analysis of number ii
Notes on the Frege-Russell theory of number, ii Philosophy 5577 - Boston College 15 November 2016 • We have now reached a general definition of number – a number is a class of equinumerous sets – sets are equinumerous if there is a one-one correspondence between them • The analysis of number had two goals: – explain how number has application – predict that numbers have the structure we assume them to have • Our definition derived from analysis of application of number – so it makes sense of application of number pretty readily: ∗ whenever a bunch of things forms a set, exactly one thing is the number of elements in the set. • But we have not yet dealt with the question of structure – we suppose numbers to satisfy such principles as ∗ 0 is a number ∗ 0 is not the successor of a number ∗ every number has a sucessor. ∗ nothing is the successor of two numbers – and indeed to support other ideas like addition, multiplication, etc. • The job of this batch of notes is to show that these statements can be proven
The definitions of zero and successor • We’re interested in proving statements about number which mention zero and successor – until we know what these are, we can hardly hope to prove interesting things about them 1
– our general approach will be to define properties of number in terms of properties of sets which have those numbers • To state things more concisely, here’s introduce a bit more notation • Since every set s belongs to exactly one equinumerosity class, it has exactly one number – we call this the cardinality of s and denote it by s. • Also, here is a notion of more general set theory – the union of sets s and t is the set whose elements are precisely the elements of at least one of s and t – write s ∪ t for the union of s and t. – then for example. . . ∗ {Venus, Neptune} ∪ {Pluto} = {Venus, Neptune, Pluto}; ∗ {{}} ∪ {} = {{}} = {{}} ∪ {{}}.
Defining zero • Let’s start by defining the cardinal number 0 • As we have seen, a number is a class of equinumerous sets • to specify any particular number N , you just need to – pick out a set n and say “N is the class of sets equinumerous with n” • So to define 0, we just need to find a set with zero elements – a set with zero elements is the empty set {} • now the definition of 0 is this: – 0 is the class of sets equinumerous with {} – in other words, 0 = {}. • How many sets are equinumerous with {}? – exactly one, namely {}! ∗ prove this yourself – therefore, 0 = {{}} ∗ note that this is not just the definition of zero ∗ it requires the assumption about sets that no two sets have the same elements
Defining successor • The definition of successor requires a little more thought
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• We will begin finding an example of a set which has ‘one more’ element than another – intuitively, if s is the result of adding a new element to t, then s has one more element than t does • in the new notation you can state our idea like this: – if x 6∈ m and n = m ∪ {x}, then n has one more element than m ∗ note that the converse is not true ∗ find an example yourself! • we now define successor like this: – a number N succeeds a number M iff some set whose cardinality is N is the result of adding a new element to a set whose cardinality is M – or in other words. . . ∗ N succeeds M iff ∃m(m = M ∧ ∃x(x 6∈ m ∧ m ∪ {x} = N ))
Proving the axioms • Let’s now prove the four basic statements about number that we noted earlier • The key thing to note is that the proofs. . . – use only the definitions of number plus much more general notions of logic • Mostly the axioms are proved with routine manipulations, so don’t get bogged down in the details • It is a little more interesting with the axiom that that every number has a successor
0 is a number 1. The definition of 0 says that 0 = {}. 2. So there’s an n such that 0 = n. 3. So by the definition of number, 0 is a number.
0 is not the successor of a number 1. There cannot be a one-one correspondence between a nonempty set and the empty one. 2. Therefore, n = 0 only if n is empty. 3. But no set of the form m ∪ {x} is empty. 3
4. Yet 0 would be the successor of a number only if m ∪ {x} = 0 for some m, x. 5. Therefore, 0 is not the successor of a number.
Nothing is the successor of two numbers So far, we are not justified in talking about the successor of a number. Even if a number has a successor, we cannot yet suppose that it has only one. Let’s fix that. 1. Suppose K is the successor both of M and of N . I’ll argue that M = N . 2. By definition of successor, there are sets m, n and objects x, y such that • m = M and x 6∈ m and m ∪ {x} = K • n = N and y 6∈ n and n ∪ {y} = K 3. So there is a one-one correspondence R : m ∪ {x} ∼ n ∪ {y} 4. If (x, y) ∈ R, then the result of removing it from R is a one-one correspondence betweeen m and n. 5. Otherwise, we have (x, a) ∈ R and (b, y) ∈ R for some a ∈ n, b ∈ m; deleting these from R and adding (a, b) to the result is a one-one correspondence between m and n. 6. So in any case, m ∼ n. 7. So M = N , as desired.
Every number has a successor • Historically, the proof of this statement was quite a bit more challenging • To see the problem, let N be a number. – the successor of N would be the cardinality of a set n ∪ {x}, where n = N and x 6∈ n – so if N has a successor, then the number of objects that exist in the universe is greater than N ! • This is a serious stumbling block • There are a few ways around it, each with tradeoffs • Russell’s approach was this: simply to assume that infinitely many objects exist. – it doesn’t really matter what the objects are ∗ e.g. maybe they are spacetime points? – he called this ‘the axiom of infinity’ • Frege’s approach is more diabolical 4
– we cannot quite reconstruct it rigorously ∗ for one thing, it requires the notions of ‘finite’ and of ‘less than’ on the finite cardinals, which we haven’t yet defined ∗ there are also some deeper foundational problems with Frege’s construction – but the idea is this. ∗ ∗ ∗ ∗
let N be a finite cardinal let n be the set of all M such that M is less than N then n = N but N 6∈ n
∗ so, n ∪ {N } is the successor of N ! • A modern approach is somewhat similar – but instead of constructing a sequence of cardinals – we just construct a sequence of sets – each set in the sequence is the set of all earlier sets, thus: {} {{}} {{}, {{}}} {{}, {{}}, {{}, {{}}}} {{}, {{}}, {{}, {{}}}, {{}, {{}}, {{}, {{}}}}} .. . – These are known as the von Neumann ordinals – People use the nth entry in the sequence as a ‘canonical’ example of a set with n elements – that is, the nth cardinal number is the cardinality of the nth von Neumann ordinal
Further arithmetical concepts A big portion of the foundation of arithmetic is now complete. We can now build up some theory.
Other numbers • So far we have defined the concept of being a number in general • we have also defined the number 0 5
• also we have shown that every number has exactly one successor • therefore, quite a few numbers are definable • to begin with, we can define 1 = the successor of 0 • About this new object we can prove the following important result: – 1 6= 0 – proof: ∗ by definition, 1 is the successor of a number ∗ But 0 is not the successor of a number ∗ Therefore, 1 6= 0. • Similarly, 2 = the successor of 1 • and it is routine to show – 2 6= 0 – 2= 6 1 • it seems like this could keep you busy for a while
Addition • Now it is time to progress from preschool to kindergarten • How to define addition? – well, if you have eight goats and four donkeys, then how many goats or donkeys do you have? ∗ should we say that if m = M and if n = N , then m ∪ n is the ‘sum’ of M and N ? · No! e.g., {a} ∪ {a} = 6 2. – instead the definition is this ∗ say that sets are disjoint if they have no elements in common ∗ K is the sum of M and N iff some set with cardinality K is the union of disjoints sets m, n with cardinalities M, N ∗ let’s write M ⊕ N for the sum of M and N • you can now establish the following facts: – – – –
M ⊕0=M 1⊕1=2 M ⊕N =N ⊕M K ⊕ (M ⊕ N ) = (K ⊕ M ) ⊕ N
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Multiplication • You have eight goats and four donkeys. You need one goat and one donkey to pull the cart. How many options do you have? • the definition is this: – say that the cartesian product of sets m, n is the set of all pairs (a, b) such that a ∈ m and b ∈ n. – K is the product of M and N iff some set with cardinality K is the cartesian product of sets m, n with cardinalities M, N – let’s write M ⊗ N for the product of M and N • You can now prove a bunch more laws, including – – – – –
M ⊗0=0 M ⊗1=M M ⊗N =N ⊗M K ⊗ (M ⊗ N ) = (K ⊗ M ) ⊗ N K ⊗ (M ⊕ N ) = (K ⊗ M ) ⊕ (K ⊗ N )
Exponentiation • You have eight goats and four donkeys. Tonight, each goat is to sleep with exactly one donkey. What is the number of possible sleeping arrangements? – more generally. . . ∗ say that a function from m to n is a many-one relation R such that · the domain of R is m · every element of the converse of R belongs to n ∗ then K is the N th power of M provided that some set with cardinality K is the set of functions from n to m, where m, n have cardinalities M, N ∗ Once again some laws:1 – – – – – – 1 The
0M = 0 M0 = 1 1M = 1 K M ⊕N = K M ⊗ K N (M ⊗ N )K = M K ⊗ N k (K M )N = K M ⊗N .
proof of the last one uses a (googleable) trick known as currying.
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Are we done? • At this point, it may look as though we have established the whole theory of arithmetic – is that so? • here is a question: – can you prove M ⊕ 1 6= M ? – the answer is no • there are cardinalities M such that M ⊕ 1 = M . – – – – – –
let m be a set which contains infinitely many objects o0 , o1 , o2 , . . .. let n be the result of adding some new object p to m then n = m ⊕ 1 but, you can put m, n into one-one correspondence (exercise!) so m = n so, m = m ⊕ 1.
• so clearly we are missing something – this is the distinction between finite and infinite
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The definitions of zero and successor • We’re interested in proving statements about number which mention zero and successor – until we know what these are, we can hardly hope to prove interesting things about them 1
– our general approach will be to define properties of number in terms of properties of sets which have those numbers • To state things more concisely, here’s introduce a bit more notation • Since every set s belongs to exactly one equinumerosity class, it has exactly one number – we call this the cardinality of s and denote it by s. • Also, here is a notion of more general set theory – the union of sets s and t is the set whose elements are precisely the elements of at least one of s and t – write s ∪ t for the union of s and t. – then for example. . . ∗ {Venus, Neptune} ∪ {Pluto} = {Venus, Neptune, Pluto}; ∗ {{}} ∪ {} = {{}} = {{}} ∪ {{}}.
Defining zero • Let’s start by defining the cardinal number 0 • As we have seen, a number is a class of equinumerous sets • to specify any particular number N , you just need to – pick out a set n and say “N is the class of sets equinumerous with n” • So to define 0, we just need to find a set with zero elements – a set with zero elements is the empty set {} • now the definition of 0 is this: – 0 is the class of sets equinumerous with {} – in other words, 0 = {}. • How many sets are equinumerous with {}? – exactly one, namely {}! ∗ prove this yourself – therefore, 0 = {{}} ∗ note that this is not just the definition of zero ∗ it requires the assumption about sets that no two sets have the same elements
Defining successor • The definition of successor requires a little more thought
2
• We will begin finding an example of a set which has ‘one more’ element than another – intuitively, if s is the result of adding a new element to t, then s has one more element than t does • in the new notation you can state our idea like this: – if x 6∈ m and n = m ∪ {x}, then n has one more element than m ∗ note that the converse is not true ∗ find an example yourself! • we now define successor like this: – a number N succeeds a number M iff some set whose cardinality is N is the result of adding a new element to a set whose cardinality is M – or in other words. . . ∗ N succeeds M iff ∃m(m = M ∧ ∃x(x 6∈ m ∧ m ∪ {x} = N ))
Proving the axioms • Let’s now prove the four basic statements about number that we noted earlier • The key thing to note is that the proofs. . . – use only the definitions of number plus much more general notions of logic • Mostly the axioms are proved with routine manipulations, so don’t get bogged down in the details • It is a little more interesting with the axiom that that every number has a successor
0 is a number 1. The definition of 0 says that 0 = {}. 2. So there’s an n such that 0 = n. 3. So by the definition of number, 0 is a number.
0 is not the successor of a number 1. There cannot be a one-one correspondence between a nonempty set and the empty one. 2. Therefore, n = 0 only if n is empty. 3. But no set of the form m ∪ {x} is empty. 3
4. Yet 0 would be the successor of a number only if m ∪ {x} = 0 for some m, x. 5. Therefore, 0 is not the successor of a number.
Nothing is the successor of two numbers So far, we are not justified in talking about the successor of a number. Even if a number has a successor, we cannot yet suppose that it has only one. Let’s fix that. 1. Suppose K is the successor both of M and of N . I’ll argue that M = N . 2. By definition of successor, there are sets m, n and objects x, y such that • m = M and x 6∈ m and m ∪ {x} = K • n = N and y 6∈ n and n ∪ {y} = K 3. So there is a one-one correspondence R : m ∪ {x} ∼ n ∪ {y} 4. If (x, y) ∈ R, then the result of removing it from R is a one-one correspondence betweeen m and n. 5. Otherwise, we have (x, a) ∈ R and (b, y) ∈ R for some a ∈ n, b ∈ m; deleting these from R and adding (a, b) to the result is a one-one correspondence between m and n. 6. So in any case, m ∼ n. 7. So M = N , as desired.
Every number has a successor • Historically, the proof of this statement was quite a bit more challenging • To see the problem, let N be a number. – the successor of N would be the cardinality of a set n ∪ {x}, where n = N and x 6∈ n – so if N has a successor, then the number of objects that exist in the universe is greater than N ! • This is a serious stumbling block • There are a few ways around it, each with tradeoffs • Russell’s approach was this: simply to assume that infinitely many objects exist. – it doesn’t really matter what the objects are ∗ e.g. maybe they are spacetime points? – he called this ‘the axiom of infinity’ • Frege’s approach is more diabolical 4
– we cannot quite reconstruct it rigorously ∗ for one thing, it requires the notions of ‘finite’ and of ‘less than’ on the finite cardinals, which we haven’t yet defined ∗ there are also some deeper foundational problems with Frege’s construction – but the idea is this. ∗ ∗ ∗ ∗
let N be a finite cardinal let n be the set of all M such that M is less than N then n = N but N 6∈ n
∗ so, n ∪ {N } is the successor of N ! • A modern approach is somewhat similar – but instead of constructing a sequence of cardinals – we just construct a sequence of sets – each set in the sequence is the set of all earlier sets, thus: {} {{}} {{}, {{}}} {{}, {{}}, {{}, {{}}}} {{}, {{}}, {{}, {{}}}, {{}, {{}}, {{}, {{}}}}} .. . – These are known as the von Neumann ordinals – People use the nth entry in the sequence as a ‘canonical’ example of a set with n elements – that is, the nth cardinal number is the cardinality of the nth von Neumann ordinal
Further arithmetical concepts A big portion of the foundation of arithmetic is now complete. We can now build up some theory.
Other numbers • So far we have defined the concept of being a number in general • we have also defined the number 0 5
• also we have shown that every number has exactly one successor • therefore, quite a few numbers are definable • to begin with, we can define 1 = the successor of 0 • About this new object we can prove the following important result: – 1 6= 0 – proof: ∗ by definition, 1 is the successor of a number ∗ But 0 is not the successor of a number ∗ Therefore, 1 6= 0. • Similarly, 2 = the successor of 1 • and it is routine to show – 2 6= 0 – 2= 6 1 • it seems like this could keep you busy for a while
Addition • Now it is time to progress from preschool to kindergarten • How to define addition? – well, if you have eight goats and four donkeys, then how many goats or donkeys do you have? ∗ should we say that if m = M and if n = N , then m ∪ n is the ‘sum’ of M and N ? · No! e.g., {a} ∪ {a} = 6 2. – instead the definition is this ∗ say that sets are disjoint if they have no elements in common ∗ K is the sum of M and N iff some set with cardinality K is the union of disjoints sets m, n with cardinalities M, N ∗ let’s write M ⊕ N for the sum of M and N • you can now establish the following facts: – – – –
M ⊕0=M 1⊕1=2 M ⊕N =N ⊕M K ⊕ (M ⊕ N ) = (K ⊕ M ) ⊕ N
6
Multiplication • You have eight goats and four donkeys. You need one goat and one donkey to pull the cart. How many options do you have? • the definition is this: – say that the cartesian product of sets m, n is the set of all pairs (a, b) such that a ∈ m and b ∈ n. – K is the product of M and N iff some set with cardinality K is the cartesian product of sets m, n with cardinalities M, N – let’s write M ⊗ N for the product of M and N • You can now prove a bunch more laws, including – – – – –
M ⊗0=0 M ⊗1=M M ⊗N =N ⊗M K ⊗ (M ⊗ N ) = (K ⊗ M ) ⊗ N K ⊗ (M ⊕ N ) = (K ⊗ M ) ⊕ (K ⊗ N )
Exponentiation • You have eight goats and four donkeys. Tonight, each goat is to sleep with exactly one donkey. What is the number of possible sleeping arrangements? – more generally. . . ∗ say that a function from m to n is a many-one relation R such that · the domain of R is m · every element of the converse of R belongs to n ∗ then K is the N th power of M provided that some set with cardinality K is the set of functions from n to m, where m, n have cardinalities M, N ∗ Once again some laws:1 – – – – – – 1 The
0M = 0 M0 = 1 1M = 1 K M ⊕N = K M ⊗ K N (M ⊗ N )K = M K ⊗ N k (K M )N = K M ⊗N .
proof of the last one uses a (googleable) trick known as currying.
7
Are we done? • At this point, it may look as though we have established the whole theory of arithmetic – is that so? • here is a question: – can you prove M ⊕ 1 6= M ? – the answer is no • there are cardinalities M such that M ⊕ 1 = M . – – – – – –
let m be a set which contains infinitely many objects o0 , o1 , o2 , . . .. let n be the result of adding some new object p to m then n = m ⊕ 1 but, you can put m, n into one-one correspondence (exercise!) so m = n so, m = m ⊕ 1.
• so clearly we are missing something – this is the distinction between finite and infinite
8
