# Positive Image of Point of Continuous Real Function implies Positive Closed Interval of Domain

## Theorem

Let $f: \R \to \R$ be a continuous real function.

Let $a \in \R$ such that $f \left({a}\right) > 0$.

Then:

$\exists k \in \R_{>0}: \exists \delta \in \R_{>0}: \forall x \in \left[{a - \delta \,.\,.\, a + \delta}\right]: f \left({x}\right) \ge k$

## Proof

Let $f \left({a}\right) = l$ where $l > 0$.

As $f$ is continuous:

$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \left\vert{y - x}\right\vert < \delta \implies \left\vert{f \left({y}\right) - f \left({x}\right)}\right\vert < \epsilon$

Let $\epsilon = \dfrac l 2 = k$.

Then:

$\exists \delta' \in \R_{>0}: \forall y \in \R: \left\vert{y - x}\right\vert < \delta' \implies \left\vert{f \left({y}\right) - f \left({a}\right)}\right\vert < \dfrac l 2$

Thus:

$\forall x \in \left({a - \delta' \,.\,.\, a + \delta'}\right): f \left({x}\right) > \dfrac l 2$

Let $\delta = \dfrac {\delta'} 2$

Then $a - \delta \in \left({a - \delta' \,.\,.\, a + \delta'}\right)$ and $a + \delta \in \left({a - \delta' \,.\,.\, a + \delta'}\right)$.

Thus:

$\left[{a - \delta \,.\,.\, a + \delta}\right] \subseteq \left({a - \delta' \,.\,.\, a + \delta'}\right)$

and hence the result.

$\blacksquare$