Unit n-Cube under Chebyshev Distance is Subspace of Real Vector Space

Theorem

Let $n \in \N$.

Let $I^n$ denote the unit $n$-cube:

$I^n = \closedint 0 1^n$

that is, the Cartesian product of $n$ instances of the closed real interval $\set {x \in \R: 0 \le x \le 1}$.

Let $d_c: I^n \times I^n \to \R$ be defined as:

$\ds \map {d_c} {x, y} = \max_{i \mathop = 1}^n \set {\size {x_i - y_i} }$

where $x = \tuple {x_1, x_2, \ldots, x_n}, y = \tuple {x_1, x_2, \ldots, x_n} \in I^n$.

Then $\struct {I_n, d_c}$ is a metric subspace of $\struct {\R^n, d_\infty}$, where $d_\infty$ is the Chebyshev distance on the real vector space $\R_n$.

Proof

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Sources

• 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $2$: Metric Spaces: $\S 7$: Subspaces and Equivalence of Metric Spaces: Example $2$