Mapping from Cartesian Product under Chebyshev Distance to Real Number Line is Continuous

From ProofWiki
Jump to navigation Jump to search


Let $M = \struct {A, d'}$ be a metric space.

Let $\displaystyle \AA = A \times A$ be the cartesian product of $A$ with itself.

Let $d_\infty: \AA \times \AA \to \R$ be the Chebyshev distance on $\AA$:

$\displaystyle \map {d_\infty} {x, y} = \max \set {\map {d'} {x_1, y_1}, \map {d'} {x_2, y_2} }$

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.


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 $\map {d_\infty} {\tuple {x, y}, \tuple {a, b}} < \dfrac \epsilon 2$.


\(\displaystyle \map d {\map {d'} {x, y}, \map {d'} {a, b} }\) \(=\) \(\displaystyle \size {\map {d'} {x, y} - \map {d'} {a, b} }\) Definition of Euclidean Metric on Real Number Line
\(\displaystyle \) \(\le\) \(\displaystyle \size {\map {d'} {x, y} - \map {d'} {a, y} } + \size {\map {d'} {a, y} - \map {d'} {a, b} }\) Triangle Inequality
\(\displaystyle \) \(\le\) \(\displaystyle \map {d'} {x, a} + \map {d'} {y, b}\) Reverse Triangle Inequality on $d'$
\(\displaystyle \) \(\le\) \(\displaystyle 2 \max \set {\map {d'} {x, a}, \map {d'} {y, b} }\)
\(\displaystyle \) \(\le\) \(\displaystyle 2 \ \map {d_\infty} {\tuple {x, y}, \tuple {a, b} }\) Definition of Chebyshev Distance
\(\displaystyle \) \(<\) \(\displaystyle 2 \cdot \frac \epsilon 2\)
\(\displaystyle \) \(=\) \(\displaystyle \epsilon\)

Therefore $d'$ is continuous.