Results 1 to 5 of 5

Math Help - Metric Space, limit

  1. #1
    Newbie
    Joined
    Nov 2008
    Posts
    23

    Metric Space, limit

    Let (X, d) be a metric space. If (x_n) is a sequence in X which converges to both a \in X and b \in X, prove that a=b.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,707
    Thanks
    1470
    Suppose x were not equal to y. Then d(x,y) is not 0. Apply the definition of convergence to both x and y with \epsilon= d(x,y)/2 to get a contradiction:
    Since x_n converges to a, there exist N_1 such that if n> N_1, then |x_n- a|< \epsilon. Since x_n converges to b, there exist N_2 such that if n> N_2 then |x_n- b|< \epsilon. Use the triangle inequality to show that those are impossible.
    Last edited by HallsofIvy; January 15th 2009 at 04:57 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Nov 2008
    Posts
    394
    Suppose to the contrary that (x_{n}) , n=1,2,.... \infty converges to two distinct limit points a and b in the metric space (X, d).

    Then d(a,b) \leq d(a, x_{n}) + d(x_{n}, b) by the the axiom of triangle inequality in the metric space.

    By definition of a convergent sequence, there exists integers N1 and N2 such that if n >= N1, then d(x_{n}, a) < \epsilon and if n >= N2, then d(x_{n}, b) < \epsilon . Let N= max (N1, N2).

    Then, if n >= N,
    d(a,b) \leq d(a, x_{n}) + d(x_{n}, b) < \epsilon + \epsilon = 2\epsilon .

    Let \epsilon = (1/2) \times d(a,b).
    Then,
    d(a,b) \leq d(a, x_{n}) + d(x_{n}, b) < \epsilon + \epsilon = 2\epsilon = d(a,b), which leads a contradiction since d(a,b) < d(a,b) is not true.

    Thus, a sequence in a metric space cannot converge to more than one limit.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,663
    Thanks
    1616
    Awards
    1
    There is an intuitive way to consider this problem.
    If \left( {x_n } \right) \to a then almost all of the terms of \left( {x_n } \right) are in any ball that contains \color{blue}a.
    So if a \not= b then there are two disjoint balls, one containing a the other containing b.
    BUT THAT IS IMPOSSIBLE. Almost all of the terms of \left( {x_n } \right) are in any ball that contains a.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    This may be similar to HallsofIvy, but it is slightly different and he/she seems to have LaTeX mistakes.. All you need is the triangle inequality

    Choose N_1 such that if n\leqslant N_1 then d(x_n,a)<\frac{\varepsilon}{2}

    Also, choose N_2 such that if n\leqslant N_2 then d(x_n,b)<\frac{\varepsilon}{2}

    Now choose N=\max\left\{N_1,N_2\right\} then for n\leqslant N we have that d(a,b)\leqslant d(x_n,a)+d(x_n,b)\leqslant\frac{\varepsilon}{2}+\f  rac{\varepsilon}{2}=\varepsilon.

    Since \varepsilon was arbitrary this implies that d(a,b)=0 which by the axioms of metric spaces is only true if a=b\quad\blacksquare
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. limit points of metric space
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: October 1st 2010, 03:21 AM
  2. Limit of function from one metric space to another metric space
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: September 17th 2010, 02:04 PM
  3. L(U) set of limit points of a set U in a metric space X is closed
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: July 7th 2010, 06:23 AM
  4. Limit points in a metric space
    Posted in the Differential Geometry Forum
    Replies: 5
    Last Post: April 7th 2009, 03:57 PM
  5. Replies: 1
    Last Post: January 16th 2009, 01:00 PM

Search Tags


/mathhelpforum @mathhelpforum