TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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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.
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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

Statement

Let $A \subseteq R$ , let $f, g : A \to R$ , let $α \in R$ and let $c \in A$ . Suppose $f$ and $g$ are continuous at $c$ (Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point). Then:

$f + g$ , $α f$ and $f g$ are continuous at $c$ ;
$∣ f ∣$ , the function $x \mapsto ∣ f (x) ∣$ , is continuous at $c$ ;
$\max {f, g}$ and $\min {f, g}$ , defined pointwise by $x \mapsto \max {f (x), g (x)}$ and $x \mapsto \min {f (x), g (x)}$ (Maximum and minimum of a set), are continuous at $c$ ;
if $g (c) \neq 0$ then, writing $A_{0} : = {x \in A : g (x) \neq 0}$ , the point $c$ lies in $A_{0}$ and the quotient $(f / g) ∣_{A_{0}} : A_{0} \to R$ , $x \mapsto f (x) / g (x)$ , is continuous at $c$ as a function on $A_{0}$ .

Moreover, with no hypothesis at all:

every constant function $A \to R$ and the identity $i d : A \to R$ , $x \mapsto x$ , are continuous on $A$ ; hence so is $x \mapsto x^{n}$ for every $n \in N$ (Integer powers $a^{m}$ ), and hence so is every polynomial function $x \mapsto a_{0} + a_{1} x + \dots + a_{n} x^{n}$ with real coefficients.

Consequently, if $f$ and $g$ are continuous on $A$ then so are $f + g$ , $α f$ , $f g$ , $∣ f ∣$ , $\max {f, g}$ and $\min {f, g}$ , and $(f / g) ∣_{A_{0}}$ is continuous on $A_{0}$ .

Claim 4 is stated on $A_{0}$ because $f / g$ is not defined where $g$ vanishes, and $g$ may well vanish at points of $A$ far from $c$ . The hypothesis is $g (c) \neq 0$ , not " $g$ nowhere zero"; what it buys is that $c$ itself lies in the smaller domain, which is what makes continuity there mean anything.

Nothing here is proved through a sequence. Claims 1 and 4 are read off from Sums, scalar multiples, products and quotients of function limits, the quotient under the hypothesis that the denominator limit is nonzero, which is itself proved from $ε$ and $δ$ , and claims 2, 3 and 5 are proved directly below. So no choice principle is used anywhere in this item.

Statement references

Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point

Maximum and minimum of a set

Integer powers $a^{m}$

Sums, scalar multiples, products and quotients of function limits, the quotient under the hypothesis that the denominator limit is nonzero

Facts & Assumptions

Given: A set $A \subseteq R$ , functions $f, g : A \to R$ , a real $α$ , a point $c \in A$ at which $f$ and $g$ are continuous, and, for claim 4, the hypothesis $g (c) \neq 0$ together with $A_{0} : = {x \in A : g (x) \neq 0}$ .

[L1]

Continuity at $c$ : for every real $ε > 0$ there is a real $δ > 0$ such that every $x \in A$ with $∣ x - c ∣ < δ$ satisfies $∣ f (x) - f (c) ∣ < ε$ (Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point).

[L2]

A point of $A$ is either a limit point of $A$ or an isolated point of $A$ , and never both; at an isolated point of its domain every function is continuous; at a limit point $c$ of $A$ , continuity of $f$ at $c$ is exactly the statement that the limit of $f$ at $c$ exists and equals $f (c)$ (Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point, Limit point, isolated point, adherent point, derived set, and dense subset of $R$ , The $ε$ -neighbourhood and the punctured $ε$ -neighbourhood of a point of $R$ , The $ε$ - $δ$ limit $\lim_{x \to c} f (x) = L$ of $f : A \to R$ at a limit point $c$ of $A$ ).

[L3]

Algebra of function limits at a limit point $c$ of $A$ : if the limits of $f$ and $g$ at $c$ exist with values $L$ and $M$ , then the limits of $f + g$ , $α f$ and $f g$ at $c$ exist with values $L + M$ , $α L$ and $L M$ ; and if $M \neq 0$ then $c$ is a limit point of $A_{0}$ , and the limit of $(f / g) ∣_{A_{0}}$ at $c$ exists and equals $L / M$ (Sums, scalar multiples, products and quotients of function limits, the quotient under the hypothesis that the denominator limit is nonzero).

[L4]

Sign preservation: if the limit of $g$ at a limit point $c$ of $A$ exists and is nonzero, then $c$ is a limit point of $A_{0} = {x \in A : g (x) \neq 0}$ (If $\lim_{x \to c} f (x) = L \neq 0$ then $∣ f ∣ > ∣ L ∣ / 2$ on a punctured neighbourhood of $c$ ; in particular if $L > 0$ then $f > L / 2 > 0$ there).

[L5]

Reverse triangle inequality: $∣ ∣ u ∣ - ∣ v ∣ ∣ \leq ∣ u - v ∣$ (The reverse triangle inequality); and $∣ u ∣ \geq 0$ , $∣ u ∣ = 0$ exactly when $u = 0$ , $∣ u v ∣ = ∣ u ∣ ∣ v ∣$ (Basic properties of the absolute value).

[L6]

Maximum and minimum of a two-element set of reals exist (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set), and for all reals $u, v$ one has $\max {u, v} = \frac{1}{2} (u + v + ∣ u - v ∣)$ and $\min {u, v} = \frac{1}{2} (u + v - ∣ u - v ∣)$ .

[L7]

Ordered-field arithmetic in $R$ : trichotomy and totality of the order, the field identities, and $0 < 1$ so that $2 \neq 0$ and $t / 2$ is defined (Ordered field, Field).

[L8]

Integer powers: $a^{0} = 1$ and $a^{n + 1} = a^{n} \cdot a$ (Integer powers $a^{m}$ ).

Proof

technique · direct

1.1L5L6L7

Justification of the identity in [L6]. Let $u, v \in R$ . By totality either $u \geq v$ or $v \geq u$ . If $u \geq v$ then $u - v \geq 0$ , so $∣ u - v ∣ = u - v$ and $\frac{1}{2} (u + v + ∣ u - v ∣) = \frac{1}{2} (2 u) = u = \max {u, v}$ , while $\frac{1}{2} (u + v - ∣ u - v ∣) = v = \min {u, v}$ . If $v \geq u$ the same computation with the roles exchanged applies, since $∣ u - v ∣ = v - u$ there.

1.2L2

The isolated case. Suppose $c$ is an isolated point of $A$ , say $N_{η} (c) \cap A = {c}$ with $η > 0$ real. Then every function on $A$ is continuous at $c$ by [L2], which gives claims 1, 2 and 3 at once. For claim 4, assume $g (c) \neq 0$ ; then $c \in A_{0}$ , and $N_{η} (c) \cap A_{0} \subseteq N_{η} (c) \cap A = {c}$ with $c$ in the left-hand side, so $c$ is an isolated point of $A_{0}$ and every function on $A_{0}$ , in particular $(f / g) ∣_{A_{0}}$ , is continuous at $c$ .

1.3L1L5

Claim 2, at any point of $A$ . Let a real $ε > 0$ be given and let $δ > 0$ be as in [L1] for $f$ and this $ε$ . For $x \in A$ with $∣ x - c ∣ < δ$ the reverse triangle inequality gives $∣ ∣ f (x) ∣ - ∣ f (c) ∣ ∣ \leq ∣ f (x) - f (c) ∣ < ε$ . So $∣ f ∣$ is continuous at $c$ , and no case distinction was needed.

1.4L1L5L7

Claim 5, constants and the identity. If $f$ is constant then $∣ f (x) - f (c) ∣ = 0 < ε$ for every $x \in A$ and every real $ε > 0$ , so any $δ > 0$ serves. For the identity, given a real $ε > 0$ take $δ : = ε$ : every $x \in A$ with $∣ x - c ∣ < δ$ has $∣ i d (x) - i d (c) ∣ = ∣ x - c ∣ < ε$ . Both are continuous at every point of $A$ .

1.5L2L3

The limit-point case, claim 1. Suppose $c$ is a limit point of $A$ . By [L2] the limits of $f$ and of $g$ at $c$ exist and equal $f (c)$ and $g (c)$ . By [L3] the limits of $f + g$ , $α f$ and $f g$ at $c$ exist and equal $f (c) + g (c)$ , $α f (c)$ and $f (c) g (c)$ , which are exactly the values of those three functions at $c$ ; by [L2] again, each of them is continuous at $c$ .

1.6L2L3L4

The limit-point case, claim 4. Suppose $c$ is a limit point of $A$ and $g (c) \neq 0$ . Then $c \in A_{0}$ , and by [L4] the point $c$ is a limit point of $A_{0}$ . By [L3] the limit of $(f / g) ∣_{A_{0}}$ at $c$ exists and equals $f (c) / g (c)$ , which is the value of $(f / g) ∣_{A_{0}}$ at $c$ ; by [L2] applied on the domain $A_{0}$ , that function is continuous at $c$ .

2.1step 1.2step 1.5step 1.6L2

Claims 1 and 4 in general. By [L2] the point $c$ is either isolated in $A$ or a limit point of $A$ ; step 1.2 settles the first case and steps 1.5 and 1.6 the second. So claims 1 and 4 hold as stated.

3.1step 1.1step 1.3step 2.1L6L7

Claim 3. By claim 1 the function $f - g = f + (- 1) g$ is continuous at $c$ , by step 1.3 so is $∣ f - g ∣$ , and by claim 1 again so are $f + g + ∣ f - g ∣$ and its scalar multiple by $1 / 2$ . By step 1.1 that scalar multiple is the function $x \mapsto \max {f (x), g (x)}$ , so the maximum is continuous at $c$ ; the same argument with $- ∣ f - g ∣$ gives the minimum.

3.2step 1.4step 2.1L8

Claim 5, powers and polynomials. The map $x \mapsto x^{0}$ is the constant $1$ and $x \mapsto x^{1}$ is the identity, both continuous on $A$ by step 1.4; and if $x \mapsto x^{n}$ is continuous on $A$ then so is $x \mapsto x^{n + 1} = x^{n} \cdot x$ , being a product of two functions continuous on $A$ by step 2.1. By induction on $n$ , $x \mapsto x^{n}$ is continuous on $A$ for every $n \in N$ . A polynomial function $a_{0} + a_{1} x + \dots + a_{n} x^{n}$ is obtained from these by finitely many scalar multiplications and additions, each of which preserves continuity by step 2.1.

4.1step 1.3step 2.1step 3.1step 3.2∎

Claims 1 to 5 are proved, all of them at an arbitrary point $c$ of $A$ and therefore, applied at every point, on the whole of $A$ ; and no sequence and no choice principle was used.

Remarks

Why the two-case shape, and why it is not an inconvenience. Continuity is defined at every point of the domain, including isolated points, where no limit exists (Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point). The algebra of limits therefore cannot be applied blindly; but at an isolated point every function is continuous, so the case is settled before it is opened. Claims 2 and 5 are proved directly from $ε$ and $δ$ and need no case distinction at all.
Absolute value, maximum and minimum are not in Sums, scalar multiples, products and quotients of function limits, the quotient under the hypothesis that the denominator limit is nonzero, and the reason is that they are not needed there. They are needed here: the extreme value theorem and the one-dimensional fixed point theorem both build auxiliary functions out of maxima, minima and differences, and Rudin 4.20, the sharp converse: on a noncompact $E \subseteq R$ there is an unbounded continuous function and a bounded continuous function with no greatest value, and if $E$ is bounded there is a continuous function on $E$ that is not uniformly continuous builds its witnesses out of $∣ x - x_{0} ∣$ and quotients.
The converse of claim 2 is false: $∣ f ∣$ may be continuous while $f$ is continuous nowhere. The function equal to $1$ on $Q$ and to $- 1$ elsewhere has constant absolute value; that it is nowhere continuous follows from the argument of The indicator of $Q$ is continuous at no point of $R$ ↗ applied verbatim, since that argument uses only that the two values are distinct.

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

Statement

Statement references

Facts & Assumptions

Proof

Remarks

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