Paper VI visual guide · Visual 27 of 42 · REFINEMENT LIMITS / WHAT THE FINITE MODELS DO NOT PROVE

Small at each level is not summable across levels

Traditional Part I: Corollary 10.5 and Section 17. Summability and uniform control are additional data.

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For each \(n\ge1\), compare the two scalar operators \(A_n=0\) and \(B_n=1/n^2\) through the identity. Their ordinary defects have sizes \(1/n^2\), while their square-root defects have sizes \(1/n\):

\[ \sum_n|A_n-B_n|<\infty, \qquad \sum_n|A_n^{1/2}-B_n^{1/2}|=\infty. \]

These are independent scalar comparisons, not a constructed inverse system. They isolate an error-scale obstruction without claiming a coherent refinement tower.

Three partial sums distinguish accumulated defects from a small one-step value.
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Reading the drawing

The solid, dashed, and dotted curves are the three sums listed at right. The source note reserves the infinite-series conclusion for the exact formulas. The horizontal axis is the refinement cutoff.

Drawing labels and data 15 labels

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  • 0
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  • 12
  • 16
  • 0
  • 1
  • 2
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  • partial-sum cutoff \(N\)
  • sum of defect norms
  • solid: \(\sum_{n\le N}n^{-1}\)
  • dashed: \(\sum_{n\le N}n^{-2}\)

dotted: \(\frac12\sum_{n\le N}n^{-2}\)

The infinite-series conclusion
uses the exact formulas.

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Evidence note from the source edition

Partial sums through \(n=16\) illustrate the contrast. The convergence and divergence statements follow from the displayed series, not from the finite plot. A common positive gap gives the third, summable upper bound.

A uniform hypothesis can change the conclusion

Replace the pair by \(A_n=1\) and \(B_n=1+1/n^2\). Then

\[ \left|B_n^{1/2}-A_n^{1/2}\right| =\frac{1/n^2}{\sqrt{1+1/n^2}+1} \le\frac1{2n^2}. \]

This is the elementary form of the uniform-gap mechanism on Visual 16. Without such information, transferring a summable generator-defect budget through a fractional function can fail.

The formal release keeps three original research questions separate: realization of broader transfer semigroups, uniform stochastic-refinement control, and latent packet closure. The exact finite examples here do not settle their existence, domain, topology, or uniformity requirements.

Likewise, exact hidden-state elimination only yields the finite memory models that were supplied. It does not establish a finite latent realization for every memory law. A complete list of matrix identities can still leave an infinite-dimensional approximation problem untouched.

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A finite example can expose a missing hypothesis. It cannot supply a projective-limit construction, a uniform constant, or a continuum realization merely by passing at several levels.

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