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

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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.

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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

Statement

Let $A \subseteq R$ , let $f : A \to R$ be continuous on $A$ (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) and let $I \subseteq A$ be order-convex (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length). Then:

$f [I]$ is order-convex, hence connected (A subset of $R$ is connected if and only if it is order-convex, that is, an interval, Separated sets, disconnection, and connected subset of $R$ );
if $I = [a, b]$ with $a \leq b$ , then $f [I] = [m, M]$ where $m = \min f [I]$ and $M = \max f [I]$ (Maximum and minimum of a set) — a closed bounded interval, degenerate exactly when $f$ is constant on $[a, b]$ .

"Interval" means "order-convex" here. As A subset of $R$ is connected if and only if it is order-convex, that is, an interval records, this library proves that the connected subsets of $R$ are exactly the order-convex ones, and does not prove that every order-convex subset is one of the nine written forms of Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length. Claim 1 is therefore stated as order-convexity, which is what the intermediate value theorem delivers; claim 2 identifies the written form in the one case where the extreme value theorem supplies the endpoints.

Facts & Assumptions

Given: A set $A \subseteq R$ , a function $f : A \to R$ continuous on $A$ , and an order-convex set $I \subseteq A$ .

[L1]

Intermediate value theorem: if $u \leq v$ in $R$ , if $f$ is continuous on $[u, v]$ and if $w$ lies between $f (u)$ and $f (v)$ in either order, then $f (t) = w$ for some $t \in [u, v]$ (Intermediate value theorem, by bisection with a canonical left-half rule: a continuous function on $[a, b]$ takes every value between $f (a)$ and $f (b)$ ).

[L2]

Continuity passes to subsets of the domain: if $B \subseteq A$ then $f ∣_{B}$ is continuous on $B$ , since the defining condition quantifies over fewer points (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).

[L3]

Order-convexity: $x, z \in S$ and $x \leq w \leq z$ imply $w \in S$ ; every closed bounded interval $[u, v]$ with $u \leq v$ is order-convex and is a subset of any order-convex set containing $u$ and $v$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

[L4]

Connectedness: a subset of $R$ is connected if and only if it is order-convex (A subset of $R$ is connected if and only if it is order-convex, that is, an interval, Separated sets, disconnection, and connected subset of $R$ ).

[L5]

$[a, b]$ with $a \leq b$ is nonempty, closed and bounded (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length, Open subset of $R$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen, Lower bound, bounded below, bounded set), hence compact (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).

[L6]

Extreme value theorem: a continuous real function on a nonempty compact subset of $R$ attains a greatest and a least value on it (Extreme value theorem: a continuous real function on a nonempty compact subset of $R$ attains a greatest and a least value, Maximum and minimum of a set).

Proof

technique · direct

1.1

Claim 1. Let $u^{'}, v^{'} \in f [I]$ and let $w$ satisfy $u^{'} \leq w \leq v^{'}$ . Write $u^{'} = f (p)$ and $v^{'} = f (q)$ with $p, q \in I$ , and let $[s, t]$ be the closed bounded interval with ${s, t} = {p, q}$ and $s \leq t$ ; by [L3] and order-convexity of $I$ we have $[s, t] \subseteq I \subseteq A$ .

L3choose

1.2

Claim 2, the endpoints. Suppose $I = [a, b]$ with $a \leq b$ . By [L5] the set $[a, b]$ is nonempty and compact, so by [L6] there are $q, p \in [a, b]$ with $f (q) \leq f (x) \leq f (p)$ for every $x \in [a, b]$ ; put $m : = f (q)$ and $M : = f (p)$ , so $m = \min f [I]$ and $M = \max f [I]$ and $m \leq M$ .

L5L6choose

2.1

By [L2] the restriction of $f$ to $[s, t]$ is continuous on $[s, t]$ , and $w$ lies between $f (s)$ and $f (t)$ in one order or the other, since ${f (s), f (t)} = {u^{'}, v^{'}}$ and $u^{'} \leq w \leq v^{'}$ . By [L1] there is $c \in [s, t] \subseteq I$ with $f (c) = w$ , so $w \in f [I]$ .

step 1.1L1L2

3.1

So $f [I]$ is order-convex, and by [L4] it is connected. This is claim 1.

step 2.1L4

4.1

Claim 2, the two inclusions. Every $z \in f [I]$ satisfies $m \leq z \leq M$ by step 1.2, so $f [I] \subseteq [m, M]$ . Conversely, $m$ and $M$ lie in $f [I]$ and $f [I]$ is order-convex by step 3.1, so every $w$ with $m \leq w \leq M$ lies in $f [I]$ ; hence $[m, M] \subseteq f [I]$ . Therefore $f [I] = [m, M]$ , a closed bounded interval, and it is the single point ${m}$ exactly when $m = M$ , that is exactly when $f$ is constant on $[a, b]$ .

step 3.1step 1.2L3∎

Remarks

The two halves come from the two theorems. Order-convexity of the image is the intermediate value theorem and needs nothing else; that the image of a closed bounded interval is again closed and bounded is the extreme value theorem, and it fails for other interval forms: the continuous image of $(0, 1)$ under $x \mapsto 1 / x$ is $(1, \infty)$ , and under $x \mapsto x$ it is $(0, 1)$ , neither closed.
The converse of claim 1 is false. A function whose image on every subinterval is order-convex need not be continuous; this is the intermediate value property without continuity, and the witness for it is not available at this point in the reading order. What is true, and is proved on this page, is that a function which is monotone and has an order-convex image is continuous (A function on an interval satisfying $f (x) \leq f (y)$ whenever $x \leq y$ , whose image is order-convex, is continuous).
Claim 2 is the shape the $n$ -th-root example uses. Applying it to $x \mapsto x^{n}$ on $[0, a + 1]$ gives an interval containing $0$ and $(a + 1)^{n} \geq a$ , hence containing $a$ ; that is the second proof of the existence of $n$ -th roots recorded in The intermediate value theorem gives a second proof that every nonnegative real has an $n$ -th root, applied to $x^{n}$ on a closed bounded interval ↗ on the companion page.

Depends on

Used by

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Sources

Intermediate value theorem (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 4 (standard reference, not scraped)
J. Lebl, Basic Analysis I, §3.3 (standard reference, not scraped)
E. Zakon, Mathematical Analysis, §4.9: The Intermediate Value Property (standard reference, not scraped)