Vector Subspace of Real Vector Space under Chebyshev Metric is Metric Subspace

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $n \in \N$.

Let $A$ be the set of all ordered $n+1$-tuples $\tuple {x_1, x_2, \ldots, x_{n + 1} }$ of real numbers such that $x_{n + 1} = 0$.

Let $d: A \times A \to \R$ be the function defined as:

$\displaystyle \forall x, y \in A: \map d {x, y} = \max_{i \mathop = 1}^n \set {\size {x_i - y_i} }$

where $x = \tuple {x_1, x_2, \ldots, x_{n + 1} }, y = \tuple {y_1, y_2, \ldots, y_{n + 1} }$.


Then $\struct {A, d}$ is a metric subspace of $\struct {\R^{n + 1}, d_\infty}$ where $d_\infty$ is the Chebyshev distance on the real vector space $\R^{n + 1}$.


Proof


Sources