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Math Help - Proof problem (open sets)

  1. #1
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    Proof problem (open sets)

    Hi, having some trouble with proving that certain sets are open. Here are the questions i'm having problems with:

    1.

    d((a1; a2); (b1; b2)) = |a1 - b1| + |a2 - b2|

    show that the set U = { (a1,a2) E R2 ; a1 + a2 > 0 } is an open set in the metric space (R2, d)

    ans:

    let a := (a1, a2) E U, I need to find an open ball Br(a) in (R2,d) such that Br(a) E U.
    I define r = a1 + a2 > 0 and suppose that b=(b1,b2) E Br(a)

    Then b1 + b2 = r - d(a,b)
    = a1 + a2 - (|a1 - b1| + |a2 - b2|)
    > or equal to r -|a1 - b1| - |a2 - b2|
    > 0
    so b E U, which implies that Br(a) E U, which means U is open?

    i think this is ok but then again i'm not sure, could anyone give an alternate proof or anything, hopefully i'll understand what the hell im doing better with another take on this??

    2.

    d( (x1, x2),(y1,y2) ) = max { |x1-y1|,|x2-y2| } is a metric on R2

    show that the upper half-plane U = {(x1,x2) E R2 ; x2>0} is an open set in the metric space (R2,d)

    ans:

    suppose that x = (x1,x2) E U, then x2>0. I need to find an open ball Br(x) in (R2,d) such that Br(x) E U
    I define r = x2 >0 then suppose that y = (y1,y2) E Br(x)

    then y2 = r - d(x,y)
    y2 = x2 - max{|x1-y1|,|x2-y2|}

    not sure where to go from here

    any ideas? sorry if the notation might be a little hard to understand, i was in a rush!

    cheers
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  2. #2
    Senior Member JaneBennet's Avatar
    Joined
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    Quote Originally Posted by sss12 View Post
    I define r = a1 + a2 > 0 and suppose that b=(b1,b2) E Br(a)

    Then b1 + b2 = r - d(a,b)
    That does not follow. Also the choice of r is not good; r=\frac{a_1+a_2}2 is better.

    Then a_1-b_1+a_2-b_2\leqslant|a_1-b_1|+|a_2-b_2|<r=\frac{a_1+a_2}2

    \Rightarrow\ b_1+b_2>a_1+a_2-\frac{a_1+a_2}2=\frac{a_1+a_2}2>0

    \Rightarrow\ (b_1,b_2)\in U

    Quote Originally Posted by sss12 View Post
    I define r = x2 >0 then suppose that y = (y1,y2) E Br(x)

    then y2 = r - d(x,y)
    Again you are doing it wrong! And r=\frac{x_2}2 is better.

    Then x_2-y_2\leqslant|x_2-y_2|\leqslant\max\{|x_1-y_1|,|x_2-y_2|\}<r=\frac{x_2}2

    \Rightarrow\ y_2>x_2-\frac{x_2}2=\frac{x_2}2>0

    \Rightarrow\ (y_1,y_2)\in U
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