ExampleConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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.

$x \cdot 1_{\mathbb{Q}}(x)$ is continuous at $0$ and at no other point

Example

Let $\mathbf{1}_{\mathbb{Q}}$ be the indicator of the rationals (The indicator of $\mathbb{Q}$ is continuous at no point of $\mathbb{R}$ ) and put

$h : \mathbb{R} \to \mathbb{R}, \qquad h(x) := x \cdot \mathbf{1}_{\mathbb{Q}}(x),$

so $h(x) = x$ for rational $x$ and $h(x) = 0$ for irrational $x$ . Then:

$h$ is continuous at $0$ (Continuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$ : the $\varepsilon$ - $\delta$ condition, its agreement with $\lim_{x \to c} f(x) = f(c)$ at a limit point, and continuity at an isolated point);
$h$ is not continuous at any $c \ne 0$ .

So the set of points of continuity of a real function can be a single point. Together with The indicator of $\mathbb{Q}$ is continuous at no point of $\mathbb{R}$ , where that set is empty, this shows how little the set of continuity points is constrained by the mere existence of the function.

The point of the example. Continuity at $c$ compares $h(x)$ with $h(c)$ , and the two branches of $h$ agree only where $x = 0$ . Multiplying the indicator by $x$ damps the jump: near $0$ both branches are small, so the discrepancy is at most $|x|$ ; away from $0$ the discrepancy is at least $|c|/2$ on every neighbourhood, because each branch is realised arbitrarily close to $c$ .

Facts & Assumptions

Given: The canonical copy $\mathbb{Q} \subseteq \mathbb{R}$ of the rationals with complement $X := \mathbb{R} \setminus \mathbb{Q}$ , and $h : \mathbb{R} \to \mathbb{R}$ with $h(x) = x$ for $x \in \mathbb{Q}$ and $h(x) = 0$ for $x \in X$ .

[L1]

Continuity at $c$ , in the form of Continuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$ : the $\varepsilon$ - $\delta$ condition, its agreement with $\lim_{x \to c} f(x) = f(c)$ at a limit point, and continuity at an isolated point: for every real $\varepsilon > 0$ there is a real $\delta > 0$ with $|h(x) - h(c)| < \varepsilon$ whenever $|x - c| < \delta$ ; and it fails at $c$ as soon as some real $\varepsilon_0 > 0$ admits, for every real $\delta > 0$ , a real $x$ with $|x-c| < \delta$ and $|h(x) - h(c)| \ge \varepsilon_0$ (The $\varepsilon$ -neighbourhood and the punctured $\varepsilon$ -neighbourhood of a point of $\mathbb{R}$ ).

[L2]

Both $\mathbb{Q}$ and $X$ are dense in $\mathbb{R}$ , so every $N_{\delta}(c)$ meets each of them (Both $\mathbb{Q}$ and $\mathbb{R} \setminus \mathbb{Q}$ are dense in $\mathbb{R}$ , and every nonempty open subset of $\mathbb{R}$ is uncountable, The closure equals the set together with its limit points, equals the set of points every neighbourhood of which meets it, and is the smallest closed superset; a set is closed iff it contains its limit points).

[L3]

$X$ is by definition the complement $\mathbb{R} \setminus \mathbb{Q}$ , so every real lies in exactly one of $\mathbb{Q}$ and $X$ ; hence $h$ is a well-defined function, and $|h(x)| \le |x|$ for every real $x$ (Basic properties of the absolute value, The indicator of $\mathbb{Q}$ is continuous at no point of $\mathbb{R}$ ).

[L4]

Absolute value and order: $|u| \ge 0$ and $|u| > 0$ exactly when $u \ne 0$ (Basic properties of the absolute value); the reverse triangle inequality $\bigl||u| - |v|\bigr| \le |u - v|$ , hence $|u - v| \ge |u| - |v|$ (The reverse triangle inequality); and the minimum of a two-element set of reals exists and is one of the two (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set, Ordered field).

Verification

technique · direct

1.1

$h$ is well defined and satisfies $|h(x)| \le |x|$ for every real $x$ : on $\mathbb{Q}$ one has $|h(x)| = |x|$ and on $X$ one has $|h(x)| = 0 \le |x|$ . Also $0 \in \mathbb{Q}$ , so $h(0) = 0$ .

L3L4

1.2

Claim 2, the setup. Let $c \ne 0$ be real, put $\varepsilon_0 := |c|/2 > 0$ by [L4], and let a real $\delta > 0$ be given. Put $\delta' := \min\{\delta,\ |c|/2\} > 0$ by [L4].

L4choose

2.1

Claim 1. Let a real $\varepsilon > 0$ be given and take $\delta := \varepsilon$ . Every real $x$ with $|x - 0| < \delta$ satisfies $|h(x) - h(0)| = |h(x)| \le |x| < \varepsilon$ by step 1.1. So $h$ is continuous at $0$ .

step 1.1L1L4

2.2

By [L2] the neighbourhood $N_{\delta'}(c)$ meets $\mathbb{Q}$ and meets $X$ : fix $p \in N_{\delta'}(c) \cap \mathbb{Q}$ and $z \in N_{\delta'}(c) \cap X$ , so $|p - c| < \delta' \le \delta$ and $|z - c| < \delta' \le \delta$ , with $h(p) = p$ and $h(z) = 0$ .

step 1.2L2choose

3.1

The rational case. Suppose $c \in \mathbb{Q}$ , so $h(c) = c$ . Take $x := z$ : then $|x - c| < \delta$ and $|h(x) - h(c)| = |0 - c| = |c| \ge |c|/2 = \varepsilon_0$ .

step 1.2step 2.2L3L4

3.2

The irrational case. Suppose $c \in X$ , so $h(c) = 0$ . Take $x := p$ : then $|x - c| < \delta$ , and $|p| \ge |c| - |p - c| > |c| - |c|/2 = |c|/2$ by [L4] and step 2.2, so $|h(x) - h(c)| = |p| > |c|/2 = \varepsilon_0$ .

step 1.2step 2.2L3L4

4.1

By [L3] the two cases of steps 3.1 and 3.2 are exhaustive, so for every real $\delta > 0$ some $x$ with $|x-c| < \delta$ has $|h(x) - h(c)| \ge \varepsilon_0$ ; by [L1] the function $h$ is not continuous at $c$ . Since $c \ne 0$ was arbitrary, claim 2 holds, and with step 2.1 the set of points of continuity of $h$ is exactly $\{0\}$ .

step 2.1step 3.1step 3.2L1L3∎

Remarks

Where the damping factor does its work. The estimate $|h(x)| \le |x|$ of step 1.1 is the whole of claim 1, and it is available only because the two branches of $h$ agree at $0$ . Replacing $x$ by any function vanishing at $0$ and continuous there gives the same conclusion at $0$ ; replacing it by a nonzero constant gives The indicator of $\mathbb{Q}$ is continuous at no point of $\mathbb{R}$ back.
The choice of $\delta' = \min\{\delta, |c|/2\}$ is what makes the irrational case work. Without shrinking to $|c|/2$ the rational point $p$ near $c$ could be close to $0$ , and then $|h(p) - h(c)| = |p|$ would be small; the shrinking keeps $p$ away from $0$ by at least $|c|/2$ .
This example is choice free, for the same reason as The indicator of $\mathbb{Q}$ is continuous at no point of $\mathbb{R}$ : density is used only in the form "every neighbourhood meets the set", never to build a sequence.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 76 results over 21 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Dirichlet function (Wikipedia) (standard reference, not scraped)
Continuous function (Wikipedia) (standard reference, not scraped)
J. Lebl, Basic Analysis I, §3.2 (standard reference, not scraped)
E. Boman and R. Rogers, An Analytic Definition of Continuity (standard reference, not scraped)

x⋅1Q(x)x \cdot 1_{\mathbb{Q}}(x)x⋅1Q​(x) is continuous at 000 and at no other point