description/proof of that for Euclidean set, taking for \(2\) points, sum of absolute differences of components of points is metric
Topics
About: metric space
The table of contents of this article
Starting Context
- The reader knows a definition of Euclidean set.
- The reader knows a definition of metric.
Target Context
- The reader will have a description and a proof of the proposition that for any Euclidean set, taking for each \(2\) points, the sum of the absolute differences of the components of the points is a metric.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(d\): \(\in \mathbb{N} \setminus \{0\}\)
\(\mathbb{R}^d\): \(= \text{ the Euclidean set }\)
\(f\): \(: \mathbb{R}^d \times \mathbb{R}^d \to \mathbb{R}, (r_1, r_2) \mapsto \sum_{j \in \{1, ..., d\}} \vert {r_1}^j - {r_2}^j \vert\)
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Statements:
\(f \in \{\text{ the metrics on } \mathbb{R}^d\}\)
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2: Proof
Whole Strategy: Step 1: see that \(f\) satisfies the conditions to be a metric.
Step 1:
Let us see that \(f\) satisfies the conditions to be a metric.
Let \(r_1, r_2, r_3 \in \mathbb{R}^d\) be any.
1) \(0 \le f (r_1, r_2)\) and \(f (r_1, r_2) = 0\) if and only if \(r_1 = r_2\): \(0 \le \sum_{j \in \{1, ..., d\}} \vert {r_1}^j - {r_2}^j \vert\); if \(f (r_1, r_2) = 0\), \({r_1}^j - {r_2}^j = 0\) for each \(j\), which implies that \(r_1 = r_2\); if \(r_1 = r_2\), \({r_1}^j - {r_2}^j = 0\) for each \(j\), which implies that \(f (r_1, r_2) = 0\).
2) \(f (r_1, r_2) = f (r_2, r_1)\): \(f (r_1, r_2) = \sum_{j \in \{1, ..., d\}} \vert {r_1}^j - {r_2}^j \vert = \sum_{j \in \{1, ..., d\}} \vert {r_2}^j - {r_1}^j \vert = f (r_2, r_1)\).
3) \(f (r_1, r_3) \le f (r_1, s_2) + f (r_2, r_3)\): \(f (r_1, r_3) = \sum_{j \in \{1, ..., d\}} \vert {r_1}^j - {r_3}^j \vert = \sum_{j \in \{1, ..., d\}} \vert {r_1}^j - {r_2}^j + {r_2}^j - {r_3}^j \vert \le \sum_{j \in \{1, ..., d\}} \vert {r_1}^j - {r_2}^j \vert + \vert {r_2}^j - {r_3}^j \vert = \sum_{j \in \{1, ..., d\}} \vert {r_1}^j - {r_2}^j \vert + \sum_{j \in \{1, ..., d\}} \vert {r_2}^j - {r_3}^j \vert = f (r_1, s_2) + f (r_2, r_3)\).
