Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Usual and taxicab metric topologies

  1. Oct 19, 2008 #1
    I am trying to show that an "epsilon ball in the taxicab metric" can be contained in a "epsilon ball in the usual metric".

    Let "Up" be a taxicab ball, and "Ud" be a usual metric ball.

    How can I show that Up( (x,y) , e ) is a subset of Ud( (x,y), e ) ?



    I have drawn the picture and I understand how the taxicab ball is inside. It is a diamond that touches the usual ball on each of the diamond's four corners and is less than everywhere else.

    How can I show this algebraically?
     
  2. jcsd
  3. Oct 19, 2008 #2
    I actually figured this out from a previous thread on this forum. Yippee.

    I need to show the reverse is true though and I can't it out. I need to show that we can construct a taxicab metric that contains any usual epsilon-ball.
     
  4. Oct 19, 2008 #3

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Just use geometry again. Now you have a circle inscribed in a diamond instead of vice versa. Neither one is that hard.
     
  5. Oct 19, 2008 #4
    I have to use algebra... for instance, for the first part I said,

    Since

    [tex]|x-a| =\sqrt{(x-a)^2} \leq \sqrt{ (x-a)^2 + (y-b)^2 } = \varepsilon[/tex]

    and

    [tex]|y-b| = \sqrt{(y-b)^2} \leq \sqrt{ (x-a)^2 + (y-b)^2 } = \varepsilon[/tex]

    then

    [tex]|x-a| + |y-b| \leq 2\varepsilon[/tex]

    Let

    [tex]k= \frac{1}{2}[/tex]

    Then

    [tex]k(|x-a| + |y-b|) \leq \varepsilon = \sqrt{ (x-a)^2 + (y-b)^2 }[/tex]
     
  6. Oct 19, 2008 #5
    Well, I don't know if you can see that latex but this is what I did,

    Since |x-a| =\sqrt{(x-a)^2} <= sqrt{ (x-a)^2 + (y-b)^2 } = e

    and |y-b| = \sqrt{(y-b)^2} <= \sqrt{ (x-a)^2 + (y-b)^2 } = e

    then |x-a| + |y-b| <= 2e

    Let k= 1/2. Then k(|x-a| + |y-b|) <= e = \sqrt{ (x-a)^2 + (y-b)^2 }
     
  7. Oct 19, 2008 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Yes, k=1/2 works fine.
     
  8. Oct 20, 2008 #7
    Is showing the other way similar? I am not getting anywhere. You said use the traingle inequality. Are we still looking for a constant?
     
  9. Oct 20, 2008 #8

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Just write down the triangle inequality for the three points (a,b), (x,y) and (x,b).
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook