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

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

1. $f[I]$ is order-convex, hence connected (A subset of $\mathbb{R}$ is connected if and only if it is order-convex, that is, an interval, Separated sets, disconnection, and connected subset of $\mathbb{R}$);
2. if $I = [a,b]$ with $a \le b$, 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 $\mathbb{R}$ is connected if and only if it is order-convex, that is, an interval records, this library proves that the connected subsets of $\mathbb{R}$ 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 $\mathbb{R}$: 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.

• 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 $f(x) \le f(y)$ whenever $x \le y$, 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 $(a+1)^{n} \ge a$, 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.