definition of finite-product metric
Topics
About: metric space
The table of contents of this article
Starting Context
- The reader knows a definition of metric space.
- The reader knows a definition of product set.
Target Context
- The reader will have a definition of finite-product 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:
\( J\): \(\in \{\text{ the finite index sets }\}\)
\( \{M_j \vert j \in J\}\): \(M_j \in \{\text{ the metric spaces }\}\) with any metric, \(dist_j\)
\( \times_{j \in J} M_j\): \(= \text{ the product set }\)
\(*dist\): \(: \times_{j \in J} M_j \times \times_{j \in J} M_j \to \mathbb{R}, (\times_{j \in J} {m_1}^j, \times_{j \in J} {m_2}^j) \mapsto \sqrt{\sum_{j \in J} dist_j ({m_1}^j, {m_2}^j)^2}\)
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Conditions:
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2: Note
Let us see that \(dist\) is indeed a metric.
Let \(m_1, m_2, m_3 \in \times_{j \in J} M_j\) be any.
1) \(0 \le dist (m_1, m_2)\) and \(dist (m_1, m_2) = 0 \iff m_1 = m_2\): \(0 \le \sqrt{\sum_{j \in J} dist_j ({m_1}^j, {m_2}^j)^2} = dist (m_1, m_2)\); if \(dist (m_1, m_2) = 0\), \(dist_j ({m_1}^j, {m_2}^j) = 0\) for each \(j \in J\), so, \({m_1}^j = {m_2}^j\) for each \(j \in J\), so, \(m_1 = m_2\); if \(m_1 = m_2\), \({m_1}^j = {m_2}^j\) for each \(j \in J\), so, \(dist_j ({m_1}^j, {m_2}^j) = 0\) for each \(j \in J\), so, \(dist (m_1, m_2) = \sqrt{\sum_{j \in J} dist_j ({m_1}^j, {m_2}^j)^2} = 0\).
2) \(dist (m_1, m_2) = dist (m_2, m_1)\): \(dist (m_1, m_2) = \sqrt{\sum_{j \in J} dist_j ({m_1}^j, {m_2}^j)^2} = \sqrt{\sum_{j \in J} dist_j ({m_2}^j, {m_1}^j)^2} = dist (m_2, m_1)\).
3) \(dist (m_1, m_3) \le dist (m_1, m_2) + dist (m_2, m_3)\): \(dist (m_1, m_3)^2 = \sum_{j \in J} dist_j ({m_1}^j, {m_3}^j)^2 \le \sum_{j \in J} (dist_j ({m_1}^j, {m_2}^j) + dist_j ({m_2}^j, {m_3}^j))^2 = \sum_{j \in J} (dist_j ({m_1}^j, {m_2}^j)^2 + dist_j ({m_2}^j, {m_3}^j)^2 + 2 dist_j ({m_1}^j, {m_2}^j) dist_j ({m_2}^j, {m_3}^j)) = \sum_{j \in J} dist_j ({m_1}^j, {m_2}^j)^2 + \sum_{j \in J} dist_j ({m_2}^j, {m_3}^j)^2 + 2 \sum_{j \in J} dist_j ({m_1}^j, {m_2}^j) dist_j ({m_2}^j, {m_3}^j) = dist (m_1, m_2)^2 + dist (m_2, m_3)^2 + 2 \sum_{j \in J} dist_j ({m_1}^j, {m_2}^j) dist_j ({m_2}^j, {m_3}^j)\), while \((dist (m_1, m_2) + dist (m_2, m_3))^2 = dist (m_1, m_2)^2 + dist (m_2, m_3)^2 + 2 dist (m_1, m_2) dist (m_2, m_3) = dist (m_1, m_2)^2 + dist (m_2, m_3)^2 + 2 \sqrt{\sum_{j \in J} dist_j ({m_1}^j, {m_2}^j)^2} \sqrt{\sum_{j \in J} dist_j ({m_2}^j, {m_3}^j)^2}\), but \(\sum_{j \in J} dist_j ({m_1}^j, {m_2}^j) dist_j ({m_2}^j, {m_3}^j) \le \sqrt{\sum_{j \in J} dist_j ({m_1}^j, {m_2}^j)^2} \sqrt{\sum_{j \in J} dist_j ({m_2}^j, {m_3}^j)^2}\), by the Cauchy-Schwarz inequality for any real or complex inner-producted vectors space: think of the Euclidean vectors space, \(\mathbb{R}^{\vert J \vert}\), with the Euclidean inner product, and regard \((dist_j ({m_1}^j, {m_2}^j)), (dist_j ({m_2}^j, {m_3}^j)) \in \mathbb{R}^{\vert J \vert}\), then, the left hand side is \(\langle (dist_j ({m_1}^j, {m_2}^j)), (dist_j ({m_2}^j, {m_3}^j) \rangle\) and the right hand side is \(\sqrt{\langle (dist_j ({m_1}^j, {m_2}^j)), (dist_j ({m_1}^j, {m_2}^j)) \rangle} \sqrt{\langle (dist_j ({m_2}^j, {m_3}^j)), (dist_j ({m_2}^j, {m_3}^j)) \rangle}\), so, \(dist (m_1, m_3)^2 \le (dist (m_1, m_2) + dist (m_2, m_3))^2\), so, \(dist (m_1, m_3) \le dist (m_1, m_2) + dist (m_2, m_3)\).
