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

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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.

$\blacksquare$


Sources