CounterexampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passverified 2026-08-09 (gpt-5.6-terra-codex-subscription)

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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On the domain $\{0\} \cup [1,2]$ every real is vacuously a limit at $0$

Statement refuted

Refuted claim: for every $A \subseteq \mathbb{R}$ , every $f : A \to \mathbb{R}$ and every $c \in A$ , at most one real $L$ satisfies

$(\forall \varepsilon > 0)\ (\exists \delta > 0)\ (\forall x \in A)\ \bigl[\ 0 < |x - c| < \delta \ \Longrightarrow\ |f(x) - L| < \varepsilon\ \bigr]$

— the false statement FALSE: a function has at most one limit at every point of its domain, isolated points included.

The witness is $A := \{0\} \cup [1,2]$ (Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length), $f : A \to \mathbb{R}$ the constant $0$ , and $c := 0$ . At $c$ the displayed formula holds for every real $L$ at once, so it determines nothing.

What this item adds. It exhibits the dichotomy inside one example: at the isolated point $0$ the formula is vacuous, while at the point $1$ of the same domain — which is a limit point of $A$ — the formula is not vacuous and At a limit point of the domain a function has at most one limit applies, so the limit there exists and is unique. The same $A$ , the same $f$ , and opposite behaviour at two of its points.

Facts & Assumptions

Given: The set $A := \{0\} \cup [1,2]$ , the constant function $f : A \to \mathbb{R}$ with $f(x) := 0$ for every $x \in A$ , and the points $0$ and $1$ of $A$ .

[L1]

The $\varepsilon$ - $\delta$ formula displayed above, and the fact that The $\varepsilon$ - $\delta$ limit $\lim_{x \to c} f(x) = L$ of $f : A \to \mathbb{R}$ at a limit point $c$ of $A$ imposes it only at a limit point of the domain, where At a limit point of the domain a function has at most one limit then makes $L$ unique.

[L2]

Limit point and isolated point: $c$ is a limit point of $S$ when $N^{*}_{\rho}(c) \cap S \ne \varnothing$ for every real $\rho > 0$ ; $c \in S$ is isolated in $S$ when $N_{\rho}(c) \cap S = \{c\}$ for some real $\rho > 0$ ; and for $c \in S$ the two are exact opposites (Limit point, isolated point, adherent point, derived set, and dense subset of $\mathbb{R}$ , The $\varepsilon$ -neighbourhood and the punctured $\varepsilon$ -neighbourhood of a point of $\mathbb{R}$ ).

[L3]

Neighbourhoods: $N_{\rho}(u) = \{\, y : |y - u| < \rho \,\}$ and $N^{*}_{\rho}(u) = N_{\rho}(u) \setminus \{u\}$ (The $\varepsilon$ -neighbourhood and the punctured $\varepsilon$ -neighbourhood of a point of $\mathbb{R}$ ).

[L4]

Intervals: $[1,2] = \{\, y : 1 \le y \le 2 \,\}$ (Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length).

[L5]

Absolute value: $|u| \ge 0$ ; $|u| = 0$ exactly when $u = 0$ ; $|u| = u$ for $u \ge 0$ ; $|0| = 0$ (Basic properties of the absolute value).

[L6]

Order in $\mathbb{R}$ : trichotomy and totality; $0 < 1$ , so $2 > 0$ and $\rho/2 > 0$ with $\rho/2 < \rho$ for $\rho > 0$ ; and of two positive reals the smaller is positive (The multiplicative identity is positive, Order is preserved by adding a constant and by adding inequalities, Sign rules for products and monotonicity of multiplication, Ordered field).

Counterexample

technique · direct

1.1

$A = \{0\} \cup [1,2]$ is a subset of $\mathbb{R}$ , and $f$ is the constant $0$ on $A$ ; both $0$ and $1$ belong to $A$ .

1.2

$0$ is an isolated point of $A$ and not a limit point of $A$ : $N_{1}(0) \cap A = \{0\}$ , since an element of $A$ is either $0$ , with $|0 - 0| = 0 < 1$ , or an element of $[1,2]$ , with $|y - 0| = y \ge 1$ and hence outside $N_1(0)$ .

L2L3L4L5

1.3

The reals $0$ and $1$ are distinct.

2.1

Take $\delta := 1$ . No $x \in A$ satisfies $0 < |x - 0| < 1$ : such an $x$ would lie in $N^{*}_{1}(0) \cap A$ , which is contained in $N_1(0) \cap A = \{0\}$ and excludes $0$ , hence is empty. So for every real $L$ and every real $\varepsilon > 0$ the choice $\delta = 1$ makes the implication vacuously true, and every real $L$ satisfies the displayed formula at $c = 0$ .

step 1.2L1L3L5

2.2

By contrast $1 \in A$ is a limit point of $A$ : given a real $\rho > 0$ , let $\sigma$ be the smaller of $\rho$ and $1$ , so $\sigma > 0$ ; then $1 + \sigma/2$ satisfies $1 \le 1 + \sigma/2 \le 1 + 1/2 \le 2$ , so it lies in $[1,2] \subseteq A$ , and $0 < |(1 + \sigma/2) - 1| = \sigma/2 < \rho$ . There The $\varepsilon$ - $\delta$ limit $\lim_{x \to c} f(x) = L$ of $f : A \to \mathbb{R}$ at a limit point $c$ of $A$ applies, At a limit point of the domain a function has at most one limit gives at most one $L$ , and in fact $\lim_{x \to 1} f(x) = 0$ , since $|f(x) - 0| = |0| = 0 < \varepsilon$ for every $x \in A$ and every real $\varepsilon > 0$ .

step 1.1L1L2L4L5L6

3.1

In particular $L = 0$ and $L = 1$ both satisfy the formula at $c = 0$ , and they are distinct: more than one real satisfies it, so the claim is refuted. This is why The $\varepsilon$ - $\delta$ limit $\lim_{x \to c} f(x) = L$ of $f : A \to \mathbb{R}$ at a limit point $c$ of $A$ is stated only at a limit point, and why $\lim_{x \to 0} f(x)$ is left undefined on this domain.

step 1.3step 2.1L1L6

4.1

So on one and the same domain the formula pins down a unique value at the limit point $1$ and no value at all at the isolated point $0$ : uniqueness of the limit is a property of limit points, not of arbitrary points of the domain.

step 2.2step 3.1∎

Remarks

Nothing about $f$ is used at $0$ . Step 2.1 never evaluates the function: the implication has no instances. Any $f$ whatever on this $A$ would give the same conclusion, which is precisely why the formula carries no information there.
The alternatives are exact. At a limit point of the domain the formula has at most one solution (At a limit point of the domain a function has at most one limit), while at an isolated point every real solves it (Limit point, isolated point, adherent point, derived set, and dense subset of $\mathbb{R}$ ).
Some texts declare the limit at an isolated point to be $f(c)$ . That convention is consistent — it selects one of the many solutions — but it is a stipulation, not a theorem, and FALSE: a function has at most one limit at every point of its domain, isolated points included records why this library declines it.

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: 34 results over 14 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

Isolated point (Wikipedia) (standard reference, not scraped)
Limit of a function (Wikipedia) (standard reference, not scraped)
J. Lebl, Basic Analysis I, §3.1 (standard reference, not scraped)

On the domain {0}∪[1,2]\{0\} \cup [1,2]{0}∪[1,2] every real is vacuously a limit at 000