Download Notes on the Frege-Russell theory of number, iii

Notes on the Frege-Russell theory of number, iii
Philosophy 5577 - Boston College
29 November 2016
• we’ve now given an account of what a (cardinal) number is:
– a number is a class of equinumerous sets.
• this easily explains how every set has a number
• it also explains much of how numbers form a structure 0, 1, 2, . . .
– in particular, we defined 0 and successor, and could then derive their
basic properties
– after defining addition, multiplication, and exponentiation we could
derive their familiar algebraic laws too
• still one quite basic fact has not been established:
– can we prove that M ⊕ 1 6= M for all M ?
• in fact we better not be able to prove this, because it is false
– let m be the set of all positive numbers 1, 2, . . .
– let n result from m by adding the element 0
– then, it is easy to see that m and n are equinumerous
∗ hence, m = n
– but also, n = m ⊕ 1
– therefore, m = m ⊕ 1
– hence, there are cardinals M such that M = M ⊕ 1
• still, there surely are some numbers M for which M ⊕ 1 6= M , namely
– the finite numbers
• but what makes a number finite?
• that is the problem for this batch of notes.
– we will see that the nature of finiteness underlies a fundamental
principle of reasoning
– the principle of mathematical induction
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Frege’s analysis of the ancestral
• Our present goal is to define the class of numbers 0, 1, . . .
– the concept seems clear enough
• What is this ‘. . . ’, or ‘etc.’ or ‘and so on’?
– can it be somehow explained?
– the answer is yes
Example - biological ancestry
• suppose a martian knows who you are, and also has the concept ‘parent’;
then can you explain to the martian just who are your ancestors?
– if s is the collection of your ancestors, then clearly. . .
i. your parents belong to s
ii. all parents of anybody in s also belong to s
• is the collection of your ancestors the only one satisfying i and ii?
– no: e.g., so does the set of all past or present organisms
• however: Frege’s proposal is that . . .
– somebody is your ancestor iff they belong to every collection which
satisfies i,ii
Example - finiteness of sets
• again, suppose your martian friend understands ‘empty set’ and ‘adding
an element’; you want to explain to him/her/it the concept of finite set
• as before, if X is the class of all finite sets, then
i. the empty set belongs to X
ii. every result of adding-one-element to something in X also belongs to
X
• we say that a set is finite if it belongs to every class satisfying i,ii
The definition of finite cardinal number
• Frege’s definition of finite cardinal number follows the same strategy
• say that a collection X is inductive iff
i. 0 belongs to X, and
ii. if N belongs to X, then so does N ⊕ 1.
• now, here is the definition:
– something is a finite cardinal number iff it belongs to every inductive
collection
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Mathematical induction
• Why is this a good analysis of finite number?
• The main reason is that it justifies a fundamental method of reasoning
• Let me introduce the method with an example
Example - internal angles of a polygon
• say that a polygon is a two-dimensional shape with straight sides
– so triangles, quadrilaterals, pentagons, hexagons etc., are all polygons
• you probably know about the sum of the internal angles of polygons
– the internal angles of a triangle sum to 180
– the internal angles of a quadrilateral sum to 360, . . .
• what about the general case?
• let X be the set of all numbers N such that. . .
– the interior angles of all N ⊕ 3-sided polygons sum to (N ⊕ 1) ⊗ 180
• let’s now argue for two things
i. 0 belongs to X
ii. if N belongs to X, then so does N ⊕ 1
• regarding i. . .
– an 0 ⊕ 3-sided polygon is just a triangle;
– (0 ⊕ 1) ⊗ 180 = 180;
– and the internal angles of a triangle sum to 180 (as is well known).
• regarding iii. . .
– suppose that N belongs to X
– then, the interior angles of all N ⊕ 3-sided polygons sum to (N ⊕ 1) ⊗
180
∗ let p be a polygon with N ⊕ 3 ⊕ 1 sides
∗ given three consecutive corners of p, connect the first and third
∗ the result is a polygon q with N ⊕ 3 sides
∗ by assumption, the internal angles of q sum to (N ⊕ 1) ⊗ 180
∗ but, the sum of the internal angles of p is 180 more than the sum
of the internal angles of q
∗ so, the sum of the internal angles of p is 180 is (N ⊕ 1 ⊕ 1) ⊗ 180
– therefore, N ⊕ 1 belongs to X
•
•
•
•
now, we just proved that the set X is inductive
but by definition, a natural number belongs to every inductive set
therefore, every natural number belongs to X
so, the claim is proved
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Example - counting the subsets of a finite set
• say that t is a subset of s, or t ⊆ s, provided that every element of t is an
element of s
– thus, every set is a subset of itself
– also the empty set is a subset of every set
– the subsets of a set are not the same as its elements: e.g.,
∗ {0, 1} has four subsets, but only two elements
∗ of course, sometimes a subset of a set is also an element, as in
{{}}
• Notice that a set with 0 elements has one subset, a set with one element
has two subsets, an set with two elements has four subsets, etc.
• This suggests that a finite set with N elements has 2N subsets
– Let’s say that a set n is good if it has 2n subsets
– then, the conjecture is that every set with a finite number of elements
is good
• To prove this, it is enough to prove two things
i. any set with 0 elements is good
ii. if all sets with N elements are good, then so are all sets with N ⊕ 1
elements
• The proof of i is straightforward:
– only one set has 0 elements, namely {}, and it has exactly 1 = 20
subsets
• as for ii,
– suppose every set with N elements is good
now suppose m has N ⊕ 1 elements
then, m = n ∪ {q} for some set n with N elements
by our assumption, n has 2N subsets
but every subset of m is the result of taking a subset of n, and
either adding or not adding the new thing q
∗ so m has 2 ⊗ 2N = 2N ⊕1 elements
∗ so m is good; but m was arbitrary
∗
∗
∗
∗
– so, every set with N ⊕ 1 elements is good!
• just as in the last argument, now we can conclude that for every number
N , every set with N elements is good.
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The moral of the story - mathematical induction
• The method of the argument of the last example is quite pervasive in math
• You want to show that every number has some property P
• Now you prove two things
i. 0 has P
ii. if N has P , then so does N ⊕ 1 for all N
• it follows that the set of all things with property P is inductive
• but by definition, something is a natural number iff it belongs to all
inductive sets
• from what we’ve shown, it follows that every natural number has P
• We’ve now completed the development of the basic structure of number
• we derived
– the basic laws of 0 and successor
– the algebra of addition, multiplication and exponentiation
– the principle of mathematical induction
• together these give what are known as the Peano axioms of arithmetic
• and that is the basis of classical number theory
The successor of a finite number is a different finite number
• At least let’s prove the thing we wanted to prove
– namely, N ⊕ 1 6= N for all N
• Indeed let’s prove something stronger
– say that t is a proper subset of s provided that t is a subset of s but
t 6= s
– now suppose that n is a set with N elements; let’s show that n is not
equinumerous with any of its proper subsets
• to this end, first note this lemma:
– if s is equinumerous with t and if s contains at least one element,
then some result of removing an element from s is equinumerous to a
result of removing an element from t
∗ Proof of lemma. Suppose s is equinumerous with t, so that R is a
one-one correspondence from s to t. By assumption, s contains at
least one element, say a. Then R must contain a pair (a, b). Let
s0 , t0 be the results of removing a from s and b from t. And let
R0 be the result of removing (a, b) from R. Then R0 is a one-one
correspondence from a0 to b0 . So, a0 , b0 are equinumerous.
• now for the proof of the desired result
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• let X be the set of all numbers N such that no set with N elements is
equinumerous to any of its proper subsets
• we will argue that X is inductive
i. The set with 0 elements is not equinumerous with any of its proper
subsets, because it doesn’t have any proper subsets at all!
ii. Suppose no set with N elements is equinumerous with any of its
proper subsets. Now, let m have N ⊕ 1 elements. And suppose m
is equinumerous with one of its proper subsets. Well, m must have
at least one element. So by the lemma, some set with n elements is
equinumerous with one of its proper subsets. This contradicts the
original assumption.
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