# Thread: Show that the Euclidean plane is a metric space?

1. ## Show that the Euclidean plane is a metric space?

Metric Space (S , d) consists of a space S and a fxn d that associates a real number with any two elements of S. The properties of a metric space are:

d(x , y) = d(x , y) forall x,y in S

0 < d(x , y) < inf forall x,y in S & x does not = y

d(x , x) = 0 forall x in S

d(x , y) <= d(x , z) + d(z , y) forall x,y,z in S

I have to show that the Euclidean plane (defined by two 2-D vectors X and Y?) is a metric space.

Havin a bit of trouble with this one...

2. So define the Euclidean metric for two vectors $x=(x_1,x_2)$ and $y=(y_1,y_2)$ by $d(x,y)=\sqrt{(x_1-y_1)^2+(x_2-y_2)^2}$.

The first three properties are easy to show. The triangle inequality requires some (not much) extra thinking.

3. yes I got as far, maybe you could explain the triangle inequality and the 'extra' thinking as I am at a wall.

I worked out all the distance equations for them and squared to get rid of the root... by induction one can say that with the z values added the inequality holds but I'm having trouble showing it... help?

4. Well, the first thing I would notice is that you can write $(x_i-y_i)^2$ as $(x_i-z_i+z_i-y_i)^2$. So we have:
$d(x,y)=\sqrt{(x_1-y_1)^2+(x_2-y_2)^2}$
$=\sqrt{(x_1-z_1+z_1-y_1)^2+(x_2-z_2+z_2-y_2)^2}$
if we let $a_i=x_i-z_i$ and $b_i=z_i-y_i$, we get
$d(x,y)=\sqrt{(a_1+b_1)^2+(a_2+b_2)^2}$

What we would now like is to show that:
$\ast\quad\sqrt{(a_1+b_1)^2+(a_2+b_2)^2}\leq\sqrt{a _1^2+a_2^2}+\sqrt{b_1^2+b_2^2}$
The LHS of the inequality is $\sqrt{a_1^2+a_2^2+b_1^2+b_2^2+2(a_1b_1+a_2b_2)}$. We hope that the radicand is less than or equal to $\left(\sqrt{a_1^2+a_2^2}+\sqrt{b_1^2+b_2^2}\right) ^2$. That is, we want: $a_1^2+a_2^2+b_1^2+b_2^2+2(a_1b_1+a_2b_2)\leq a_1^2+a_2^2+b_1^2+b_2^2+2\sqrt{a_1^2+a_2^2}\sqrt{b _1^2+b_2^2}$. If we boil this down to the bare inequality, we want to show that $a_1b_1+a_2b_2\leq\sqrt{a_1^2+a_2^2}\sqrt{b_1^2+b_2 ^2}$.

The last inequality there (if you haven’t seen it before) is the Cauchy-Schwarz inequality. The easiest way to prove the triangle inequality is first to prove the C.S.I., then work backwards (which is easier to do now that you know where to go) to prove $\ast$. Show that $\ast$ is equivalent to the triangle inequality.

5. Show the Euclidean Plane is a metric space.

$x=(a_1,a_2)$
$y=(b_1,b_2)$
$S=${ $(x,y)|a_1,a_2,b_1,b_2 \in R$}

$d(x,y)=d(y,x) \Rightarrow \sqrt{(a_1-b_1)^2 + (a_2-b_2)^2}=\sqrt{(b_1-a_1)^2 + (b_2-a_2)^2}$

from the RHS

$=\sqrt{(a^2_1-2a_1b_1+b^2_1) + (a^2_2-2a_2b_2+b^2_2)}$
$=\sqrt{(a_1-b_1)^2 + (a_2-b_2)^2}$

$d(x,x)=0 \Rightarrow \sqrt{(a_1-a_1)^2+(b_1-b_1)^2}=0$
$0+0=0$

$d(x,y) \leq d(x,z)+d(z,y)$

The triangle inequality is much easier to deal with when using vector notation:
we can define $d(x,z)$ and $d(z,y)$ as some vectors $\vec{a}$ and $\vec{b}$, respectively, and $d(x,y)$ (the addition of those two vectors) as $\vec{a}+\vec{b}$

This gives us $\|\vec{a}+\vec{b}\| \leq \|\vec{a}\|+\|\vec{b}\|$

Working from the LHS:
$\|\vec{a}+\vec{b}\|^2=(\vec{a}+\vec{b})\cdot(\vec{ a}+\vec{b})$
$=\vec{a}\cdot(\vec{a}+\vec{b})+\vec{b}\cdot(\vec{a }+\vec{b})$
$=\|\vec{a}\|^2+2(\vec{a}\cdot\vec{b})+\|\vec{b}\|^ 2$

by the Cauchy-Schwartz Inequality ( $|\vec{a}|\cdot|\vec{b}| \leq \|\vec{a}\|\|\vec{b}\|$)

$\Rightarrow$ $\|\vec{a}\|^2+2(\vec{a}\cdot\vec{b})+\|\vec{b}\|^2 \leq \|\vec{a}\|^2+2(\|\vec{a}\|\|\vec{b}\|)+\|\vec{b}\ |^2$

$\|\vec{a}\|^2+2(\|\vec{a}\|\|\vec{b}\|)+\|\vec{b}\ |^2 \equiv (\|\vec{a}\|+\|\vec{b}\|)^2$
$(\|\vec{a}+\vec{b}\|)^2 \leq (\|\vec{a}\|+\|\vec{b}\|)^2$
$\Rightarrow \|\vec{a}+\vec{b}\| \leq \|\vec{a}\|+\|\vec{b}\|$ $QED$

So I get it now. I was stuck because you had to use the cuachy schwartz ineq to tie the knot at the end.

thanks for the help

6. Originally Posted by Angrypoonani
The triangle inequality is much easier to deal with when using vector notation:
Yes, exactly. A general rule of thumb is that the more abstract your setting, the easier it is to show many results. For example, it is possible to show using C.S. that any inner product induces a norm ( $\|x\|^2=\langle x,x\rangle$), and it is easy to see that any norm induces a metric ( $d(x,y)=\|x-y\|$). Then since the dot product is an inner product on $\mathbb{R}^2$, the function defined as $d(x,y)=\sqrt{(x-y)\cdot(x-y)}$ is necessarily a metric.