**Lemma** If $f: E \to R$ is [[Continuous Function (Epsilon-Delta Definition)|continuous]] at $c \in E$ and $f(c) > \alpha$ then there exists a $\delta >0$ such that $f(x) > \alpha$ for all $x \in E$ with $|x-c| < \delta$. Similarly, if $f(c) < \alpha$ then there exists a $\delta >0$ such that $f(x) < \alpha$ for all $x \in E$ with $|x-c|<\delta$. **Proof** Suppose that $f(c) > \alpha$. Apply the definition with $\epsilon=f(c)-a$. Then there exists $\delta >0$ such that if $x \in E$ and $|x-c| < \delta$ then $|f(x)-f(c)|<f(c) - \alpha$and so $f(x) = f(c)+(f(x)-f(c)) \geq f(c)- |f(x)-f(c)| > \alpha$The proof for $f(c) < \alpha$ is similar, but $\epsilon=\alpha-f(c)$ Example: ![[Preservation of Inequalities by Continuous Functions.png|500]] See **Application** [[Algebra of Continuous Real Functions]].