Mapping from Cartesian Product under Chebyshev Distance to Real Number Line is Continuous
Jump to navigation
Jump to search
Theorem
Let $M = \struct {A, d'}$ be a metric space.
Let $\ds \AA = A \times A$ be the cartesian product of $A$ with itself.
Let $d_\infty: \AA \times \AA \to \R$ be the Chebyshev distance on $\AA$:
- $\ds \map {d_\infty} {x, y} = \max \set {\map {d'} {x_1, y_1}, \map {d'} {x_2, y_2} }$
where $x = \tuple {x_1, x_2}, y = \tuple {y_1, y_2} \in \AA$.
Then the mapping:
- $d': \struct {A \times A, d_\infty} \to \struct {\R, d}$
where $d$ is the usual metric, is continuous.
Proof
From definition of continuous mapping:
We just need to show that:
- $\forall \tuple {a, b} \in A \times A: \forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall \tuple {x, y} \in A \times A: \map {d_\infty} {\tuple {x, y}, \tuple {a, b}} < \delta \implies \map d {\map {d'} {x, y}, \map {d'} {a, b}} < \epsilon$
Let $\tuple {a, b} \in A \times A$ and $\epsilon \in \R_{>0}$.
Let $\tuple {x, y} \in A \times A$.
Suppose $\map {d_\infty} {\tuple {x, y}, \tuple {a, b}} < \dfrac \epsilon 2$.
Then:
\(\ds \map d {\map {d'} {x, y}, \map {d'} {a, b} }\) | \(=\) | \(\ds \size {\map {d'} {x, y} - \map {d'} {a, b} }\) | Definition of Euclidean Metric on Real Number Line | |||||||||||
\(\ds \) | \(\le\) | \(\ds \size {\map {d'} {x, y} - \map {d'} {a, y} } + \size {\map {d'} {a, y} - \map {d'} {a, b} }\) | Triangle Inequality | |||||||||||
\(\ds \) | \(\le\) | \(\ds \map {d'} {x, a} + \map {d'} {y, b}\) | Reverse Triangle Inequality on $d'$ | |||||||||||
\(\ds \) | \(\le\) | \(\ds 2 \max \set {\map {d'} {x, a}, \map {d'} {y, b} }\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds 2 \map {d_\infty} {\tuple {x, y}, \tuple {a, b} }\) | Definition of Chebyshev Distance | |||||||||||
\(\ds \) | \(<\) | \(\ds 2 \cdot \frac \epsilon 2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \epsilon\) |
Therefore $d'$ is continuous.
$\blacksquare$
Sources
- 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $2$: Metric Spaces: $\S 5$: Limits: Exercise $9$