> [!theorem] > > Let $f: [a, b] \to \real$ be a real-valued [[Continuity|continuous]] [[Function|function]] such that $f(a) < 0 < f(b)$. Then $\exists c \in [a, b]: a < c < b$ and $f(c) = 0$. > > *Proof*. Let $L = \bracs{x \in [a, b]: f(x) \le 0}$. $L$ is a non-empty, bounded set. By [[Real Numbers|completeness]], $\sup L$ exists. Note that $a \le \sup L \le b$ by definition of [[Supremum and Infimum|supremum]] and because $L \subseteq [a, b]$. As $f(a) < 0 < f(b)$, $a < \sup L < b$ by continuity. > > Let $n \in \nat$ and consider $\sup L - \frac{1}{n}$, then > $ > \exists x_n \in L: x_n > \sup L - \frac{1}{n} > $ > which constructs a sequence $(x_n)$ with $\limv{n}{x_n} = \sup L$ and $\limv{f(x_n)} = f(\sup L)$ by continuity, therefore $f(\sup L) \le 0$. > > Since $\sup L < b$, there exists a sequence $(x_n)$ in $[a, b]$ with $x_n > \sup L$ and $\limv{n}{x_n} = \sup L$. Then $\limv{n}{f(x_n)} \ge 0$ and $\limv{n}{f(x_n)} = f\paren{\limv{n}f(x_n)} \ge 0$ by continuity. > > Therefore $f(\sup L) = 0$. > [!theorem] > > The image of an interval by a non-constant, continuous function is an interval of the similar form. Closed intervals always map to closed intervals. Open ends can be mapped to anything.