Alphabeta Math
CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26
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xk=k has xk+1−xk→0 and is not Cauchy

Statement refuted

Refuted claim: a sequence of reals whose consecutive differences tend to 0 is Cauchy (FALSE: if ∣xk+1−xk∣→0 then (xk) is Cauchy, Limits and Cauchy sequences of reals).

The witness is xk=k for k∈N. Its consecutive differences satisfy

xk+1−xk  =  k+1−k  =  1k+1+k  ⟶  0,

while the sequence itself is unbounded, hence not Cauchy (Every Cauchy sequence of reals is bounded). The refutation is carried out in full in FALSE: if ∣xk+1−xk∣→0 then (xk) is Cauchy; this item records the witness and adds the sharper statement that k diverges to +∞ (Divergence to +∞ and to −∞).

Facts & Assumptions

Given: The sequence (xk) of reals with xk:=k, where k denotes the canonical natural k⋅1R (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Square roots exist: a unique a≥0 with (a)2=a; the positives are {x2:x≠0}).

[L1]

The witness and its two properties: xk+1−xk=1/(k+1+k) tends to 0, and (xk) is unbounded and not Cauchy (FALSE: if ∣xk+1−xk∣→0 then (xk) is Cauchy).

[L3]

Powers and order: for a,b≥0 and n≥1, a<b exactly when an<bn (Monotonicity of x↦xn and of n↦an).

[L4]

Canonical naturals: positive for n≥1, and strictly increasing in the index (Canonical naturals are positive and strictly increasing); reciprocals of positives are positive and reciprocation reverses the order (Inverses of positives are positive, and reciprocation reverses order).

[L6]

Absolute value: ∣t∣=t for t≥0, and ∣t∣≥t (Basic properties of the absolute value).

[L7]

Every Cauchy sequence of reals is bounded (Every Cauchy sequence of reals is bounded).

[L8]

Divergence to +∞: for every real M there is K with xk>M for all k≥K (Divergence to +∞ and to −∞).

[L9]

Trichotomy and transitivity of the order on R (Complete ordered field (least-upper-bound property), Ordered field).

Counterexample

technique · direct
1.1

The sequence (xk) satisfies the hypothesis of the refuted claim, its consecutive differences tending to 0, and it is not Cauchy.

givenL1L7
1.2

The failure is as strong as possible: (xk) diverges to +∞. Let M∈R and put M′:=∣M∣≥M, so M′≥0. By [L5] fix a natural n≥1 with (M′)2<n.

givenL5L6
2.1

It therefore refutes the claim: having null consecutive differences does not make a sequence Cauchy.

step 1.1L1
3.1

For every k≥n: (xk)2=k≥n>(M′)2 with xk≥0 and M′≥0, so xk>M′≥M. Since M was arbitrary, xk→+∞.

step 1.2L2L3L4L8L9∎

Remarks

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

51 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources