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

Naïve set theory 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 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 z 1,...,z n :  z 1...  z n  a  x [x  a  P(x)]

Naïve set theory 15.1.b 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]  z 1  z 2  a  x [x  a  (x=z 1  x=z 2 )] 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

Naïve set theory 15.1.c Proposition 1. For each wff P(x), the following is true:  z 1...  z n  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. “  a!Q(a)” abbreviates “  a[Q(a)  b(Q(b)  b=a)]”. I.e.,  a!Q(a) says that 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.

The empty set, singletons and unordered pairs 15.2 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].

Subsets 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. 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.

Subsets 15.3.b 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)].

Intersection and union 15.4.a Let a and b be sets. The intersection a  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 union a  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:

Intersection and union 15.4.b 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:  z 1...  z n  !c  x [x  c  P(x)]. The above two formulas are just instances of this.

Intersection and union 15.4.c Proposition 8. 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

Sets of sets 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}}

Ordered tuples 15.5.b An ordered pair: ; an ordered triple: ; 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 2 +y 2 =1} The main property of ordered pairs: =  (x=u  y=v) Generally, =  (x 1 =y 1 ...  x n =y n ) In set theory, is modeled (defined) as {{x},{x,y}}. Next, is modeled as >. Generally, is understood as >.

Modeling relations in set theory 15.6.a Relations between elements of a domain D are modeled as sets of ordered tuples: Larger: { | x,y  D and x is larger than y} Between: { | x,y,z  D and x is between y and z} etc. So, R(x,y) can be understood as an abbreviation of  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 ]

Modeling relations in set theory 15.6.b The inverse R -1 of a binary relation R is defined by R -1 = { |  R} LeftOf -1 = < -1 = HusbandOf -1 = SameRow -1 = Observation: R=R -1 iff

Modeling relations in set theory 15.6.c A binary relation R is 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-equivalence class [x] R of x by [x] R = {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  R 3. [x]=[y] iff [x]  [y] 

Functions 15.7 A function is a binary relation R on a set D satisfying the condition of Functionality:  x   1 yR(x,y) Here   1 y 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  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 be defined on x iff x is in the domain of f. g is said to be an extension 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).

The powerset of a set 15.8.a 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.

The powerset of a set 15.8.b Proposition 13. For any set b, the Russell set for 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 of b.

Russell’s Paradox 15.9.a The universal 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.

Russell’s Paradox (reproduced in Fitch) 15.9.b 1.  x  y(y  x   y  y) 2. c  y(y  c   y  y) 3. c  c   c  c  Elim: 2 4.  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!

Zermelo Frankel set theory ZFC a 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 {{{...}}}.

Zermelo Frankel set theory ZFC b 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 , {  }, {{  }}, {{{  }}}, Axiom of Replacement: Given any set c and any function f with domain c, there is a set {f(x) | x  c}.

Zermelo Frankel set theory ZFC c 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.

15.10.d Sizes of sets The size |S| of a finite set S is just the number of its elements. But how about the sizes of infinite sets? Their sizes (called cardinalities) can be compared as well! Let N={0,1,2,3,…} and E={0,4,6,8,…}. Do we have |E|<|N|? No. |N| is not bigger than |E| because one can assign, to each element of N, a unique element of E. E.g., this way: ….      …. The sets of cardinality |N|, along with finite sets, are said to be countable. Generally, given two sets S and T, we say that |S|  |T| iff there is an injective function (injection) f: S  T. “Injective” means that f assigns different elements of T to different elements of S. We say that |S|<|T| iff |S|  |T| but not |T|  |S|. And we say that |S|=|T| iff we have both |S|  |T| and |T|  |S|. Thus, |E|=|N|.

15.10.e Do all infinite sets have the same size? Is the set {… -3, -2, -1, 0, 1, 2, 3, …} countable? … … Yes:        … … How about the set of all (positive) rational numbers? Still countable: /1 2/1 3/1 4/1 1/2 2/2 3/2 4/2 1/3 2/3 3/3 4/3 1/4 2/4 3/4 4/ /1 5/2 5/3 5/4 5/5 4/5 3/5 2/51/5 1/1 2/1 3/1 4/1 1/2 2/2 3/2 4/2 1/3 2/3 3/3 4/3 1/4 2/4 3/4 4/4 5/1 5/2 5/3 5/4 5/5 4/5 3/5 2/51/5

15.10.f An uncountable set But there are sets properly bigger than N (uncountable). For instance, the set R of all rational numbers, or even just the set R [0,1) of all rational numbers r with 0  r<1 Every number from R [0,1) can be written as an infinite binary numeral of the form 0. x 1 x 2 x 3 x 4 … where each x i is either 0 or 1. Assume R [0,1) is countable. Then there should be a list of all elements of it, for instance, 0: : … 2: … 3: … 4: … 5: … etc. Take the “diagonal” and change all of its floating point bits, thus getting the number … This number cannot be on the list because it differs from each list item #n in the nth floating point bit. Contradiction! So, R [0,1) is uncountable.

15.10.g Cantor’s Theorem |N| is the smallest of all infinite cardinalities. According to the celebrated continuum hypothesis, the next-smallest cardinality is |R|. But there is no “biggest” cardinality. This follows from Cantor’s Theorem: For any set S, we have |S| < |  (S)| (i.e., the cardinality of S is strictly smaller than that of the powerset of S).