Understanding of the Metric Space axioms - (axiom 2 only)

[chwala]

Homework Statement

See below

Relevant Equations

Metric spaces

Am refreshing on Metric spaces been a while...

Consider the axioms below;
1. $d(x,y)≥0$ ∀ $x, y ∈ X$ - distance between two points
2. $d(x,y) =0$ iff $x=y$, ∀ $x,y ∈ X$
3.$d(x,y)=d(y,x)$ ∀$x, y ∈ X$ - symmetry
3. $d(x,y)≤d(x,z)+d(z,y)$ ∀$x, y,z ∈ X$ - triangle inequality

The proofs are clear to me, i just read on that. I wanted to check how to show that axiom $2$ holds...
My take is given set $R$ with usual metric si defined by,
$d_1(x,y)$=$|x-y|$, ∀ $x, y ∈ X$, then $d_1(x,y)$= $\sqrt {(x-x)^2+(x-x)^2}$ since $x=y$

 

[Orodruin]

Axioms cannot be proven by definition.

What you can do is to prove that some construction satisfies some set of axioms.

 

[PeroK]

The proofs are clear to me, i just read on that. IU wanted to ask if we could prove axiom $2$ with
given set $R$ with usual metric,
$d_1(x,y)$=Modulus $x-y$ $x, y ∈ X$, then $d_1(x,y)$= $\sqrt {(x-x)^2+(x-x)^2}$ since $x=y$

I'm not sure what you are doing there. Axiom 2 for $\mathbb R$ says:
$$|x - y| = 0 \ \text{iff} \ \ x = y$$That can be proved from the definition of the modulus.

Hint: without loss of generality assume $x \ge y$.

 

[chwala]

Trying to simply show or state that if $x=y$, then the distance between the two points in a $2D$ plane is equal to 0. You're saying that is wrong?

 

[chwala]

Axioms cannot be proven by definition.

What you can do is to prove that some construction satisfies some set of axioms.

That's what I meant...learning point...I may need to amend thread title...

 

[PeroK]

Trying to simply show or state that if $x=y$, then the distance between the two points in a 2D plane is equal to 0. You're saying that is wrong?

You're confusing $d(x, y)$, where $x, y \in \mathbb R$ and $d(r_1, r_2)$ where $r_1 = (x_1, y_1)$ etc. are points in the plane.

 

[chwala]

You're confusing $d(x, y)$, where $x, y \in \mathbb R$ and $d(r_1, r_2)$ where $r_1 = (x_1, y_1)$ etc. are points in the plane.

I've seen that...let me look at it again...you are right. Thanks Perok.

 

[PeroK]

Note that:$$d(x,y) = | x - y|$$ and$$d(r_1, r_2) = ||r_1 - r_2|| = \sqrt{(x_1 - x_2)^2 + (y_1-y_2)^2}$$

 

[chwala]

But if indeed $x=y$, then it follows that $x$ and $y$ are one and same point...we then have $(x_1, y_1)= (x_2,y_2)$ whose Modulus is equal to 0...clarify on this. Thanks.

 

[PeroK]

But if indeed $x=y$, then it follows that $x$ and $y$ are one and same point...we then have $(x_1, y_1)= (x_2,y_2)$ whose Modulus is equal to 0...clarify on this. Thanks.

Yes, but it's "if and only if". You need to show that if $|x - y| = 0$, then $x = y$.

 

[chwala]

Yes, but it's "if and only if". You need to show that if $|x - y| = 0$, then $x = y$.

That is exactly what i wanted to state from post $1$,
$d_1(x,y)$=$|x-y|$ ∀$x, y ∈ X$,
then if $x=y$, and given that $x=(m_1,n_1)$, then $y=m_1,n_1$. It follows that
$d_1(x,y)$= $\sqrt {(m_1-m_1)^2+(n_1-n_1)^2}$=$\sqrt {(0)^2+(0)^2}=0$

 

[PeroK]

That is exactly what i wanted to state from post $1$,
$d_1(x,y)$=$|x-y|$ ∀$x, y ∈ X$,
then if $x=y$, and given that $x=(m_1,n_1)$, then $y=m_1,n_1$. It follows that
$d_1(x,y)$= $\sqrt {(m_1-m_1)^2+(n_1-n_1)^2}$=$\sqrt {(0)^2+(0)^2}=0$

You still haven't shown the converse.

 

[chwala]

You still haven't shown the converse.

You mean for $y=x$, then we shall have,
$d_1(y,x)$= $\sqrt {(n_1-n_1)^2+(m_1-m_1)^2}$=$\sqrt {(0)^2+(0)^2}=0$

implying property on Commutativity holds...

 

[PeroK]

You mean for $y=x$, then we shall have,
$d_1(y,x)$= $\sqrt {(n_1-n_1)^2+(m_1-m_1)^2}$=$\sqrt {(0)^2+(0)^2}=0$

implying property on Commutativity holds...

No, I mean that you must show that $d(x, y) = 0 \ \Rightarrow \ x = y$.

Take $d(x, y) = \sin^2(x - y)$. Clearly, $d(x,x) = 0$, but $d$ is not a metric, as $\sin^2(x-y) = 0 \not \Rightarrow \ x = y$.

 

[chwala]

No, I mean that you must show that $d(x, y) = 0 \ \Rightarrow \ x = y$.

Take $d(x, y) = \sin^2(x - y)$. Clearly, $d(x,x) = 0$, but $d$ is not a metric, as $\sin^2(x-y) = 0 \not \Rightarrow \ x = y$.

If i am getting you right by converse we are trying to establish the fact that the axiom only holds for $(x,y)$ if and only if $d$ is a Metric (the distance function) ... otherwise it won't hold...

 

[PeroK]

If i am getting you right by converse we are trying to establish the fact that the axiom only holds for $(x,y)$ if and only if $d$ is a Metric ...otherwise it won't hold...

It might be best if you accept that $|x-y|$ is a metric and not try to prove it.

 

[chwala]

It might be best if you accept that $|x-y|$ is a metric and not try to prove it.

OK...let me refresh on this...Pure Maths is not for the faint hearted...its
many years since i looked at this...Ring theory, Real Analysis etc ...time to look at them.
Cheers Perok!

 

[Hall]

@PeroK Can you please give us a little big picture of relation between General Relativity and Metric Spaces?

 

[PeroK]

@PeroK Can you please give us a little big picture of relation between General Relativity and Metric Spaces?

Different sort of metric!

 

[chwala]

@PeroK Can you please give us a little big picture of relation between General Relativity and Metric Spaces?

Two different world's...you are talking of tensors man...