then ∃x φ is true… ’

• P. 23, ¶ 2.3.5, item (iii): ‘(where a is allowed to appear in σ)’: change σ to φ.

• P. 23, ¶ 2.3.5: ‘this rule allows us to obtain the sentence τ… ’

• P. 24: allow Fig. 2.2 to float to the top of a page?

• P. 30, ¶ 3.2.3: replace ‘However, ⋃ a is a set’ with ‘However, the union of a set is a set’.

• ¶ 3.5.9: the meaning of greater than should perhaps have been given explicitly in ¶ 3.5.5.

• P. 36, ¶ 3.6.3: In the formula displayed over two lines, the terminal & on the first line should be repeated on the second (as this is the convention I use elsewhere).

• In the following line, replace to with (in)to.

• Pp. 38 f., ¶ 3.7.1: change E to C.

• P. 40: Exercise (3) should follow (5).

• P. 45, ¶ 4.1.8: slant chain as a technical term.

• Last line of text, but two: ‘… will be (in ¶ 5.1.3) another example… ’

• ¶ 4.2.2 can be broken into 3 paragraphs.

• ¶ 4.3.1: ‘This means by ¶ 4.2.1… ’; in item (ii): delete from C; afterwards: ‘The five numbered conditions here for being an arithmetic structure are sometimes… ’

• ¶ 4.4.1, last line but one: C should be D.

• p. 55, ¶ 5.1.1, just before (5.1): ‘Meanwhile we have’. In (5.2) and (5.3), the functions F are really sets and should be written that way. (Actually they are variables…)

• ¶ 5.1.3 (iv): change C to a (both times).

• ¶ 5.1.5: Give the numerical reference (4.3.5) for the Recursion Theorem.

• ¶ 5.3.7: Add: ‘For all a and b in A, and all m and n in ℕ’; in (ii), replace x ↦ x^{a} with x ↦ x^{m}.