Defining a Metric

  • Thread starter TylerH
  • Start date
  • #1
724
0

Main Question or Discussion Point

How does one redefine the metric of a given set, such as the reals? I thought it would be an interesting concept to have a metric defined like so:
[tex]d_X:X^n \times X^n \to \Re[/tex]
[tex](x_1, x_2, x_3, \cdots, x_n), (y_1, y_2, y_3, \cdots, y_n) \mapsto \sum^{n}_{i=1}{\frac{x_i+y_i}{2}}[/tex]
Does it have to be consistent with common sense? For example, does 5 have to be closer to 4 than 0?

Think about how messed up a graphical representation of that would look.
 

Answers and Replies

  • #2
1,384
2
Your formula doesn't define a metric. For example, consider R2, and let x = (1,0), and y = (-1,0). Then

d(x,y) = (1-1+0+0)/2 = 0.

But x does not equal y. And

d(y,y) = (-1-1+0+0)/2 = -1.

Or have I misunderstood? When you say "redefine the metric", do you mean (as I guessed) use a different metric from the standard one with a given set, or do you mean "redefine the concept of a metric", that is, attach the name "metric" to some other concept?
 
  • #3
Landau
Science Advisor
905
0
How does one redefine the metric of a given set, such as the reals?
I don't understand the question. Redefine the metric of a given set?
I thought it would be an interesting concept to have a metric defined like so:
[tex]d_X:X^n \times X^n \to \Re[/tex]
[tex](x_1, x_2, x_3, \cdots, x_n), (y_1, y_2, y_3, \cdots, y_n) \mapsto \sum^{n}_{i=1}{\frac{x_i+y_i}{2}}[/tex]
You said that X is just a set. So the addition x_i+y_i does not make any sense. It seems you mean that X is a subset of R?

A metric is a function with some special properties. One of them is that its image is the non-negative reals. Your function does not satisfy this condition, unless X consists only of non-negative numbers itself.
 
  • #4
radou
Homework Helper
3,115
6
As Landau pointed out, a metric must satisfy certain properties. One is often used to standard metrics, which have obvious geometrical interpretations and agree with what you'd call "common sense", but there are no limitations on this one - if it's defined properly, then it's a metric, no matter if it's "intuitive" or not.
 
  • #5
724
0
Yeah, that was a dumb mistake. I forgot the most basic property of metrics, that d(a,a) MUST be 0.

What I meant by "redefining the metric of a set" is, for example, to use the real numbers, but to redefine it's metric so it is not longer a flat space. To continue the example, say I wanted a 1 dimensional space(or, number line), distorted along x^2. I would redefine the metric as(the set with the redefined metric being X):
[tex]d_X:\Re \times \Re \to \Re[/tex]
[tex]\left( x_1, x_2 \right) \mapsto \int_{x_1}^{x_2} \sqrt{1+4x^2}dx[/tex]
 
  • #6
Landau
Science Advisor
905
0
Seeing you talk about 'flat space', I am starting to suspect that you are talking about a Riemannian metric instead of a metric in the sense of metric spaces?
 
  • #7
724
0
lol No, I don't think so. It'll take me a couple weeks to understand what a Riemann manifold even is! The only metric I had learned of before your last post was that described on the second Wikipedia page.

I understand the metric to be the distance between two points in a set. The part that interested me was imagining the distorted spaces that would be created if you were to define a function of N dimensions, and graph a N-1 dimension function on it. Like with my previous example, one graph y=x^2, and "bend" it straight, while preserving the scale. Such that, due to the distortion, the distance from 0 to 1 is [tex]\int_0^1 \sqrt{1+4x^2}dx[/tex] instead of 1, which it would be if x was distorted along y=x(which would be none at all).
 

Related Threads on Defining a Metric

  • Last Post
Replies
4
Views
4K
  • Last Post
Replies
8
Views
5K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
4
Views
2K
Replies
2
Views
1K
  • Last Post
Replies
1
Views
993
Replies
8
Views
3K
Top