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Finite thresholds rule out an r-independent VC covering bound

Statement

For each positive integer m, let N:=16m be the natural number obtained by multiplying 16 by itself m times. Equip Xm:={1,…,N} with the uniform probability Qm({i})=1/N. For k=0,…,N, define the threshold function fk:=1{1,…,k}, with f0≡0. Put Fm:={f0,…,fN}.

These are nested threshold indicators. Their VC-subgraph index is V=2, while for r=m and ϵ=1/64,

N(ϵ,Fm,Lm(Qm))=N+1=16m+1.

Thus a purported entropy bound whose right side has no dependence on r is false even when the VC index is fixed.

Facts & Assumptions

Given: A positive integer m, the finite uniform probability space, and the threshold class in the Counterexample.

[F1]

A class of functions is VC-subgraph when its measurable strict subgraphs have finite VC index (VC-subgraph classes, envelopes, and the first-unshattered index); for set classes, the index is one plus the VC dimension (Shattering, VC dimension, and first-unshattered index for set classes).

[F2]

The Lr(Qm) distance is the rth-root of the integral of the absolute difference to power r, and covers use centers in the class (Covering, packing, bracketing numbers, and metric entropy).

[F4]

A finite uniform law assigns mass 1/N to each of its N points and is a probability measure by finite additivity (Probability measures and probability spaces).

[F5]

The natural powers 16m grow beyond every fixed real bound: (1+15)m≥1+15m by the binomial theorem, and the Archimedean property supplies arbitrarily large m (Integer powers am, The binomial theorem in R: (x+y)n=∑k<n+1ι ⁣(nk) xky n−k, Every complete ordered field is Archimedean).

Proof

Given: The data in the Counterexample.

1.1F3givenalgebrainduction

The real-power definition gives 161=exp⁡(log⁡16)=16. Its exponent law gives 16j+1=16j16 for every natural j, so induction identifies the real power 16m with the natural product defining N. By [F3], N−1/m=(16m)−1/m=16−1=1/16.

1.2F1given

The subgraphs Cfk={(i,t):t<fk(i)} are nested in increasing k. A point (1,1/2) belongs to CfN and not to Cf0, so a singleton is shattered. No two-point set is shattered by a nested family, since shattering would require both opposite singleton traces, which nested sets cannot realize. Therefore the VC-subgraph dimension is exactly one and its index is V=2.

2.1F2F4step 1.1algebra

For k≠ℓ, the functions differ on exactly ∣k−ℓ∣ atoms. By [F2] and [F4], ∥fk−fℓ∥Qm,m=(∣k−ℓ∣N)1/m≥N−1/m=1/16.

3.1F2step 2.1algebra

Since 1/16>2ϵ=1/32, a closed Lm(Qm) ball of radius ϵ centered at a class member cannot contain two distinct class members: the triangle inequality would make their distance at most 2ϵ. Thus every cover needs at least N+1 centers; taking each member as its own center gives a cover with N+1. This proves the exact covering number.

4.1F5step 3.1∎

Fix any proposed finite right side that depends only on ϵ, the VC index, and constants independent of r. At ϵ=1/64 and V=2 it is a fixed finite number. By [F5], choose m so large that 16m+1 exceeds that number. Step 3.1 then contradicts the proposed bound, so this family rules out a VC covering bound whose right-hand side is independent of r.

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