Preimage of Subset under Mapping/Examples/Identity Function with Discontinuity

Example of Preimage of Subset under Mapping

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

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

Let the following subsets of $\R$ be defined:

$\ds A$

$=$

$\ds \openint \gets 0$

$\ds B$

$=$

$\ds \openint 0 \to$

$\ds C$

$=$

$\ds \openint 0 1$

$\ds D$

$=$

$\ds \openint {-1} 1$

$\ds E$

$=$

$\ds \openint {-2} {-1}$

$\ds F$

$=$

$\ds \openint {\dfrac 1 2} 2$

$f^{-1} \sqbrk A = \set {x \in \R: \map f x < 0} = \openint \gets 0$

$f^{-1} \sqbrk B = \set {x \in \R: \map f x > 0} = \openint 0 \to$

$f^{-1} \sqbrk C = \set {x \in \R: 0 < \map f x < 1} = \openint 0 1$

$f^{-1} \sqbrk D = \set {x \in \R: -1 < \map f x < 1} = \openint {-1} 0 \cup \openint 0 1$

$f^{-1} \sqbrk {E \cup F} = \openint {-2} {-1} \cup \openint {\dfrac 1 2} 2 \cup \set 0$

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 $2.3.15$