# Continuity Test for Real-Valued Functions/Necessary Condition

## Theorem

Let $\struct {S, \tau}$ be a topological space.

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

Let $x \in S$.

Let $f$ be continuous at $x$

Then:

$\forall \epsilon \in \R_{>0} : \exists U \in \tau : x \in U : \map {f^\to} U \subseteq \openint {\map f x - \epsilon} {\map f x + \epsilon}$

## Proof

Let $f$ be continuous at $x$.

Then by definition:

For every neighborhood $N$ of $\map f x$ in $\R$, there exists a neighborhood $M$ of $x$ in $S$ such that $\map {f^\to} M \subseteq N$.

From:

Open Ball in Real Number Line is Open Interval
Open Ball of Metric Space is Open Set
Set is Open iff Neighborhood of all its Points

it follows that:

For every $\epsilon \in \R_{>0}$, there exists a neighborhood $M$ of $x$ in $S$ such that $\map {f^\to} M \subseteq \openint {\map f x - \epsilon} {\map f x + \epsilon}$

By definition of a neighborhood in $\struct {S, \tau}$:

for all neighborhoods $M$ there exists $U \in \tau$ such that $z \in U \subseteq M$

Thus:

For every $\epsilon \in \R_{>0}$, there exists $U \in \tau$ with $x \in U$ such that $\map {f^\to} U \subseteq \openint {\map f x - \epsilon} {\map f x + \epsilon}$

$\blacksquare$