# Continuous Mapping (Metric Space)/Examples/Identity Function with Discontinuity

## Example of Non-Continuous Mapping

Let $f: \R \to \R$ be the real function defined as:

$\forall x \in \R: \map f x = \begin {cases} x & : x \ne 0 \\ 1 & : x = 0 \end {cases}$

Then $f$ is not continuous.

## Proof

Let $U \subseteq \R$ be the open set $\openint 0 \to$.

Then:

$f^{-1} \sqbrk U = \hointr 0 \to$

Note that we have $0 \in f^{-1} \sqbrk U$.

Let $\epsilon \in \R_{>0}$ be arbitrary.

Then:

 $\ds \openint {-\epsilon} 0$ $\nsubseteq$ $\ds U$ $\ds \leadsto \ \$ $\, \ds \not \exists \epsilon \in \R_{>0}: \,$ $\ds \map {B_\epsilon} 0$ $\subseteq$ $\ds U$

and so $f^{-1} \sqbrk U$ is not open in $\R$.

Thus an open set does not necessarily have an open preimage.

Hence $f$ is not continuous by definition.

$\blacksquare$