TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28

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$f : A \to \mathbb{R}$ is continuous at $c \in A$ if and only if $\omega_f(c) = 0$

Statement

Let $A \subseteq \mathbb{R}$ , let $f : A \to \mathbb{R}$ and let $c \in A$ . Then

$f \text{ is continuous at } c \quad \Longleftrightarrow \quad \omega_f(c) = 0$

(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, The oscillation $\omega_f(S) = \sup\{\,|f(x) - f(y)| : x, y \in S\,\}$ of $f$ on a set and the oscillation $\omega_f(c) = \inf_{\delta > 0} \omega_f(A \cap N_\delta(c))$ at a point, both taken in the extended reals).

Since $\omega_f(c) \ge 0$ always (The oscillation $\omega_f(S) = \sup\{\,|f(x) - f(y)| : x, y \in S\,\}$ of $f$ on a set and the oscillation $\omega_f(c) = \inf_{\delta > 0} \omega_f(A \cap N_\delta(c))$ at a point, both taken in the extended reals), the equivalent form of the right-hand side is: for every real $\varepsilon > 0$ there is a real $\delta > 0$ with $\omega_f(A \cap N_\delta(c)) < \varepsilon$ .

This is the tool that converts a pointwise condition into a set condition. Continuity at $c$ is a statement about $f$ near $c$ with a quantifier over $\varepsilon$ ; $\omega_f(c) = 0$ is the vanishing of a single extended real attached to the point. The change of form is what makes the discontinuity set accessible: the sets $\{\,x : \omega_f(x) \ge \varepsilon\,\}$ are closed (For every real $\varepsilon > 0$ the set $\{\,x \in A : \omega_f(x) \ge \varepsilon\,\}$ is the intersection with $A$ of a closed subset of $\mathbb{R}$ ; in particular it is closed in $\mathbb{R}$ when $A = \mathbb{R}$ ) and their union over $\varepsilon = 1, 1/2, 1/3, \dots$ is the discontinuity set (For $f : A \to \mathbb{R}$ the set of points of $A$ at which $f$ is discontinuous is the intersection with $A$ of an $F_\sigma$ subset of $\mathbb{R}$ , and the set of points at which $f$ is continuous is the intersection with $A$ of a $G_\delta$ subset; for $A = \mathbb{R}$ the two sets are $F_\sigma$ and $G_\delta$ outright).

Facts & Assumptions

Given: $A \subseteq \mathbb{R}$ , a function $f : A \to \mathbb{R}$ , and a point $c \in A$ .

[L1]

$f$ is continuous at $c$ exactly when for every real $\varepsilon > 0$ there is a real $\delta > 0$ with $|f(x) - f(c)| < \varepsilon$ for every $x \in A$ with $|x - c| < \delta$ ; equivalently for every $x \in A \cap N_\delta(c)$ (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, The $\varepsilon$ -neighbourhood and the punctured $\varepsilon$ -neighbourhood of a point of $\mathbb{R}$ ).

[L2]

$\omega_f(S) = \sup\{|f(x) - f(y)| : x, y \in S\}$ and $\omega_f(c) = \inf\{\omega_f(A \cap N_\delta(c)) : \delta > 0\}$ , both in $\overline{\mathbb{R}}$ ; $0 \le \omega_f(c) \le \omega_f(A \cap N_\delta(c))$ for every real $\delta > 0$ , and $c \in A \cap N_\delta(c)$ (The oscillation $\omega_f(S) = \sup\{\,|f(x) - f(y)| : x, y \in S\,\}$ of $f$ on a set and the oscillation $\omega_f(c) = \inf_{\delta > 0} \omega_f(A \cap N_\delta(c))$ at a point, both taken in the extended reals, The extended real line $\overline{\mathbb{R}} = \mathbb{R} \cup \{-\infty, +\infty\}$ , its order, and the arithmetic that is left undefined).

[L3]

In $\overline{\mathbb{R}}$ every subset has a least upper bound and a greatest lower bound; a supremum is at most an extended real $u$ exactly when $u$ bounds every member of the set, and an infimum is at least an extended real $\ell$ exactly when $\ell$ bounds every member from below (Every subset of $\overline{\mathbb{R}}$ has a least upper bound and a greatest lower bound in $\overline{\mathbb{R}}$ , agreeing with the real supremum and infimum on nonempty sets bounded in $\mathbb{R}$ ).

[L4]

$|u - w| \le |u - v| + |v - w|$ and $|u| \ge 0$ for reals $u, v, w$ (Basic properties of the absolute value).

Proof

technique · direct

1.1

Suppose $f$ is continuous at $c$ and let $\varepsilon > 0$ be real. Take $\delta > 0$ with $|f(x) - f(c)| < \varepsilon/2$ for every $x \in A \cap N_\delta(c)$ .

1.2

Conversely, suppose $\omega_f(c) = 0$ and let $\varepsilon > 0$ be real. Not every member of $\{\omega_f(A \cap N_\delta(c)) : \delta > 0\}$ can be $\ge \varepsilon$ , for then $\varepsilon$ would be a lower bound of that set and the infimum $\omega_f(c) = 0$ would satisfy $0 \ge \varepsilon$ . So there is a real $\delta > 0$ with $\omega_f(A \cap N_\delta(c)) < \varepsilon$ .

L2L3

2.1

For $x, y \in A \cap N_\delta(c)$ with $\delta$ as in step 1.1, $|f(x) - f(y)| \le |f(x) - f(c)| + |f(c) - f(y)| < \varepsilon/2 + \varepsilon/2 = \varepsilon$ ; so $\varepsilon$ is an upper bound of the set whose supremum is $\omega_f(A \cap N_\delta(c))$ , and therefore $\omega_f(A \cap N_\delta(c)) \le \varepsilon$ .

step 1.1L2L3L4

2.2

With $\delta$ as in step 1.2 and any $x \in A \cap N_\delta(c)$ : both $x$ and $c$ lie in $A \cap N_\delta(c)$ , so $|f(x) - f(c)|$ is one of the values whose supremum is $\omega_f(A \cap N_\delta(c))$ and therefore $|f(x) - f(c)| \le \omega_f(A \cap N_\delta(c)) < \varepsilon$ .

step 1.2L2L3

3.1

Hence $0 \le \omega_f(c) \le \varepsilon$ for every real $\varepsilon > 0$ . If $\omega_f(c)$ were not $0$ it would satisfy $0 < \omega_f(c) \le 1$ , hence be a positive real, and taking $\varepsilon := \omega_f(c)/2$ would give $\omega_f(c) \le \omega_f(c)/2$ , which is false for a positive real. So $\omega_f(c) = 0$ .

step 2.1L2L3

4.1

Since $\varepsilon > 0$ was arbitrary in step 1.2, the continuity condition holds at $c$ , and $f$ is continuous at $c$ . Together with step 3.1 this proves the equivalence.

step 3.1step 2.2L1∎

Remarks

Where the extended reals are used, and where they are not. The definition of $\omega_f$ needs them, because $f$ may be unbounded near $c$ and the supremum may then be $+\infty$ ; the proof needs only the order relations, and the two directions never compare an infinite value with a real except through the inequality $0 \le \omega_f(c) \le \varepsilon$ , which already forces $\omega_f(c)$ to be real.
The oscillation measures how badly continuity fails, not merely whether it does. The theorem uses only whether $\omega_f(c)$ vanishes, but the number itself carries more: the oscillation of Thomae's function at a point $c$ is proved below to be exactly $t(c)$ (The Dirichlet function is continuous at no point of $\mathbb{R}$ , and Thomae's function is continuous at every irrational and at no rational, so its set of continuity points is exactly the set of irrationals and its oscillation at $c$ equals $t(c)$ ), so it is $1/q$ at a rational with least denominator $q$ and $0$ at every irrational.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 41 results over 12 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

Oscillation (mathematics) (Wikipedia) (standard reference, not scraped)
Real Function is Continuous at Point iff Oscillation is Zero (ProofWiki) (standard reference, not scraped)

f:A→Rf : A \to \mathbb{R}f:A→R is continuous at c∈Ac \in Ac∈A if and only if ωf(c)=0\omega_f(c) = 0ωf​(c)=0