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Help with metrics

  1. Nov 24, 2005 #1
    We're asked to show that the following mappings [itex]d_i:M_i\times{M_i}\rightarrow{\mathbb{R}}[/itex] are metrics on the sets [itex]M_i[/itex]. Moreover, we should think about what the corresponding open neighbourhood [itex]B_{\epsilon}(x)=\{y\in{M_i}|d_i(x,y)<\epsilon\}[/itex] for a point [itex]x\in{M_i}[/itex] and [itex]\epsilon>0[/itex] looks like.

    1) Discrete metric: for a set [itex]M_1[/itex]
    [itex]d_1:M_1\times{M_1}\rightarrow{\mathbb{R}},\displaystyle d_1(x,y)=\begin{cases}0&\text{if }x=y\\1&\text{otherwise} \end{cases}.[/itex]

    2) The French Railway Metric: For [itex]M_2={\mathbb{R}}^2[/itex]
    [itex]d_2:M_2\times{M_2}\rightarrow{\mathbb{R}},\displaystyle d_2(x,y)=\begin{cases}||x-y||&\text{if }y=tx(t\in{\mathbb{R}})\\||x||+||y||&\text{otherwise} \end{cases}.[/itex]

    3) p-adic Metric on [itex]\mathbb{Z}[/itex]
    Let [itex]M_3=\mathbb{Z}[/itex] and p is a prime number:

    Please, give me at least a hint about what the structure of the proof might be...how to start and how to end.

    ...and tell me is it normal if you do such things in the first three weeks of your undergraduate maths study?...I mean are those concepts something that you should "mellow" to be able to accumulate and understand clearly?
    Last edited: Nov 24, 2005
  2. jcsd
  3. Nov 24, 2005 #2

    matt grime

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    As ever in this subforum, what have you done to try to prove them? You are only required to verify some small number of rules, so what can you verify? What can't you do?

    As for "is this normal" then who knows... Metric spaces as an abstract concept are not taught in the first three weeks of the first year of an undergraduate course at any place I know of in the UK or in the USA. They aren't very hard, but there are usually more pressing things to be taught in that time
  4. Nov 24, 2005 #3
    Yes, you're right Matt. I can't help thinking one should be mature enough to study such things, so that afterwards they leave a clear picture of the whole concept in your mind. We're just not accustomed yet to absorb them, I think. In our course it's all about generalising everything to a high degree of abstraction. I now got the fact that we're going to get the concept of limits and convergence presented in the frame of arbitrary metric spaces. I mean you can always put it down to one-dimensional case and work with sequences on the real line, but I'm pretty sure that this will come up in the exams, if you take into account just how much attention we've paid to these topics so far (we also treated closed and open sets, balls and whatsoever)...so frankly speaking I'm a bit scared. :frown:
    Well it's also worth to say that I'm a real beginner, although my nickname might make one think the converse (I just had no idea of which nickname to choose, so being coincidentally on MathWorld I picked a random entry:smile: )...So I'd like to be treated as a beginner (at least by those who have read the post, of course), that is as a simple student who wishes to get things on track. :smile:

    Ok, back to the tasks:
    I think (though unsure and careful) that, in order to show that a mapping is a metric, one should check whether the given set together with the distance function defined on it is a metric space.
    There are those 3 axioms for a metric space:
    A1) [itex]d(x,y)>0&\text{ when }x\neq{y}&\text{ and }d(x,y)=0&\text{ iff }x=y[/itex]
    A2) [itex]d(x,y)=d(y,x)[/itex] and A3)[itex]d(x,y)\le{d(x,z)+d(z,y)}[/itex]

    1) So for the discrete metric we've got:
    - [itex]d_1(x,y)=0[/itex] iff x=y fulfills A1, and [itex]d_1(x,y)=1\ge{0}[/itex] fulfills A1 too.
    - since both relations "=" and "[itex]\neq[/itex]" are reflexive we've got:
    d(x,y)=d(y,x) in both cases, which agrees with A2
    - Let [itex]z\in{M_1}[/itex] be such that:
    [itex]\display d_1(x,z)=\begin{cases}0&\text{if }x=z\\1&\text{if }x\neq{z} \end{cases}.[/itex]
    and [itex]\display d_1(z,y)=\begin{cases}0&\text{if }z=y\\1&\text{if }z\neq{y} \end{cases}.[/itex], then
    [itex]0=d_1(x,y)\le{d_1(x,z)+d_1(z,y)}=0[/itex] in case x=y,
    and [itex]1=d_1(x,y)\le{d_1(x,z)+d_1(z,y)}=2[/itex] in case [itex]x\neq{y}[/itex]. Both cases agree with A3.
    Thus we showed that [itex](M_1,d)[/itex] is a metric space if we define the metric function [itex]d_1(x,y)[/itex] as we did it, so it is indeed a metric on the given set.

    Ok, the neighbourhood is something like this:
    For ease let x=(0,0) and [itex]\epsilon=1[/itex];
    Then for the first case we have [itex]B_1((0,0))=\{(0,0)\}[/itex] with [itex]d_1(x,y)=0<1=\epsilon[/itex], so the one-element set containing the point (0,0) is an open ball.
    For the second case the set [itex]B_1((0,0))=M_1\backslash\{(0,0)\}}[/itex] with [itex]d_1(x,y)=1[/itex] is not an open ball ,since [itex]d_1(x,y)=1\not<1[/itex]. It follows that this set is a closed ball and it can be open only if [itex]\epsilon>1[/itex], but then the set [itex]B_1((0,0))=M_1[/itex] is also an open ball.
    The overall picture is:
    For [itex]0<\epsilon\le{1}[/itex]: [itex]B_1(x)=\{x\}[/itex]
    and for [itex]\epsilon>1[/itex]: [itex]B_1(x)=M_1[/itex]

    Is it correct? What about the argumentation?

    Last edited: Nov 24, 2005
  5. Nov 25, 2005 #4

    matt grime

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    Why are you letting x=(0,0)? Firstly the neighbourhood is something in M, not MxM, and secondly these are metric spaces, NOT vector (normed) spaces, so there is no way to assume such thing as zero in M.

    Let x be arbitrary, let e be arbitrary. What y's satisfy d(x,y)<e? If e<1 then y=x is the only option, if e=>1 then all y's are in B_e(x).

    The argument is sound, certainly good for a beginner. You'll eventually learn what you can and can't omit from the proof.

    I would have said:

    given d(x,y)=1 if x=/=y 0 if x=y, then

    d is positive and zero only when x=y by definition,

    it satisfies d(x,y)=d(y,x) since it is symmetric in its arguments, {I think you used reflexive or something here, which is not accurate, isn't reflexive, for a relation wRw?}

    it satisfies the triangle inequality becuase the only way it could fail is if the RHS is 0 and the LHS is 1, but then that implies x=z=y and x=/=y which is a contradiction.
    Last edited: Nov 25, 2005
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