# Question on a property of a function which is not a metric

• B
• trees and plants

#### trees and plants

We have the function d from VxV to another set(not necessarily R) for which the following properties are to be satisfied:
i) d(x,y)=0<=>x=y
ii)d(x,y)=d(y,x)
iii)d(x,z)≤(d2(x,y)+d2(y,z))1/2
∀ x,y,z ∈ V.
What do you say? Would this function have interesting properties on a set and theorems to be studied? I think perhaps it is not in the math literature. It somehow contradicts the Euclidean distance but also allows for it if someone wants. Sorry if i should not start a thread like this, because it is not perhaps in the math literature. Is it wrong if i write things like these in physics forums? What kind of things should i write about? What is allowed?

What is d2 ? If not d( d(x,y) ), then what ?

It is d squared,d times d but now that i think of it because it does not belong necessarilly to R then it belongs to a set where multiplication and powers of numbers are allowed. But perhaps that set needs more properties for the addition and multiplication at least.

It is a superset of R at least this set.

It is d squared,d times d
Then don't write d2(x,y) but ( d(x,y) )2

But perhaps that set needs more properties
Yes, definitely:

Well, it can't be ##\mathbb {C}## either ( no ##\le## ).

This function d with the set V is not a metric space, because if you take the third property i wrote and square both sides, it does not give a metric space or an example of a metric space. Could it be a topological space? How can i test this? I know the definition of a topological space i think but how do i do it? Could some examples of it or the general case for each point have a neighborhood homeomorphic to R? Then it could allow for someone to do differential calculus on it.

This function i defined, in only the negative case of the inequality, is not a property of a perpendicular triangle in euclidean space if we think of the pythagorean theorem. The slope of a straight line i think is related i think to the derivative of a function at least in special cases. In some shapes with a slope a triangle is shown that is related to the slope of the straight line.

So could somehow although this function is not for < of the inequality a superset of R, tangent vectors be applied to perhaps the "abstract shapes" it has and allow for it locally to be homeomorphic to Euclidean space and allow for differential calculus for a person to do on it?

Well, i thought of that third property of function d. It should satisfy the condition d(x,y)d(y,z)=0 (1) , but with d(x,y)≠0≠d(y,z) (2). Because of the first property i wrote before, the fist property can not hold so should not be considered. Otherwise we have a contradiction.

d(x,y)≠0≠d(y,z) means that by putting y=x or z=y we can not have the first property i wrote before, it is a contradiction. It goes against what we know from R because if we have a=b=0<=>ab=0 (3) for real numbers a,b.Also, it goes against what we know for a metric where having the same points as inputs in the metric we get 0. It is not a metric. Remember that equivalence (3) holds also for complex numbers.

If we consider d(z,y) in the third property, then it does not contradict the second property because d(z,x)<(d2(z,y)+d2(y,x))1/2. By using the second identity for z,y and y,x, which means d2(z,y)=d(2(y,z) and d2(y,x)=d2(x,y). I think that those two properties do not contradict each other.

So, it is about if someone finds it interesting or not to study a function like this with a set V which belongs to a set let us call it NRNMS for not real not metric space. NRNMS is not like complex numbers, complex numbers contain imaginary numbers which are not real, but NRNMS does not contain complex numbers and is not a metric space.

For NRNMS, a mathematical object can be defined like the open ball of radius r, BNRNMS(x,r)={z∈V: d(x,z)< (d2(z,y)+d2(y,x))1/2<r} with r≠0. If r is not real number and ≠0, then it belongs to NRNMS. It can not be smaller than 0 or bigger as we say in the real numbers, because then it would be a real number and we reach a contradiction.

So, perhaps the < we have which is not inequality we have for real numbers needs more studying.Could it be a total order? It satisfies i think the connexity condition. I want to check whether V has or should have some of the properties of a field or if it satisfies some axioms of the real numbers.

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... Would this function have interesting properties on a set and theorems to be studied?...
The impression that people get about mathematics, from courses and textbooks, is often that it is about giving definitions and proving theorems about the terms defined. But that is only the way maths is presented, because it is the fastest and shortest way. Usually mathematics is not done that way. It is about solving problems. So you need to start with a good problem and try to solve it. The solution may lead to new concepts, or not. And before you can solve problems that are original and interesting, you need to learn enough of what is already done, and most of all you need to practice problem solving.

• ChinleShale, PhDeezNutz and S.G. Janssens
The impression that people get about mathematics, from courses and textbooks, is often that it is about giving definitions and proving theorems about the terms defined. But that is only the way maths is presented, because it is the fastest and shortest way. Usually mathematics is not done that way. It is about solving problems. So you need to start with a good problem and try to solve it. The solution may lead to new concepts, or not. And before you can solve problems that are original and interesting, you need to learn enough of what is already done, and most of all you need to practice problem solving.
Someone can make their own questions in mathematics or answer questions other people have made. Even making definitions could be from a question like: what do i want to define? A question is like an end. Reaching that end answers the question. You give the information needed to answer the question correctly.

I think more experience in answering questions or problems helps, it is that i do not want to answer problems already answered by others so much and i prefer unanswered questions or problems to try to answer.

Perhaps sometimes in math or the sciences what is considered interesting has to do with the acceptance from others of your work?

We have the function d from VxV to another set(not necessarily R) for which the following properties are to be satisfied:
i) d(x,y)=0<=>x=y
ii)d(x,y)=d(y,x)
iii)d(x,z)≤(d2(x,y)+d2(y,z))1/2
∀ x,y,z ∈ V.
What do you say? Would this function have interesting properties on a set and theorems to be studied?
If you're going to make up your own problems, you at least need to fill in the important information, such as:
What is set V?
How is your distance metric defined? You have some of its properties, but nowhere do you say how the metric d actually is defined.
So you need to start with a good problem and try to solve it.
Agreed.
And before you can solve problems that are original and interesting, you need to learn enough of what is already done, and most of all you need to practice problem solving.
Also agreed. To the OP -- the problem in this thread is pretty half-baked. It would be better to look in an analysis textbook to see some problems that are interesting before attempting to write your own problems.
I think more experience in answering questions or problems helps, it is that i do not want to answer problems already answered by others so much and i prefer unanswered questions or problems to try to answer.
So you want to "do mathematics" but you're not interested in doing the "grunt work" associated with this field?
Perhaps sometimes in math or the sciences what is considered interesting has to do with the acceptance from others of your work?
It's not just sometimes ...

• BvU
I would chime in on @martinbn 's post #12 to say that much of problem solving in mathematics leads to developing theories that explain large amount mathematical phenomena. One compelling example is the theory of Riemann surfaces. Another is the theory of Characteristic Classes of vector bundles. A third is the idea of intrinsic differential geometry which partly arose in an attempt to understand what it means to compare measurements in different places on a manifold.

Problem solving can also branch off into entire new areas of study. A good example is the attempts to solve the Poincare conjecture.

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But if you look at the third property iii)##d(x,z)≤\sqrt{d^2(x,y)+d^2(y,z)}## and square it like ##d^2(x,z)≤d^2(x,y)+d^2(y,z)## you get an inequality for the pythagorean theorem for orthogonal triangles in ##\mathbb{R^2}## i think, so could it provide another space with this function with these properties i mentioned earlier?

-2<1 so that ##(-2)^2<1^2## ?

-2<1 so that ##(-2)^2<1^2## ?
The set that ##d(x,y)## belongs to is not ##R## but another set like it.I think perhaps i did not say it but it has properties like it, but does not share all of the properties of ##R##. This set is not ##R## and i do know if its numbers can be approximated by ##R##.

The set that belongs to is not but another set like it.I think perhaps i did not say it but it has properties like it, but does not share all of the properties of .
it is too smart for me

it is too smart for me
It could be a set with defined properties like we have in ##R##, axioms and theorems could then be proved from it. I do not think it is that difficult.

I just do not know if this set is worth studying it.Or if it is interesting.For me it is interesting , i do not know for others. Remember that we have non euclidean geometries which have alternative parallel postulates or could be defined i think with different metrics or the curvatures they have as spaces are different.

In this case that we have the third property which is different than the pythagorean theorem perhaps something different could be produced. In the pythagorean theorem an equality holds, in this property it as an inequality as an analogy of the pythagorean theorem that could be defined.

This function does not have a graphical representation as we know in ##R## or ##R^2## . If d(x,y) belonged to ##R## then it could, but ##d(x,y)\notin R##

Thread locked. There is too little information here for a constructive discussion.