First-order Set Theory

1. Language, Proof and Logic First-order Set Theory Chapter 15

2. 15.1.a Two binary (and infix) predicate symbols in the language:  and =. Two sorts of variables (so that the language is many-sorted): a,b,c,... , ranging over sets x,y,z,... , ranging over everything --- ordinary objects as well as sets Naïve set theory Two axioms (premises to all arguments): 1. The Axiom of Extensionality: ab [x(xa  xb)  a=b] 2. The Axiom (axiom scheme) of Comprehension, for each formula P(x): ax[xa  P(x)] More generally, if P(x) contains additional free variables z1,...,zn: z1...znax [xa  P(x)]

3. 15.1.b Naïve set theory Instances of the Axiom of Comprehension. What do they say? ax [xa  StudentOfThisClass(x)] ax [xa  Likes(x,x)] ax [xa zLikes(x,z)] zax [xa  Likes(z,x)] ax [xa  x=x] ax [xa  xx] z1z2ax [xa  (x=z1  x=z2)] There is a set of all students of this class There is a set of those who like themselves There is a set of those who don’t like anyone For everyone, there is a set of those who he or she likes There is a set of all objects There is an empty set For any pair of objects, there is a set consisting of precisely those objects

4. 15.1.c Naïve set theory “a!Q(a)” abbreviates “a[Q(a)b(Q(b)  b=a)]”. I.e.,a!Q(a) says that Proposition 1. For each wff P(x),the following is true: z1...zna!x [xa  P(x)]. Proof. That there is at least one such a is guaranteed by the Axiom of Comprehension. And that any two such a’s would be equal is guaranteed by the Axiom of Extensionality. In informal contexts, we use: Brace notation: { x | P(x)} --- “the set of objects satisfying P(x)’’ E.g. {x | zLikes(x,z)}, {x | x does not like anyone}, etc. List notation: Just list the elements of the set between braces. E.g. {a,b,c}, {0,2,4,6,...}, etc.

5. 15.2 The empty set, singletons and unordered pairs The empty set: {x | xx} Informally, denoted by , or {}. The singleton set of x: {x}. The unordered pair of x and y: {x,y}. Proposition 1 guarantees that all these sets exist and are unique. Namely: a!z [za  zz]; xa!z [za  z=x]; x ya!z [za  z=xz=y].

6. 15.3.a a is a subset of b (ab) iff every element of a is also an element of b. ab can and will be understood as an abbreviation of An alternative approach to : Treat it as a legal predicate symbol of the language of set theory, and add ab[a  bx(xa  xb)] as a (one more) axiom to the system. Subsets Proposition 4.a(aa). I.e., ax(xa  xa). Proof: By Reiteration, we can derive  from . So, by  Intro, we get. Now, by  Intro,we get x(x  x). And, again by  Intro,we get ax(xa  xa), i.e. a(aa), as desired.

7. 15.3.b Subsets Proposition 5.ab [a=b (ab  ba)]. I.e., Proof: Consider any sets a and b. Assume a=b. By Proposition 4, aa and hence, of course, we also have aa aa. Then, by = Elim, ab  ba. Now assume ab  ba. This means ... a and b have the same elements. But then, by the Axiom of Extensionality, a=b. Thus, a=b (ab  ba). Since a and b were arbitrary, we conclude that ab [a=b (ab  ba)].

8. 15.4.a Intersection and union Let a and b be sets. The intersectionab of a and b is the set whose members are just those objects in both a and b. Here is a definition of ab, which can be used as an axiom in proving things about intersection: The unionab of a and b is the set whose members are just those objects in either a or b or both. Here is a definition of ab, which can be used as an axiom in proving things about union:

9. 15.4.b Intersection and union Propositions 6 and 7. Any pair of sets a and b has a unique intersection and a unique union: Proof. According to Proposition 1, for each wff P(x),the following is true: z1... zn!cx [xc  P(x)]. The above two formulas are just instances of this.

10. 15.4.c Intersection and union • Proposition8. Let a, b and c be any sets. • 1. ab =ba • 2. ab =ba • 3. ab=b iff ba • 4. ab=b iff ab • 5. a  (bc) = (ab)  (ac) • 6. a  (bc) = (ab)  (ac) You try it, p. 427

11. 15.5.a Using the FO language of set theory, say that: 1. a = {0} 2. a = {0,1} 3. a =  4. a = {} 5. a = {{}} 6. a = {,{}} 7. a = {{0},{1}} 8. a = {{0},{0,1}} Sets of sets

12. 15.5.b • An ordered pair: <x,y>; an ordered triple: <x,y,z>; etc. • How do these differ from {x,y}, {x,y,z}, etc.? • Important structures! E.g., points on the plane are understood • as ordered pairs of real numbers, and points in the space as ordered • triples of real numbers. • E.g., circle of radius 1 is the set of ordered pairs defined by • {<x,y> | x2+y2=1} • The main property of ordered pairs: • <x,y>=<u,v> (x=u  y=v) • Generally, <x1,...,xn>=<y1,...,yn> (x1=y1 ...  xn=yn) • In set theory, <x,y> is modeled (defined) as • {{x},{x,y}}. • Next, <x,y,z> is modeled as <x,<y,z>>. • Generally, <x1,x2,...,xn> is understood as <x1,<x2,...,xn>>. Ordered tuples

13. 15.6.a Modeling relations in set theory Relations between elements of a domain D are modeled as sets of ordered tuples: Larger: {<x,y> | x,yD and x is larger than y} Between: {<x,y,z> | x,y,zD and x is between y and z} etc. So, R(x,y) can be understood as an abbreviation of <x,y>R. Important possible properties of a binary relation R: Transitivity: xyz [(R(x,y)  R(y,z)) R(x,z)] Reflexivity: xR(x,x) Irreflexivity: xR(x,x) Symmetry: xy[R(x,y)  R(y,x)] Asymmetry: xy[R(x,y) R(y,x)] Antisymmetry: xy[(R(x,y)  R(y,x)) x=y]

14. 15.6.b Modeling relations in set theory • The inverseR-1of a binary relation R is defined by • R-1 = {<x,y> | <y,x>R} • LeftOf-1 = • <-1= • HusbandOf-1= • SameRow-1= Observation: R=R-1 iff

15. 15.6.c Modeling relations in set theory • A binary relation Ris said to be an equivalence relation iff R has • each of the following three properties: • Reflexive • Symmetric • Transitive • Given an equivalence relation R on a set D and an object xD, • we define the R-equivalenceclass[x]R of x by • [x]R = {y | <x,y>R} • Proposition 9. Let R be an equivalence relation on a set D. Then, for • all x,yD, we have: • 1. x[x] • 2. [x]=[y] iff <x,y>R • 3. [x]=[y] iff [x][y]

16. 15.7 Functions A function is a binary relation R on a set D satisfying the condition of Functionality: x1yR(x,y) Here 1y means “there is at most one y such that...” Such a function R is said to be total iff it satisfies the condition of Totality: xyR(x,y) A common practice is to denote functions by f,g,...,and write f(x)=y rather than <x,y>f. The domain of a function f is the set {x |y(f(x)=y)} The range of a function f is the set {y |x(f(x)=y)} f is said to bedefined on x iff x is in the domain of f. g is said to be anextension of f iff the domain of f is a subset of the domain of g and, for every x from the domain of f, g(x)=f(x).

17. 15.8.a The powerset of a set Proposition 10.For any set b there is a unique set whose members are just the subsets of b. In symbols:b!cx(xc  xb). Proof: Immediately from Proposition 1, as always. The set whose existence is claimed in Proposition 10 is called the powerset of b and is denoted by (b). Proposition 11. Let a and b be any sets. 1. b (b). 2. (b). 3. a  b iff (a) (b). Proof: Obvious.

18. 15.8.b The powerset of a set Proposition 13.For any set b, theRussell setfor b, the set {x | xb  xx} is a subset of b but not a member of b. Proof: Let c be the Russell set for b. That cb is obvious. Now, for a contradiction, assume cb. But do we then have cc or cc? Neither is possible! Proposition 12. For any set b, it is not the case that (b)b. Proof: An immediate corollary of Proposition 13. Specifically, the Russel set for b is a member of (b)but not a member ofb.

19. 15.9.a Russell’s Paradox Theuniversal set V is one containing everything, and can be defined by V = {x | x=x}. The existence of such a set is guaranteed by the axiom of comprehension. Since V contains everything, every set is a subset of it. Thus, we have (V)V. But this is in contradiction with Proposition 12 !!!!!!!!!!!!!! NAIVE SET THEORY is INCONSISTENT!!!!!!!!!!!!!!!!!!!! The above discovery is called Russell’s Paradox. It can be reproduced without referring to powersets or the universal set. Russell’s Paradox made easy: Consider the set Z = {x | xx}(Z is nothing but the Russell set for V) Do we have ZZ? Neither “yes” nor “no” is possible.

20. 15.9.b Russell’s Paradox (reproduced in Fitch) • 1. xy(yx yy) • c y(yc yy) • cc cc Elim: 2 • Taut Con: 3 • 5. Elim: 1, 2-4 From the above, by  Intro, we can further derive xy(yx yy). This means that the Axiom of Comprehension is not only false, it is in fact logically (FO) false, no matter what  means!

21. 15.10.a Zermelo Frankel set theory ZFC • ZFC is obtained from naive set theory by leaving untouched the • 1. Axiom of Extensionality, • then, for the purpose of ruling out “too large” sets, weakening the • Axiom of Comprehension to the • 2. Axiom of Separation: abx[xb  (xa  P(x))]; • This, however, throws out the baby with the bath water. To save the • baby, the following five axioms 3-7 are added back, each being an • easy consequence of the old Axiom of Comprehension. • Plus, the (controversial) Axiom of Choice is added, • plus the Axiom of regularity, which rules out non-cumulative • (formed before their members were formed) sets such as {{{...}}}.

22. 15.10.b Zermelo Frankel set theory ZFC 3. Unordered Pair Axiom: For any two objects there is a set that has both as elements. 4. Union Axiom: Given any set c of sets, the union of all the members of c is also a set. 5. Powerset Axiom: Every set has a powerset. 6. Axiom of Infinity: There is a set of all natural numbers. Here natural numbers can be understood as , {}, {{}}, {{{}}},... 7. Axiom of Replacement: Given any set c and any function f with domain c, there is a set {f(x) | xc}.

23. 15.10.c Zermelo Frankel set theory ZFC 8. Axiom of Choice: If b is a set of non-empty sets, then there is a function g with domain b such that, for each cb, g(c)c. That is, g “chooses” an element from each member c of b. 9. Axiom of Regularity: No set has a nonempty intersection with each of its elements. Rationale for the Axiom of Regularity: Consider any set c. Pick an element b of c which was formed at least as early as any other element of c. Assume now that x is in both b and c. Since x is in b, b was formed later than x. But it was our assumption that no element of c was formed earlier than b.