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Points in 2- and 3-dimensional space

  1. Nov 24, 2005 #1
    I've got a question concerning neighbourhoods of points in 2- and 3-dimensional space.

    How can we explicitly show, using only the definition of a given metric, that the according neighbourhood is some figure?

    For example the three metrics are given:
    1) [itex]d(x,y)=\sqrt{\sum_{i=1}^{n}{(x_i-y_i)^2}}[/itex];
    2) [itex]d(x,y)=\displaystyle\max_i|x_i-y_i|[/itex];
    3) [itex]d(x,y)=\sum_{i=1}^{n}|x_i-y_i|[/itex]

    If [itex]a\in{\mathbb{R}}^2[/itex] and the metric is 1), then it's clear:
    [itex]d(a,x)=\sqrt{(a_1-x_1)^2+(a_2-x_2)^2}<\epsilon[/itex] and it's clear that it's a circle...because one can rewrite:[itex](a_1-x_1)^2+(a_2-x_2)^2<{\epsilon}^2[/itex]

    But I have problems to see a picture when looking at the other metrics:
    [itex]d(a,x)=|a_1-x_1|+|a_2-x_2|<\epsilon[/itex] tell me nothing at the moment about what the according neighbourhood might look like.

    Could you please enlighten me on this case? :smile:
    Last edited: Nov 24, 2005
  2. jcsd
  3. Nov 24, 2005 #2


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    A good way to reduce confusion is to take y to be the origin: (0,0).
    (It also keeps me from getting confused about whether "y" refers to a different point or the y-component of a single point!) Once you see a neighborhood of (0,0), all neighborhoods of other points look the same. It might also help to take [itex]\epsilon[/itex] to be 1.
    That way, for example, N1((0,0)) using metric 1, the "usual" metric, is the set of points (x,y) such that x2+ y2< 1, the disk with center (0,0) and radius 1 (strictly speaking the neighborhood is not a circle- the boundary of the neighborhood is a circle). All neighborhoods in that metric are disks.

    With metric 2, we must have max(|x|,|y|)< 1. It boundary is given by max(|x|,|y|)= 1. If x< y, then that is just |y|= 1 so y= 1 or y= -1. Draw the two lines y= 1 and y= -1. If y< x, then that equation is just |x|=1 so x=1 or x=-1. Draw the two lines x= 1 and x= -1. Stare at that picture until it dawns on you!

    With metric 3, we must have |x|+ |y|< 1. The boundary of that is |x|+ |y|= 1. Standard method of working with absolute values is to separate "positive" from "negative". If x and y are both positive (first quadrant), the equation is just x+ y= 1. Draw that line. If x and y are both negative, the equation is just -x- y= 1 or x+ y= -1. Draw that line.
    I'll let you do the other two cases: x>0, y<0 and x<0, y>0. Draw those for lines and stare at the picture!
  4. Nov 24, 2005 #3
    Oh, Jesus Christ, HallsofIvy, I didn't think it would be that easy!!! Thank you very much!:smile: ...Why didn't it come?!...I'm about to bounce my head against the wall for that!
    It's all about that kick that doesn't come when one needs it most!
    I think it's another situation, especially for a beginner, when maths looks so abstract that one gives up the hope to find a door back to reality or to one's previous knowledge, although it is just beside you. :redface:

    Ok, for the first metric it's a square (without the boundary) with diagonal=[itex]2\sqrt{2}\epsilon[/itex].
    For the second metric we have:
    for x>0 and y<0: y=x-1; for x<0 and y>0: y=x+1...so it's again a square without the boundary, but turned 45° around the given point and whose diagonal=[itex]2\epsilon[/itex]
    If I got it right, to get another square/neighbourhood different from the second above, i.e. turned for example to the angle 0°<angle<45°, we must construct another metric, right? So for a slightly transformed figure, the metric is completely different, isn't it? Is there a(n) rule/algorithm for constructing metrics of "topologically" equivalent figures/objects (in an arbitrary space)?
  5. Nov 24, 2005 #4

    matt grime

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    Topological equivalence in no way depends on the *shape* of a given basic figure. What matters is when sets are open in both topologies. Personally I think you're s concentrating the wrong thing. What the things look like is not important.
  6. Nov 25, 2005 #5


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    No, the shape is not important but its an simple, interesting exercise. It also helps to show that the shape is not important!

    It is easy to see that, given any disk, we can find a square and a diamond about the center that are completely contained in that disk.
    Given any square we can find a disk and a diamond that are completely contained in that square.
    Finally, given any diamond, we can find a disk and a square that are completely contained in that diamond.

    That means that if a point is an interior point of a set using any one of those metrics it is also an interior point of the set using either of the other two. That in turn means that the three metrics have exactly the same open sets. Since everything in topology can be phrased in terms of open sets, the three metrics give exactly the same topological results- we are free to choose which ever is easiest to work with.

    That is true, by the way for any Rn, not just 2 and 3 dimensions. It is not, however, true for infinite dimensional spaces.

    If, for example we look at function spaces, say the set of functions defined on [a,b], the corresponding metrics are:
    [tex]d(f,g)= \sqrt{\int_a^b (f-g)^2 dx}[/tex]
    the L2 metric.
    [tex]d(f,g)= max(|f(x)|,|g(x)|)[/tex] for x in [a,b]
    the "uniform" metric.
    [tex]d(f,g)= \int_a^b |f(x)- g(x)|dx[/tex]
    the L1 metric.

    Those give quite different topologies- in fact, the function sets for which they are defined are different.
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