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

• chwala
In summary: Very advanced math ...I have read that GR is based on Riemannian Geometry, which is a branch of differential geometry. Metric spaces are a generalization of Euclidean spaces that deals with the distance between points. They are used in mathematics to study properties of spaces and functions. In general relativity, the curvature of space-time is described by a metric tensor, which is a mathematical object that measures the distance and angles between points in space-time. This allows us to understand how gravity affects the motion of objects and the structure of the universe. So, while metric spaces are used to study general properties of spaces, they are also a crucial tool in understanding the fundamental principles of general relativity. In summary, metric spaces and general relativity are
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
chwala said:
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?

Orodruin said:
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...

chwala said:
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.

PeroK said:
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.

chwala said:
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##.

PeroK said:
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##

chwala said:
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.

PeroK said:
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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chwala said:
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
PeroK said:
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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chwala said:
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.

PeroK said:
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?

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

Hall said:
@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

## 1. What is the second axiom of the Metric Space?

The second axiom of the Metric Space states that the distance between any two points in the space must be a positive real number. This means that the distance cannot be negative or zero.

## 2. How does the second axiom affect the definition of a Metric Space?

The second axiom is one of the three axioms that define a Metric Space. Without this axiom, the space would not be considered a Metric Space as it would not satisfy the properties of a distance function.

## 3. Can the second axiom be violated in a Metric Space?

No, the second axiom must hold true for all points in the space. If the distance between any two points is not a positive real number, then the space cannot be considered a Metric Space.

## 4. Why is the second axiom important in the study of Metric Spaces?

The second axiom ensures that the distance function in a Metric Space is well-defined and allows for the use of important mathematical tools and techniques in the study of these spaces. It also helps to establish the basic properties of distance, such as the triangle inequality.

## 5. Are there any other axioms in the definition of a Metric Space?

Yes, there are two other axioms in addition to the second axiom that define a Metric Space. The first axiom states that the distance between any two points must be non-negative, and the third axiom states that the distance between two points must be equal to zero if and only if the two points are the same. Together, these three axioms form the foundation of the Metric Space concept.

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