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Weak Mixing and the Chacon Transformation — Examples

1 · Prerequisites

2 · Summary

The tower recurrence gives the first three heights 1,4,13 and the closed formula (3r+11)/2. A separate calculation tracks spacer widths, finite geometric sums and the reservoir tending to zero. These computations fix the indexing used in the construction.

The correlation example keeps one interval fixed while the return times vary. Its positive gap 2/81 shows why the failure of strong mixing is a failure of a limit for a single pair of sets. It does not rely on selecting a different witness at every stage. The source obligations these examples inherit from the construction are the ones recorded as resolved for the companion page's batch coverage, not a fresh claim about those sources.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-adaptedVerification: AI-adaptedjudge pass (gpt-5.6-terra)Open item page →

First three chacon tower heights

Example

The first three formal Chacon tower heights are h0=1, h1=4, h2=13. For every integer r0, hr=(3r+11)/2.

Facts & Assumptions

[F1]

The finite tower recursion is h0=1, hr+1=3hr+1 Chacon three cut one spacer towers. Its finite combinatorial clause is choice-free.

Verification

Given: The formal recursion of F1.

1.1

Direct substitution gives h1=31+1=4 and h2=34+1=13. The extra one counts the spacer level, whereas the three copies count the old levels in the three columns.

F1
2.1

Set br=(3r+11)/2. Then b0=(31)/2=1 and 3br+1=(3r+23+2)/2=(3r+21)/2=br+1. Induction on r using the same initial value and recurrence proves hr=br. At r=1,2 this gives (91)/2=4 and (271)/2=13, agreeing with step 1.1. Only finite arithmetic and induction are used; no measure existence or choice assumption is used.

F1step 1.1
ExampleConstruction: AI-adaptedVerification: AI-adaptedjudge pass (gpt-5.6-terra)Open item page →

Chacon spacer measure budget

Example

Assume AC for the Lebesgue-measure interpretation. The initial Chacon column has mass 2/3. The first two spacers have masses 2/9 and 2/27. All spacers together have mass 1/3; after stage r0 the unused reservoir has mass 3(r+1)0. Spacers are retained, so it is the unused tail, not total spacer mass, that tends to zero.

Facts & Assumptions

[F1]

Spacer Jj has width 2/3j+2, the initial column has width 2/3, and all the physical spacers are disjoint and retained Chacon three cut one spacer towers.

[F2]

Assume AC for the measure assertions in F1 The Axiom of Choice.

Verification

Given: The finite normalized Chacon construction.

1.1

At the first cut the spacer has width 2/32=2/9, and at the second it has width 2/33=2/27. After stage r the retained spacer mass is the finite sum Sr=j=0r12/3j+2. The empty sum at r=0 is zero. For r1, multiplying the sum by 11/3 cancels all interior terms, giving (2/3)Sr=(2/9)(13r), hence Sr=1/33(r+1). The formula also gives zero at r=0.

F1F2
2.1

Adding the initial mass yields 2/3+Sr=13(r+1), whose missing mass is exactly the reservoir. As r, Sr1/3 and the reservoir tends to zero. Countable additivity on the disjoint spacer intervals identifies their union's measure with this sum; their union is also the physical interval [2/3,1), since their adjacent endpoints tend to one. The numerical geometric identities themselves are choice-free; AC is used only through F1's Lebesgue-measure interpretation.

F1F2step 1.1
CounterexampleConstruction: AI-adaptedVerification: AI-adaptedjudge pass (gpt-5.6-terra)Open item page →

Chacon correlation subsequence prevents mixing

Statement refuted

The normalized Chacon transformation is strongly mixing.

Facts & Assumptions

[F1]

For Chacon, A=[0,2/9) has measure 2/9 and μ(AThrA)2/27 for all r1 Chacon tower height correlations obstruct mixing. The heights satisfy hr=(3r+11)/2 Chacon three cut one spacer towers.

[F2]

Assume AC, as in F1 The Axiom of Choice.

Counterexample

Given: The Chacon probability system under AC.

1.1

Use the fixed pair (A,A) from F1. The product of its measures is (2/9)2=4/81, while its correlations at the stated heights are at least 2/27=6/81. Their difference is therefore at least 6/814/81=2/81>0 for every r1.

F1F2
2.1

A sequence tending to zero must eventually have absolute value below 1/81 on every subsequence whose indices tend to infinity. The correlations minus their product violate this necessary condition along hr. Thus the strong-mixing conclusion refuted by F1 fails, without any need to assert that the correlations have a subsequential limit. The witness set is fixed, not chosen anew at each height.

F1step 1.1

Sources