TheoremBase

Existence comes from recursion on N with a step that multiplies by the next term below n and does nothing from n on, restricted to [n]; uniqueness is proved by induction on the set of k that exceed n or where the two maps agree.

Proof

Indices. By Intervals of Natural Numbers: Initial Segments, Adding One Element, Splitting and Shifting §segment, [n]={k∈N:k≤n}[n]=\{k\in\mathbb{N}:k\le n\}. Since 1≤n1\le n by Arithmetic and Order of the Natural Numbers §least, 1∈[n]1\in[n]. Let k∈Nk\in\mathbb{N} with k<nk<n. Then k+1≤nk+1\le n: otherwise n<k+1n<k+1 by Arithmetic and Order of the Natural Numbers §trichotomy and Arithmetic and Order of the Natural Numbers §partial-order, and k<n<k+1k<n<k+1 contradicts Arithmetic and Order of the Natural Numbers §successor. So k+1∈[n]k+1\in[n], and ak+1a_{k+1} is defined.

Existence. Let g:N×X→Xg:\mathbb{N}\times X\to X be given by g(k,u)=u∗ak+1g(k,u)=u\ast a_{k+1} if k<nk<n and g(k,u)=ug(k,u)=u otherwise, a map by Maps and Relations Given by Formulas §binary. By Recursion on the Natural Numbers Starting at One §recursion, applied with the set XX, the element a1a_{1} and the map gg, there is a map f:N→Xf:\mathbb{N}\to X with f(1)=a1f(1)=a_{1} and f(k+1)=g(k,f(k))f(k+1)=g(k,f(k)) for every k∈Nk\in\mathbb{N}. Let s=f∣[n]:[n]→Xs=f|_{[n]}:[n]\to X. Then s(1)=f(1)=a1s(1)=f(1)=a_{1}, and for k∈[n]k\in[n] with k<nk<n we have k+1∈[n]k+1\in[n], so

s(k+1)=f(k+1)=g(k,f(k))=f(k)∗ak+1=s(k)∗ak+1.s(k+1)=f(k+1)=g(k,f(k))=f(k)\ast a_{k+1}=s(k)\ast a_{k+1}.

Uniqueness. Let s,s′:[n]→Xs,s':[n]\to X both have the stated properties, and let AA be the set of k∈Nk\in\mathbb{N} such that either n<kn<k, or k≤nk\le n and s(k)=s′(k)s(k)=s'(k). Then 1∈A1\in A, since 1≤n1\le n and s(1)=a1=s′(1)s(1)=a_{1}=s'(1). Let k∈Ak\in A. If n<k+1n<k+1, then k+1∈Ak+1\in A. Otherwise k+1≤nk+1\le n by Arithmetic and Order of the Natural Numbers §trichotomy, and since k<k+1k<k+1 by Arithmetic and Order of the Natural Numbers §successor, Arithmetic and Order of the Natural Numbers §partial-order gives k<nk<n; in particular n<kn<k fails by trichotomy, so s(k)=s′(k)s(k)=s'(k) because k∈Ak\in A, and

s(k+1)=s(k)∗ak+1=s′(k)∗ak+1=s′(k+1),s(k+1)=s(k)\ast a_{k+1}=s'(k)\ast a_{k+1}=s'(k+1),

so again k+1∈Ak+1\in A. By Arithmetic and Order of the Natural Numbers §induction, A=NA=\mathbb{N}. For k∈[n]k\in[n] we have k≤nk\le n, so n<kn<k fails by trichotomy, and k∈Ak\in A gives s(k)=s′(k)s(k)=s'(k). Hence s=s′s=s' by Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction §equality.

Citations

Loading…

Dependencies

Uses0

Loading…

Comments

Log in to comment.

Loading…