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

• chwala

#### chwala

Gold Member
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##

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Axioms cannot be proven by definition.

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

• chwala
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##.

• Hall and 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?

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...

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.

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.

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.

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##.

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##

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.

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...

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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...

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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.

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!

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

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

@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...

• PeroK