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
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $2$: Continuity generalized: metric spaces: $2.3$: Open sets in metric spaces: Example $2.3.15$