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Reciprocals of the positive integers decrease to zero

Statement

In the complete ordered field R, the sequence ϵn:=1/(n+1), n∈N, is positive, strictly decreasing, and converges to zero.

Facts & Assumptions

[F1]

Complete ordered field (least-upper-bound property): every nonempty subset of an ordered field that is bounded above has a least upper bound.

[F2]

The natural numbers N (von Neumann), Addition of natural numbers, Addition is commutative, Left identity for addition, and Order on the natural numbers: N is the von Neumann set of naturals, its successor is σ(n)=n+1, and m≤n means m+k=n for some k∈N. Since 1+n=n+1=σ(n), every successor is at least 1. Natural indices remain in N; in field expressions the successor is carried into R by the canonical natural map of The canonical natural ι(n)=n⋅1F of a field.

[F3]

The canonical natural ι(n)=n⋅1F of a field and Canonical naturals are positive and strictly increasing: write ιN:N→R for the canonical natural map, with ιN(0)=0 and ιN(σ(n))=ιN(n)+1. It is positive on positive naturals and strictly increases there. Thus the displayed denominator n+1 means ιN(σ(n)), not a von Neumann set used as a real number.

[F4]

The unique embedding of ℚ into an ordered field: the rational embedding into R preserves order, so every rational q>0 has positive real image q^. The multiplicative identity is positive, Inverses of positives are positive, and reciprocation reverses order, and Sign rules for products and monotonicity of multiplication supply 1>0, positivity of the inverse of a positive real, and reversal of order by reciprocation on positive reals.

[F5]

Limits and Cauchy sequences of reals: a real sequence converges to 0 when, for every rational q>0, its absolute value is eventually less than q^.

Proof

Given: [F1]–[F5].

1.1F1F2F3F4algebra

The set NR:={ιN(n):n∈N} is nonempty. If it were bounded above, completeness [F1] would give s=sup⁡NR. Since 1>0 by [F4], s−1<s, so s−1 is not an upper bound; choose n∈N with ιN(n)>s−1. By the successor rule in [F3], ιN(σ(n))=ιN(n)+1>s, while σ(n)∈N, contradicting that s is an upper bound. Hence NR is unbounded above.

2.1F2F3F4F5step 1.1algebra∎

For every n∈N, σ(n)=1+n by natural-addition commutativity and the successor rule, so 1≤σ(n) by the definition of natural order; hence ιN(σ(n))>0 by [F3]. The successor recursion gives ιN(σ(σ(n)))=ιN(σ(n))+1>ιN(σ(n)), since 1>0. Reciprocation reverses this strict inequality by [F4], hence 0<ϵn and ϵn+1<ϵn. Now fix rational q>0 and let q^>0 be its real image. By step 1.1 choose N∈N with ιN(σ(N))>1/q^. If n≥N, then σ(N)≤σ(n) by [F2] and Order is compatible with addition, and the increasing-map clause of [F3] gives ιN(σ(n))≥ιN(σ(N)). Therefore 0<ϵn≤1/ιN(σ(N))<q^. This is convergence to zero by [F5].

Depends on

Dependency tree · 2 levels

Sources