# Non-Zero Value of Continuous Real-Valued Function has Neighborhood not including Zero

## Theorem

Let $M = \struct {A, d}$ be a metric space.

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

Let $\map f a > 0$ for some $a \in M$.

Then there exists $\delta \in \R_{>0}$ such that:

$\forall x \in \map {B_\delta} a$

where $\map {B_\delta} a$ denotes the open $\delta$-ball of $a$ in $M$.

## Proof

Let $\epsilon \in \R_{>0}$ such that $\epsilon < \map f a$.

Let $\map {N_\epsilon} {\map f a} = \openint {\map f a - \epsilon} {\map f a + \epsilon}$ be the $\epsilon$-neighborhood of $\map f a$.

Then $\map {N_\epsilon} {\map f a}$ is an open interval of $\map f a$ such that:

$\forall y \in \map {N_\epsilon} {\map f a}: y > 0$

We are given that $f$ is continuous.

So, by definition:

$\exists U \subseteq A: \forall x \in U: \map f x \in \map {N_\epsilon} {\map f a}$

where $U$ is an open set of $M$.

As $a \map {N_\epsilon} {\map f a}$, it follows that:

$a \in U$

By definition of open set:

$\exists \delta \in \R_{>0}: \map {B_\delta} a \subseteq U$

Hence:

$\forall x \in \map {B_\delta} a: \map f x \in \map {N_\epsilon} {\map f a}$

As we have already noted that:

$\forall y \in \map {N_\epsilon} {\map f a}: y > 0$

the result follows.

$\blacksquare$