False statementConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21

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FALSE: every continuous function on a compact interval has a rectifiable graph

Statement

False claim: if $f : [a, b] \to R$ is continuous on a compact interval, then its graph path $x \mapsto (x, f (x))$ is rectifiable.

Facts & Assumptions

Given: The universal claim in the Statement.

[L1]

A function has bounded variation on $[a, b]$ exactly when its finite partition-variation sums are bounded above (Bounded variation and total variation on an interval).

[L2]

The harmonic series $\sum_{k \geq 1} 1 / k$ diverges (For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$ , case $p = 1$ ).

[L3]

The shift formulas give $\sin ((2 k + 1) π / 2) = (- 1)^{k}$ for $k \in N$ (Quarter-turn values and shifts by pi/2 and pi).

[L4]

For every real $u$ , $∣ \sin u ∣ \leq 1$ (Parity and the Pythagorean identity for sine and cosine).

[L5]

A path in $R^{n}$ is rectifiable if and only if each coordinate function has bounded variation (A path in $R^{n}$ is rectifiable exactly when every coordinate has bounded variation).

[L6]

For every real $ε > 0$ , there is a positive integer $N$ with $1 / N < ε$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ).

[L7]

Every closed bounded interval in $R$ is compact (Heine-Borel by bisection: every closed bounded interval $[a, b]$ is compact).

[L8]

The number $π = 2 γ$ is positive because $γ \in (0, 2)$ (Pi as twice the smallest positive zero of cosine, Cosine has a smallest positive zero, lying strictly between zero and two).

[L9]

Sine is continuous, the reciprocal is continuous away from zero, and algebraic combinations and composites of continuous real functions are continuous (The derivatives of sine and cosine are cosine and minus sine, A function differentiable at $c$ is continuous at $c$ , Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function, A composite of continuous functions is continuous, with no side hypothesis of the kind the composition of limits needs).

[L10]

The identity real function is continuous, and a map into a product is continuous if each of its component maps is continuous (Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function, claim 5, A map into a product is continuous iff each of its components is; the projections are continuous and open; and each projection is surjective when every factor is nonempty, which for an infinite index set uses the Axiom of Choice, claim 2).

Refutation

technique · contradiction

1.1

Suppose, for contradiction, that every continuous real function on a compact interval has a rectifiable graph.

assume-contra

1.2

Define $f (0) : = 0$ and $f (x) : = x \sin (1 / x)$ for $0 < x \leq 1$ . By [L4], $∣ f (x) ∣ \leq x$ , so $f$ is continuous at zero; [L9] gives continuity elsewhere.

L4L9algebra

1.3

Put $x_{k} : = 2 / ((2 k + 1) π)$ . By [L6] and [L8], $x_{k} ↓ 0$ , so choose $K$ with $x_{K} \leq 1$ . By [L3], $f (x_{k}) = (- 1)^{k} x_{k}$ .

L3L6L8chooseconstructalgebra

2.1

For $N > K$ , use the partition with points $0, x_{N}, x_{N - 1}, \dots, x_{K}, 1$ , omitting a duplicate endpoint if needed. Consecutive oscillatory nodes contribute $∣ f (x_{k}) - f (x_{k + 1}) ∣ = x_{k} + x_{k + 1} \geq x_{k}$ to its variation sum.

step 1.3L1constructalgebra

3.1

Since $x_{k} = 2 / ((2 k + 1) π) \geq 1 / (π (k + 1))$ , the variation sums in step 2.1 dominate partial tails of a fixed positive multiple of the harmonic series. They are unbounded by [L2], so [L1] says $f$ does not have bounded variation.

step 2.1L1L2algebra

4.1

The identity coordinate is continuous by [L10], and $f$ is continuous by step 1.2, so the same fact makes $γ (x) = (x, f (x))$ a path. Its second coordinate is not of bounded variation by step 3.1. The forward implication in [L5], read contrapositively, therefore shows that $γ$ is not rectifiable.

step 1.2step 3.1L5L10

5.1

The interval $[0, 1]$ is compact by [L7] and $f$ is continuous by step 1.2, so step 1.1 would make its graph rectifiable, contradicting step 4.1. The universal claim is false.

step 1.1step 1.2step 4.1L7discharge-contradiction∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

80 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

Thomas Lam, 21-236 Recitation Notes, §4.3 (standard reference, not scraped)