Results 1 to 4 of 4

Math Help - xn->a, yn->b imply d(xn, yn)->d(a,b)

  1. #1
    Senior Member
    Joined
    Jan 2009
    Posts
    404

    xn->a, yn->b imply d(xn, yn)->d(a,b)

    Let (x_n) and (y_n) be sequences in a metric space X. Suppose that x_n converges to a and y_n converges to b. Prove that d(x_n, y_n) converges to d(a, b).
    ================================

    |d(x_n, y_n) - d(a,b)|
    We need to relate this to the hypothesis, so I think perhaps subtracting and adding d(x_n, a) would help...but still I can't figure out how to prove the result...

    Any help is appreciated!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by kingwinner View Post
    Let (x_n) and (y_n) be sequences in a metric space X. Suppose that x_n converges to a and y_n converges to b. Prove that d(x_n, y_n) converges to d(a, b).
    ================================

    |d(x_n, y_n) - d(a,b)|
    We need to relate this to the hypothesis, so I think perhaps subtracting and adding d(x_n, a) would help...but still I can't figure out how to prove the result...

    Any help is appreciated!
    d(x_n,y_n)-d(x,y)\leqslant d(x_n,x)+d(y_n,y)-d(x,y)\leqslant d(x_n,x)+d(y_n,y)+d(x,y)-d(x,y)=d(x_n,x)+d(y_n,y).

    Also, d(x,y)-d(x_n,y_n)\leqslant d(x,x_n)+d(y,x_n)-d(x_n,y_n)\leqslant d(x,x_n)+d(y,y_n)+d(x_n,y_n)-d(x_n,y_n)=d(x_n,x)+d(y_n,y).

    Combinging these two gives \left|d(x_n,y_n)-d(x,y)\right|\leqslant d(x_n,x)+d(y_n,y).

    Conclude.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Jan 2009
    Posts
    404
    Quote Originally Posted by Drexel28 View Post
    d(x_n,y_n)-d(x,y)\leqslant d(x_n,x)+d(y_n,y)-d(x,y)\leqslant d(x_n,x)+d(y_n,y)+d(x,y)-d(x,y)=d(x_n,x)+d(y_n,y).

    Also, d(x,y)-d(x_n,y_n)\leqslant d(x,x_n)+d(y,x_n)-d(x_n,y_n)\leqslant d(x,x_n)+d(y,y_n)+d(x_n,y_n)-d(x_n,y_n)=d(x_n,x)+d(y_n,y).

    Combinging these two gives \left|d(x_n,y_n)-d(x,y)\right|\leqslant d(x_n,x)+d(y_n,y).

    Conclude.
    Why is the first inequality (d(xn,yn) d(xn,x)+d(yn,y) ) true?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by kingwinner View Post
    Why is the first inequality (d(xn,yn) d(xn,x)+d(yn,y) ) true?
    Should be d(x_n,x)+d(y_n,x) and the next step gets all the x_n's with the x's and the y_n's with the ys
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. how to show xn>0 need not imply lin xn>a
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: February 21st 2011, 07:59 AM
  2. p=2^k+1 and (a/p)=-1 imply a is primitive modulo p
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: July 11th 2010, 07:34 AM
  3. Semigroups: Do the following 4 equations imply x=y?
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: June 2nd 2010, 12:37 AM
  4. Pointwise convergence does not imply L1
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: April 13th 2010, 06:30 AM
  5. equations imply equation
    Posted in the Algebra Forum
    Replies: 14
    Last Post: July 11th 2006, 04:35 AM

Search Tags


/mathhelpforum @mathhelpforum