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Math Help - Real Anaylsis help

  1. #1
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    Real Anaylsis help

    Ok, So I am having some trouble with my real analysis class and homework, and was wanting some help...


    1) Show that if (x_n) and (y_n) are convergent sequences, then the sequences (u_n) and (v_n) defined by (u_n):=max{ x_n , y_n) and (v_n):=min{ x_n , y_n } are also convergent.
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  2. #2
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    Quote Originally Posted by osudude View Post
    1) Show that if (x_n) and (y_n) are convergent sequences, then the sequences (u_n) and (v_n) defined by (u_n):=max{ x_n , y_n) and (v_n):=min{ x_n , y_n } are also convergent.
    Suppose that \left( {x_n } \right) \to X\;\& \;\left( {y_n } \right) \to Y.
    There are to cases to consider: X = Y\;\& \;X \ne Y.

    Here a hint on the second one.
    Suppose that (WLOG) Y < X then Z=\frac{X+Y}{2} so that Y<Z<X.
    Thus almost all of the terms of \left( {x_n } \right) are greater than Z.

    And almost all of the terms of \left( {y_n } \right) are less than Z.

    What does that tell you about \max\text{ and }\min?
    Last edited by Plato; November 12th 2009 at 12:13 PM.
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  3. #3
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    that they coverge both to z since the max of y is almost z and same for the min of x??
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  4. #4
    Senior Member Shanks's Avatar
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    Relation between max(min) and absolute value

    In fact , we have u_n=max\{x_n,y_n\}=\frac {1}{2}(x_n+y_n+|x_n-y_n |), similarly v_n=min \{x_n,y_n\}=\frac {1}{2}(x_n+y_n-|x_n-y_n |).
    since x_n and y_n converges, so is u_n and v_n.
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