LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-26

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

The map $x \mapsto x + c (x)$ is a homeomorphism from $[0, 1]$ onto $[0, 2]$

Statement

Let $c : [0, 1] \to R$ be the Cantor function. Define

$ψ : [0, 1] \to R, ψ (x) : = x + c (x) .$

Then $ψ$ is a homeomorphism from $[0, 1]$ onto $[0, 2]$ .

Facts & Assumptions

Given: The Cantor function $c : [0, 1] \to R$ and the map $ψ (x) = x + c (x)$ .

[L1]

The Cantor function is continuous on $[0, 1]$ , is nondecreasing, and satisfies $c (0) = 0$ and $c (1) = 1$ (The Cantor function is continuous on $[0, 1]$ ).

[L2]

The image of an interval under a continuous real function is order-convex, hence an interval, and the image of a closed bounded interval is a closed bounded interval (The image of an interval under a continuous real function is order-convex, hence an interval, and the image of a closed bounded interval is a closed bounded interval).

[L3]

A continuous bijection from a compact space to a Hausdorff space is a homeomorphism (A continuous image of a compact space is compact; a continuous real-valued map on a nonempty compact space attains a maximum and a minimum; and a continuous bijection from a compact space to a Hausdorff space is a homeomorphism, claim 3).

[L4]

A subset of $R$ is compact if and only if it is closed and bounded (A subset of $R$ is compact if and only if it is closed and bounded, Open cover, subcover, compact subset of $R$ (every open cover has a finite subcover), and sequentially compact subset).

Proof

technique · direct

1.1

The map $ψ$ is continuous on $[0, 1]$ , being the sum of the identity map and the continuous function $c$ .

L1algebra

1.2

If $0 \leq x < y \leq 1$ , then $c (x) \leq c (y)$ by [L1], so $ψ (y) - ψ (x) = (y - x) + (c (y) - c (x)) > 0$ . Therefore $ψ$ is strictly increasing and hence injective.

L1algebra

2.1

By [L1], $ψ (0) = 0$ and $ψ (1) = 2$ . Also $0 \leq x \leq 1$ and $0 \leq c (x) \leq 1$ for every $x \in [0, 1]$ , so $ψ ([0, 1]) \subseteq [0, 2]$ . Since step 1.1 makes $ψ$ continuous, [L2] makes $ψ ([0, 1])$ an interval, and the same step shows that this interval contains the endpoints $0$ and $2$ , so it is exactly $[0, 2]$ .

step 1.1L1L2

3.1

The interval $[0, 1]$ is compact by [L4], and $[0, 2]$ is a Hausdorff subspace of $R$ . Steps 1.1, 1.2 and 2.1 therefore make $ψ$ a continuous bijection from a compact space to a Hausdorff space, so [L3] gives that $ψ$ is a homeomorphism.

step 1.1step 1.2step 2.1L3L4∎

Depends on

Used by

Dependency tree · two levels

70 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

John K. Hunter, Measure Theory (UC Davis lecture notes), Example 2.22 (standard reference, not scraped)
Cantor function (Wikipedia) (standard reference, not scraped)

The map x↦x+c(x)x↦x+c(x) is a homeomorphism from [0,1][0,1] onto [0,2][0,2]