---
tag: 0ATVXC3NCC
kind: statement
role: theorem
name: Boundedness theorem
msc: [26A15]
uses: [0P27W9RW0G, 0KDJD249DT]
sources:
  - url: https://en.wikipedia.org/wiki/Extreme_value_theorem
    locator: Proof of the boundedness theorem
---

## Statement

Let $a \le b$ and let $f \colon [a, b] \to \mathbb{R}$ be [[continuous|0P27W9RW0G]].
Then $f$ is bounded. That is, there is $K$ with $|f(x)| \le K$ for all $x \in [a, b]$.

## Remarks

Suppose not. Then there are points $x_n \in [a, b]$ with $|f(x_n)| > n$. By the [[Bolzano–Weierstrass theorem|0KDJD249DT]], a subsequence converges to some $c \in [a, b]$. Continuity at $c$ gives a contradiction.
The interval must be closed and bounded. The function $1/x$ on $(0, 1]$ is continuous but not bounded.
Mathlib states a form for compact sets, as `IsCompact.bddAbove_image`.

## Explanations

### L1 human

On a closed bounded interval, a continuous function cannot shoot off to infinity.
Its graph stays between two horizontal lines.