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Metric Space/Open Ball

  1. Sep 3, 2011 #1
    Hi guys, two problems, first one I understand for the most part, the second one, I do not know how to set up and solve.

    1. The problem statement, all variables and given/known data
    Let X = R[itex]^{n}[/itex] for x = (a[itex]_{1}[/itex],...,a[itex]_{n}[/itex]) and y = (b[itex]_{1}[/itex],...,b[itex]_{n}[/itex]), define
    d[itex]_{\infty}[/itex](x,y) = max {|a[itex]_{1}[/itex]-b[itex]_{1}[/itex]|,...,|a[itex]_{n}[/itex]-b[itex]_{n}[/itex]|}. Prove that this is a metric.


    2. Relevant equations
    Just the triangle inequality part for this one.


    3. The attempt at a solution
    I've proven the first 3 properties, not quite sure on the last part.
    My attempt was to break it up into n cases by supposing a single
    difference as the max in each case but with ellipsis but I wasn't
    sure on the exactly how. Here's what I attempted:

    The distance between the two points is the largest of the n cases.
    Let z = (z[itex]_{1}[/itex],...,z[itex]_{n}[/itex]). Then we have n cases to check.
    Case 1: d(x,y) = |a[itex]_{1}[/itex]-b[itex]_{1}[/itex]|
    Notice that |a[itex]_{1}[/itex]-z[itex]_{1}|[/itex]|[itex]\leq[/itex]max{|a[itex]_{1}[/itex]-z[itex]_{1}[/itex]|,
    (I stopped here because I wasn't sure how to proceed,
    should I write out to n cases with the ellipsis or two cases?)
    What I was going to show after this was that the triangle inequality
    holds for a[itex]_{1}[/itex], a[itex]_{2}[/itex], ..., a[itex]_{n}[/itex].


    1. The problem statement, all variables and given/known data
    Prove that the set S = {(x[itex]_{1}[/itex],y[itex]_{1}[/itex] : x[itex]_{1}[/itex] + y[itex]_{1}[/itex] > 0}
    is an open subset of R[itex]^{2}[/itex] in the Euclidean metric.


    2. Relevant equations
    Euclidean metric, Schwarz Inequality?, Open Ball.


    3. The attempt at a solution
    I'm not sure how to proceed with this one at all. Picture-wise,
    it'd be the region above the y=-x line, I'm guessing,
    and I'm guessing I have to pick some arbitrary point in that
    region and calculate a strict inequality to show that this is open.
    But I do not know how to proceed at all, I'm looking for some hand-holding
    at this point really because I want to understand it step by step.

    Thanks for any help!
     
  2. jcsd
  3. Sep 3, 2011 #2

    micromass

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    Let's prove this one like this. Can you first show that

    [tex]|a_k-b_k|\leq \max \{|a_1-z_1|,...,|a_n-z_n|\}+\max \{|z_1-b_1|,...,|z_n-b_n|\}[/tex]

    ??

    OK, what does open mean for you?? You need to show for al (x,y) in the set that there exists a ball around(x,y) that stays in the set. How would you choose that ball??
    Does it makes sense what I'm saying?
     
  4. Sep 3, 2011 #3
    [itex]|a_k-z_k+z_k-b_k|[/itex][itex]\leq |a_k-z_k|+|z_k-b_k|\leq[/itex] max[itex]{|a_1-z_1|,...,|a_n-z_n|}[/itex]+max{|[itex]z_1-b_1|,...,|z_n-b_n[/itex]|}?


    I guess for me right now, open for is something without its boundary points. I would have to pick an r such that for some [itex](x_2, y_2), (x_1, y_1) then (x_2-x_1)^{2} + (y_2-y_1)^{2} \leq r^{2}[/itex]?
     
    Last edited: Sep 3, 2011
  5. Sep 3, 2011 #4

    micromass

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    OK, and now that the maximum of the left-hand side.

    What is your definition of open? Saying that it's something without boundary isn't good enough unless you specify what a boundary point is.
     
  6. Sep 3, 2011 #5
    The max's are the distances, gotcha, thanks.

    In this problem, I'd say the boundary is the y = -x line.
     
  7. Sep 3, 2011 #6

    micromass

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    Yes, but what is the definition of a boundary??
     
  8. Sep 3, 2011 #7
    Set of points that encloses a space, endpoints? I don't know what you mean.
     
  9. Sep 3, 2011 #8

    micromass

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    Please, look up in your course the definition of "open set" and such things. We can't really do much unless we have some definitions to work with...
     
  10. Sep 3, 2011 #9
    It's just the open ball centered at x

    B(x,r) = { y in X: d(x,y) strictly less than r}
     
  11. Sep 3, 2011 #10

    micromass

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    What is just the open ball??

    Could you please quote the exact definition of an open set?? Yes, it has something to do with open balls.
     
  12. Sep 3, 2011 #11
    Just quoting the definition from my notes:

    Given a metric space (X,d) a subset S[itex]\subseteq[/itex]X is called open provided that whenever x[itex]\epsilon[/itex]S, then there is an r>0 such that B(x;r)[itex]\subseteq[/itex]S.
     
  13. Sep 3, 2011 #12

    micromass

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    Exactly!!

    So for our [itex]S=\{(x,y)~\vert~x+y>0\}[/itex], we need to find for each [itex](x,y)\in S[/itex], a r such that [itex]B(x,r)\subseteq S[/itex].

    So, let's pick an arbitrary (x,y) in S. What do you think we should take as our r?
     
  14. Sep 3, 2011 #13

    micromass

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    By the way, did you see the open set characterization of continuity yet?? That might come in very handy!!
     
  15. Sep 3, 2011 #14
    Something greater than 0 but less than x+y? I don't know how we go about picking the r.

    I'm just starting out in real analysis, haven't come across it yet, we've only gotten past closed sets which was a bit easier to digest.
     
  16. Sep 3, 2011 #15

    micromass

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    OK, take an arbitrary point in S. What is the distance between that point and the line y=-x??
     
  17. Sep 3, 2011 #16
    [itex]\sqrt{(y_{o}-y)^{2}+(x_{0}-x})^2[/itex], since y=-x, do I make that substitution into the formula?
     
  18. Sep 3, 2011 #17

    micromass

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    What is [itex](x_0,y_0)[/itex] and (x,y)??
     
  19. Sep 3, 2011 #18
    (x,y),(xo,yo) in S
     
  20. Sep 3, 2011 #19

    micromass

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    But why would you do that?? Why would you pick both of them in S??

    I want to find the least distance between a point (x,y) in S and the line y=-x. This is just a calculus/geometry question... How would you solve that??
     
  21. Sep 3, 2011 #20
    I was thinking we had to use the eclidean distance formula for some reason. |x+y|/sqrt(2)
     
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