1. ## Metric Space

For $0 show that $d_p(x,y)=(|x_1-y_1|^p+|x_2-y_2|^p)^{\frac{1}{p}}$ is not a metric on $\mathbb{R}^2$.
I think i've done this correctly, but i'm new to this module so I just want some confirmation that i've done it right (it's one of those modules where questions are unassessed so I could just be doing all the questions wrong!)

There are two axioms that need to be satisfied:

$d(x,y)=0$ iff $x=y$
$d(x,z) \leq d(x,y)+d(y,z)$

It satisfies the first axiom since $d(x,z)=0$ gives $|x_1-y_1|^p+|x_2-y_2|^p=0$ and this is only possible if $x_1=y_1$ and $x_2=y_2$.

I'm aiming to show that the second inequality does not hold.

I've decided to go with a proof by contradiction and try to contradict the first axiom.

So suppose $d(x,z) \leq d(x,y)+d(y,z)$.

Then $d(x,z) \leq (|x_1-y_1|^p+|x_2-y_2|^p)^{\frac{1}{p}}+(|y_1-z_1|^p+|y_2-z_2|^p)^{\frac{1}{p}}$

So when $z=x$ we have $d(x,z) \leq 2(|x_1-y_1|^p+|x_2-y_2|^p)^{\frac{1}{p}}$ so provided that $x_1 \neq y_1$ and $x_2 \neq y_2$, we have that $d(x,z)$ is not necessarily 0. This contradicts the first axiom.

Is this correct? After writing it down my proof by contradiction seems a bit odd =S

2. Originally Posted by Showcase_22
I think i've done this correctly, but i'm new to this module so I just want some confirmation that i've done it right (it's one of those modules where questions are unassessed so I could just be doing all the questions wrong!)

There are two axioms that need to be satisfied:

$d(x,y)=0$ iff $x=y$
$d(x,z) \leq d(x,y)+d(y,z)$

It satisfies the first axiom since $d(x,z)=0$ gives $|x_1-y_1|^p+|x_2-y_2|^p=0$ and this is only possible if $x_1=y_1$ and $x_2=y_2$.

I'm aiming to show that the second inequality does not hold.

I've decided to go with a proof by contradiction and try to contradict the first axiom.

So suppose $d(x,z) \leq d(x,y)+d(y,z)$.

Then $d(x,z) \leq (|x_1-y_1|^p+|x_2-y_2|^p)^{\frac{1}{p}}+(|y_1-z_1|^p+|y_2-z_2|^p)^{\frac{1}{p}}$

So when $z=x$ we have $d(x,z) \leq 2(|x_1-y_1|^p+|x_2-y_2|^p)^{\frac{1}{p}}$ so provided that $x_1 \neq y_1$ and $x_2 \neq y_2$, we have that $d(x,z)$ is not necessarily 0. This contradicts the first axiom.

Is this correct? After writing it down my proof by contradiction seems a bit odd =S
But it can be 0, you didn't prove it can't be 0. So this doesn't contradict the first axiom :O

I don't have time to find how to do the question though, sorry

3. Originally Posted by Showcase_22
I think i've done this correctly, but i'm new to this module so I just want some confirmation that i've done it right (it's one of those modules where questions are unassessed so I could just be doing all the questions wrong!)

There are two axioms that need to be satisfied:

$d(x,y)=0$ iff $x=y$
$d(x,z) \leq d(x,y)+d(y,z)$

It satisfies the first axiom since $d(x,z)=0$ gives $|x_1-y_1|^p+|x_2-y_2|^p=0$ and this is only possible if $x_1=y_1$ and $x_2=y_2$.

I'm aiming to show that the second inequality does not hold.

I've decided to go with a proof by contradiction and try to contradict the first axiom.

So suppose $d(x,z) \leq d(x,y)+d(y,z)$.

Then $d(x,z) \leq (|x_1-y_1|^p+|x_2-y_2|^p)^{\frac{1}{p}}+(|y_1-z_1|^p+|y_2-z_2|^p)^{\frac{1}{p}}$

So when $z=x$ we have $d(x,z) \leq 2(|x_1-y_1|^p+|x_2-y_2|^p)^{\frac{1}{p}}$ so provided that $x_1 \neq y_1$ and $x_2 \neq y_2$, we have that $d(x,z)$ is not necessarily 0. This contradicts the first axiom.

Is this correct? After writing it down my proof by contradiction seems a bit odd =S

Why do you think this question belongs in abstract/linear algebra and not, say in Analysis or at least Calculus? Anyway:

Take $x=(0,0)\,,\,y=(1,1)\,,\,z=(1,0)$ and now check directly that $d_p(x,y) \nleq d_p(x,z)+d_p(z,y)\,\,\,when\,\,\,0

Tonio