> [!NOTE] Theorem > Let $f$ be a [[Continuous Real Function|continuous real function]] on a [[Closed Real Interval|closed real interval]] $[a,b].$ Then $f$ is [[Bounded Real Function|bounded]] on $[a,b].$ **Proof**: Suppose that $f$ is not bounded above on $[a,b]$. Then for each $n \in \mathbb{N}$, there exists $x_{n} \in [a,b]$ such that $f(x_{n}) \geq n$. Also the sequence $(x_{n})_{n\in \mathbb{N}}$ is [[Bounded Sequence|bounded]] since $x_{n}\in[a,b].$ By [[Bolzano-Weierstrass Theorem (Sequential Compactness of The Reals)]], that it has a convergent subsequence, $x_{n_{j}} \to y$ for some $y\in [a,b]$ since [[Limits of Real Sequence Preserve Weak Inequalities]]. Since $f$ is continuous at $y$, it follows that $f(x_{n_{j}}) \to f(y)$ but by assumption, $f(x_{n_{j}}) \to \infty$ while $f(y) < \infty.$ This yields a contradiction so $f$ is bounded above on $[a,b].$ For the lower bound consider the function $x \mapsto -f(x)$. **Proof**: Consider $X = \{ x \in [a,b]: f \text{ is bounded on } [a,x] \}$. The set $X$ is non-empty since $a \in X,$ and bounded above since $x \leq b$ for all $x \in X$. So by [[Real numbers|LUBA]], $\exists c:= \sup X.$ Suppose that $c<b.$ Then $f$ is bounded on $[a,c]$. Using continuity with $\varepsilon=1$, there exists a $\delta>0$ such that $|x-c|<\delta \implies |f(x)-f(c)|<1$. In particular $f$ is bounded on $\left[ c, c +\frac{\delta}{2} \right]$ which implies that $c + \frac{\delta}{2} \in X$ so $c$ was not an upper bound, a contradiction. So $f$ is bounded on $[a,b]$.