Use the whole infinite real periodic sequence with nonempty period_values=[25, -26, 27/2], starting at index0 and repeating that full block forever. This is a declared infinite law, not an observed finite prefix. Use residue=0 and subsequence parameter j=25≥0. Give (residue, j, k_j, value, constant_subsequence_certificate, certificate).
G001 — Use the whole infinite real periodic sequence with nonempty period_values=[25, -26, 27/2], starting at index0 and repeating that full block forever. This is a declared infinite law, not an observed finite prefix. Use residue=0 and subsequence parameter j=25≥0. Give (residue, j, k_j, value, constant_subsequence_certificate, certificate).
Definition and scope
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ORIGINAL period index=0;value=25;absolute value=25
After
prefix absolute maximum after this original entry=25
Conditions: none
Scan absolute values over the full period rather than the maximum signed value.A negative entry may determine the bound;retain original positions and duplicates.
ORIGINAL period index=1;value=-26;absolute value=26
After
prefix absolute maximum after this original entry=26
Conditions: none
Scan absolute values over the full period rather than the maximum signed value.A negative entry may determine the bound;retain original positions and duplicates.
ORIGINAL period index=2;value=27/2;absolute value=27/2
After
prefix absolute maximum after this original entry=26
Conditions: none
Scan absolute values over the full period rather than the maximum signed value.A negative entry may determine the bound;retain original positions and duplicates.
ORIGINAL NONEMPTY full period=[25, -26, 27/2];period length=3
After
for EVERY n>=0,unique n=3q+r with0<=r<3 gives a_n=period_values[r];therefore |a_n|<=26
Conditions: none
Whole-sequence boundedness follows from the declared infinite periodic law and division with remainder,not observed future behavior.This proves the stated model,not the general Bolzano-Weierstrass theorem from finite samples.
ORIGINAL residue=0,period=3;k(j)=0+3j for EVERY j>=0;displayed j=25,k_j=75
After
for ALL j,k(j+1)-k(j)=3>0;remainder always0;a_k(j)=25;for EVERY epsilon>0 error0<epsilon
Conditions: none
Each index advances by a positive full period,so this is an infinite strictly increasing subsequence.Constant values have zero error for every epsilon.The displayed j is only an evaluation,not the universal proof.Existence of a convergent subsequence forces neither convergence of the original sequence nor a unique subsequential limit.
The full substitutions and premise checks give the original answer; source identity,format and status are unchanged.
This question review and source binding
Review date: 2026-10-26
Original questions, answers, step expressions, numeric ASTs and historical records are retained. Source candidate labels remain historical and are not overwritten as formal proof.