Euclidean Metric on Real Vector Space is Metric

From ProofWiki
Jump to navigation Jump to search

Theorem

The Euclidean metric on a real vector space $\R^n$ is a metric.


Proof 1

The Euclidean metric on $\R^n$ is a special case of the $p$-product metric.

The result follows from $p$-Product Metric on Real Vector Space is Metric.

$\blacksquare$


Proof 2

Consider the Euclidean space $M = \struct {\R^n, d_2}$ where $d_2$ is the distance function given by:

$\ds \map {d_2} {x, y} = \paren {\sum_{i \mathop = 1}^n \paren {x_i - y_i}^2}^{\frac 1 2}$

where $x = \tuple {x_1, x_2, \ldots, x_n}$ and $y = \tuple {y_1, y_2, \ldots, y_n}$.


Proof of $\text M 1$

\(\displaystyle \map {d_2} {x, x}\) \(=\) \(\displaystyle \paren {\sum_{i \mathop = 1}^n \paren {x_i - x_i}^2}^{\frac 1 2}\) Definition of $d_2$
\(\displaystyle \) \(=\) \(\displaystyle \paren {\sum_{i \mathop = 1}^n 0^2}^{\frac 1 2}\)
\(\displaystyle \) \(=\) \(\displaystyle 0\)

So axiom $\text M 1$ holds for $d_2$.

$\Box$


Proof of $\text M 2$

It is required to be shown:

$\map {d_2} {x, y} + \map {d_2} {y, z} \ge \map {d_2} {x, z}$

for all $x, y, z \in \R^n$.


Let:

$(1): \quad z = \tuple {z_1, z_2, \ldots, z_n}$
$(2): \quad$ all summations be over $i = 1, 2, \ldots, n$
$(3): \quad x_i - y_i = r_i$
$(4): \quad y_i - z_i = s_i$.

Thus we need to show that:

$\ds \paren {\sum \paren {x_i - y_i}^2}^{\frac 1 2} + \paren {\sum \paren {y_i - z_i}^2}^{\frac 1 2} \ge \paren {\sum \paren {x_i - z_i}^2}^{\frac 1 2}$


We have:

\(\displaystyle \map {d_2} {x, y} + \map {d_2} {y, z}\) \(=\) \(\displaystyle \paren {\sum \paren {x_i - y_i}^2}^{\frac 1 2} + \paren {\sum \paren {y_i - z_i}^2}^{\frac 1 2}\) 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 $2$
\(\displaystyle \) \(=\) \(\displaystyle \paren {\sum \paren {x_i - y_i + y_i - z_i}^2}^{\frac 1 2}\) Definition of $r_i$ and $s_i$
\(\displaystyle \) \(=\) \(\displaystyle \paren {\sum \paren {x_i - z_i}^2}^{\frac 1 2}\)
\(\displaystyle \) \(=\) \(\displaystyle \map {d_2} {x, z}\) Definition of $d_2$

So axiom $\text M 2$ holds for $d_2$.

$\Box$


Proof of $\text M 3$

\(\displaystyle \map {d_2} {x, y}\) \(=\) \(\displaystyle \paren {\sum \paren {x_i - y_i}^2}^{\frac 1 2}\) Definition of $d_2$
\(\displaystyle \) \(=\) \(\displaystyle \paren {\sum \paren {y_i - x_i}^2}^{\frac 1 2}\)
\(\displaystyle \) \(=\) \(\displaystyle \map {d_2} {y, x}\) Definition of $d_2$

So axiom $\text M 3$ holds for $d_2$.

$\Box$


Proof of $\text M 4$

\(\displaystyle x\) \(\ne\) \(\displaystyle y\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \exists k \in \closedint 1 n: x_k\) \(\ne\) \(\displaystyle y_k\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle x_k - y_k\) \(\ne\) \(\displaystyle 0\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {\sum \paren {x_k - y_k}^2}^{\frac 1 2}\) \(>\) \(\displaystyle 0\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \map {d_2} {x, y}\) \(>\) \(\displaystyle 0\) Definition of $d_2$

So axiom $\text M 4$ holds for $d_2$.

$\blacksquare$


Also see


Sources