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

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Theorem

Let $n \in \N$.

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

$I^n = \left[{0 \,.\,.\, 1}\right]^n$

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


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

$\displaystyle d_c \left({x, y}\right) = \max_{i \mathop = 1}^n \left\{ {\left\lvert{x_i - y_i}\right\rvert}\right\}$

where $x = \left({x_1, x_2, \ldots, x_n}\right), y = \left({x_1, x_2, \ldots, x_n}\right) \in I^n$.


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


Proof


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