Set Theory And Foundations Of Mathematics: An Introduction To Mathematical Logic - Volume I: Set Theory. Douglas Cenzer

Set Theory And Foundations Of Mathematics: An Introduction To Mathematical Logic - Volume I: Set Theory - Douglas Cenzer


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by ϵ. Σ (or sometimes Σ) denotes the set ⋃n∈ Σn and Σω, or Σ, denotes the set of infinite sequences.

      A constant string σ of length n consisting of the symbol k will be denoted kn. For m < |σ|, σm is the string (σ(0), . . . , σ(m − 1)); σ is an initial segment of τ (written στ) if σ = τm for some m. Initial segments are also referred to as prefixes. Similarly, τ is said to be a suffix of σ if |τ| ≤ |σ| and, for all i < |τ|, σ(|σ| − |τ| + i) = τ(i). The concatenation σ⌢τ (or sometimes στ or just στ) is defined by σ⌢τ = (σ(0), σ(1), . . . , σ(m−1), τ(0), τ(1), . . . , τ(n−1)), where |σ| = m and |τ| = n; in particular we write σ⌢a for σ⌢(a) and a⌢ σ for (a)⌢σ. Thus we may also say that σ is a prefix of τ if and only if τ = σ⌢ρ for some ρ and that τ is a suffix of σ if and only if σ = ρ⌢τ for some ρ.

      For any x ∈ Σ and any finite n, the initial segment xn of x is (x(0), . . . , x(n − 1)). We write σx if σ = xn for some n. For any σ ∈ Σn and any x ∈ Σ, we have σ⌢x = (σ(0), . . . , σ(n − 1), x(0), x(1), . . . ). The lexicographic, or dictionary ordering on Σ, is defined so that σ

τ if either στ or if σ(n) < τ(n), where n is the least such that σ(n) ≠ τ(n).

      Proposition 2.6.2. The relationis a partial ordering.

      The proof is left as an exercise.

      Finite strings in {0, 1} can be viewed as representing dyadic rational numbers.

      Definition 2.6.3. For σ ∈ {0, 1}, let = ∑i<|σ| 2−i−1σ(i).

      For example, q1101 represents

Note here that the number 13 in base two is just 1101, that is, 13 = 8 + 4 + 1.

      A similar definition can be given for numbers in base k when the alphabet Σ = {0, 1, . . . , k − 1}. We can now define the lexicographic order on {0, 1}.

      Definition 2.6.4. Let σ

τ if and only if either στ or < .

      The following facts are left as exercises.

      Fact 2.6.5. For any σ, τ ∈ {0, 1},

      • = qτ if and only if either στ or τσ;

      • if σ

τ, then qσ.

      There is another way to state that < .

      Lemma 2.6.6. For any σ, τ ∈ {0, 1}, < qτ if and only if there exists n such that σn = τn and σ(n) < τ(n).

      Proof. Suppose first that σn = τn and σ(n) < τ(n) (that is, σ(n) = 0 and τ(n) = 1), and let ρ = σn. Let |σ| = k. Then

      This can be used to define the lexicographic order over any well-ordered alphabet Σ, by having σ

τ if and only if either στ or there exists n such that σn = τn and σ(n) < τ(n).

      Proposition 2.6.7. The lexicographic order on Σ is a linear ordering.

      Proof. Reflexive: σσ, so that σ

σ.

      Antisymmetric: Suppose that σ

τ and τ
σ. There are two cases.

      Case 1: στ. In this case, = , so that τ σ implies that τσ. Therefore σ = τ, since ⊑ is antisymmetric.

      Case 2: < . In this case, τ

σ is not possible.

      Transitive: Suppose that σ

τ and τ
ρ. There are two cases.

      Case 1: στ. In this case, = . Now, there are two subcases.

      • In the first case, if τρ, then σρ and therefore σ

ρ.

      • In the second case, if < , then < and thus σ

ρ as desired.

      Case 2: < . Now τ

ρ implies that and therefore < . Thus σ
ρ.

      Total: This is left as an exercise.

      A tree T over Σ is a set of finite strings from Σ which is closed under initial segments. Then τT is an immediate successor of a string σT if τ = σ⌢a for some a ∈ Σ.

      We will generally assume that T

. That is, the nodes of T are finite sequences of natural numbers. Such a tree defines a subset [T] of the so-called Baire space Скачать книгу