Euclidean Metric is Metric/Proof 2
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Theorem
Let $M_{1'} = \left({A_{1'}, d_{1'}}\right), M_{2'} = \left({A_{2'}, d_{2'}}\right), \ldots, M_{n'} = \left({A_{n'}, d_{n'}}\right)$ be metric spaces.
Let $\displaystyle \mathcal A = \prod_{i \mathop = 1}^n A_{i'}$ be the cartesian product of $A_{1'}, A_{2'}, \ldots, A_{n'}$.
The Euclidean metric on $\mathcal A$ is a metric.
Proof
We have that the Euclidean metric on $\mathcal A$ is defined as:
- $\displaystyle \map {d_2} {x, y} = \paren {\sum_{i \mathop = 1}^n \paren {\map {d_{i'} } {x_i, y_i} }^2}^{\frac 1 2}$
where $x = \tuple {x_1, x_2, \ldots, x_n}, y = \tuple {y_1, y_2, \ldots, y_n} \in \mathcal A$.
Proof of $M1$
| \(\displaystyle \map {d_2} {x, x}\) | \(=\) | \(\displaystyle \paren {\sum_{i \mathop = 1}^n \paren {\map {d_{i'} } {x_i, x_i} }^2}^{\frac 1 2}\) | Definition of $d_2$ | ||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \paren {\sum_{i \mathop = 1}^n 0^2}^{\frac 1 2}\) | as $d_{i'}$ fulfils axiom $M1$ | ||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle 0\) |
So axiom $M1$ holds for $d_2$.
$\Box$
Proof of $M2$
Let:
- $(1): \quad z = \tuple {z_1, z_2, \ldots, z_n}$
- $(2): \quad$ all summations be over $i = 1, 2, \ldots, n$
- $(3): \quad \map {d_{i'} } {x_i, y_i} = r_i$
- $(4): \quad \map {d_{i'} } {y_i, z_i} = s_i$.
Thus we need to show that:
- $\displaystyle \paren {\sum \paren {\map {d_{i'} } {x_i, y_i} }^2}^{\frac 1 2} + \paren {\sum \paren {\map {d_{i'} } {y_i, z_i} }^2}^{\frac 1 2} \ge \paren {\sum \paren {\map {d_{i'} } {x_i, z_i} }^2}^{\frac 1 2}$
We have:
| \(\displaystyle \map {d_2} {x, y} + \map {d_2} {y, z}\) | \(=\) | \(\displaystyle \paren {\sum \paren {\map {d_{i'} } {x_i, y_i} }^2}^{\frac 1 2} + \paren {\sum \paren {\map {d_{i'} } {y_i, z_i} }^2}^{\frac 1 2}\) | Definition of $d_2$ | ||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \paren {\sum r_i^2}^{\frac 1 2} + \paren {\sum s_i^2}^{\frac 1 2}\) | |||||||||||
| \(\displaystyle \) | \(\ge\) | \(\displaystyle \paren {\sum \paren {r_i + s_i}^2}^{\frac 1 2}\) | Minkowski's Inequality for Sums: index $2$ | ||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \paren {\sum \paren {\map {d_{i'} } {x_i, y_i} + \map {d_{i'} } {y_i, z_i} }^2}^{\frac 1 2}\) | Definition of $r_i$ and $s_i$ | ||||||||||
| \(\displaystyle \) | \(\ge\) | \(\displaystyle \paren {\sum \paren {\map {d_{i'} } {x_i, z_i} }^2}^{\frac 1 2}\) | as $d_{i'}$ fulfils axiom $M2$ | ||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \map {d_2} {x, z}\) | Definition of $d_2$ |
So axiom $M2$ holds for $d_2$.
$\Box$
Proof of $M3$
| \(\displaystyle \map {d_2} {x, y}\) | \(=\) | \(\displaystyle \paren {\sum \paren {\map {d_{i'} } {x_i, y_i} }^2}^{\frac 1 2}\) | Definition of $d_2$ | ||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \paren {\sum \paren {\map {d_{i'} } {y_i, x_i} }^2}^{\frac 1 2}\) | as $d_i$ fulfils axiom $M3$ | ||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \map {d_2} {y, x}\) | Definition of $d_2$ |
So axiom $M3$ holds for $d_2$.
$\Box$
Proof of $M4$
| \(\displaystyle x\) | \(\ne\) | \(\displaystyle y\) | |||||||||||
| \(\displaystyle \leadsto \ \ \) | \(\displaystyle \exists k \in \closedint 1 n: x_k\) | \(\ne\) | \(\displaystyle y_k\) | ||||||||||
| \(\displaystyle \leadsto \ \ \) | \(\displaystyle \map {d_k} {x_k, y_k}\) | \(>\) | \(\displaystyle 0\) | as $d_k$ fulfils axiom $M4$ | |||||||||
| \(\displaystyle \leadsto \ \ \) | \(\displaystyle \paren {\sum \paren {\map {d_{i'} } {x_i, y_i} }^2}^{\frac 1 2}\) | \(>\) | \(\displaystyle 0\) | ||||||||||
| \(\displaystyle \leadsto \ \ \) | \(\displaystyle \map {d_2} {x, y}\) | \(>\) | \(\displaystyle 0\) | Definition of $d_2$ |
So axiom $M4$ holds for $d_2$.
$\blacksquare$
