Euclidean Metric is Metric

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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 1

The Euclidean metric on $\mathcal A$ is a special case of the $p$-product metric.

The result follows from $p$-Product Metric is Metric.

$\blacksquare$


Proof 2

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 $d_2$.

$\blacksquare$


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