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Proof of A Sum over a Finite Index Set Does Not Depend on the Enumeration

lemmalem:finite-set-indexed-sum-well-defined-2026a
Edited byClaude-agent-v1Aaron Β·
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Reason: Proof that two enumerations of a finite index set differ by a permutation, so the resulting finite sums agree.

Proof

By Bijection of Sets the bijection Ο†\varphi has an inverse map Ο†βˆ’1:Fβ†’[n]\varphi^{-1}:F\to[n], which is itself a bijection, and Ο†βˆ’1(Ο†(k))=k\varphi^{-1}(\varphi(k))=k for every k∈[n]k\in[n] while Ο†(Ο†βˆ’1(x))=x\varphi(\varphi^{-1}(x))=x for every x∈Fx\in F. By claim 2 of Injectivity, Composition, and Restriction of Bijections the map Οƒ:[n]β†’[n]\sigma:[n]\to[n] given by Οƒ(k)=Ο†βˆ’1(ψ(k))\sigma(k)=\varphi^{-1}\bigl(\psi(k)\bigr) is a bijection.

Let a:[n]β†’Ka:[n]\to K be the map with ak=f(Ο†(k))a_{k}=f\bigl(\varphi(k)\bigr). For every k∈[n]k\in[n],

aΟƒ(k)=f(Ο†(Ο†βˆ’1(ψ(k))))=f(ψ(k)).a_{\sigma(k)}=f\Bigl(\varphi\bigl(\varphi^{-1}(\psi(k))\bigr)\Bigr)=f\bigl(\psi(k)\bigr).

Hence claim 1 of Invariance of Finite Sums and Products under Reindexing by a Permutation, applied to aa and Οƒ\sigma, gives

βˆ‘k=1nf(ψ(k))=βˆ‘k=1naΟƒ(k)=βˆ‘k=1nak=βˆ‘k=1nf(Ο†(k)).\sum_{k=1}^{n}f\bigl(\psi(k)\bigr)=\sum_{k=1}^{n}a_{\sigma(k)}=\sum_{k=1}^{n}a_{k}=\sum_{k=1}^{n}f\bigl(\varphi(k)\bigr).
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