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Math Help - Proof linking continuity and convergence

  1. #1
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    Proof linking continuity and convergence

    i have no background in analysis, and have absolutely no idea how to write proofs, need some help proving this theorem using epsilon:

    if sequence x (x_n) approaches "a", and sequence y (y_n) approaches b and sequence x (x_n) is greater than or equal to sequence y (y_n) for all n, then "a" is greater than or equal to "b".

    thanks
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  2. #2
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    Quote Originally Posted by ml692787 View Post
    i have no background in analysis, and have absolutely no idea how to write proofs, need some help proving this theorem using epsilon:

    if sequence x (x_n) approaches "a", and sequence y (y_n) approaches b and sequence x (x_n) is greater than or equal to sequence y (y_n) for all n, then "a" is greater than or equal to "b".

    thanks
    If look here.

    As a result we have a simple result.

    Theorem 1: Let (x_n) be a convergent sequence of non-negative terms. Then its limits is non-negative. (This should be obvious from the link I gave you).

    Theorem 2: Let x_n\geq y_n for all n. And (x_n) \mbox{ and }(y_n) are convergent sequences. Then \lim \ x_n \geq \lim \ y_n.

    Proof: Define a sequence z_n = x_n - y_n. Now this means that z_n \geq 0. And furthermore (z_n) is convergent since it is a difference of two convergent sequences. Thus, \lim \ z_n \geq 0. Thus, \lim \ x_n - \lim \ y_n \geq 0 because the limit of z_n is the difference of the two individial sequences. Thus, \lim \ x_n \geq \lim \ y_n by Theorem 1. Q.E.D.
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    If look here.

    As a result we have a simple result.

    Theorem 1: Let (x_n) be a convergent sequence of non-negative terms. Then its limits is non-negative. (This should be obvious from the link I gave you).

    Theorem 2: Let x_n\geq y_n for all n. And (x_n) \mbox{ and }(y_n) are convergent sequences. Then \lim \ x_n \geq \lim \ y_n.

    Proof: Define a sequence z_n = x_n - y_n. Now this means that z_n \geq 0. And furthermore (z_n) is convergent since it is a difference of two convergent sequences. Thus, \lim \ z_n \geq 0. Thus, \lim \ x_n - \lim \ y_n \geq 0 because the limit of z_n is the difference of the two individial sequences. Thus, \lim \ x_n \geq \lim \ y_n by Theorem 1. Q.E.D.
    Hey, TPH, could we have said the following?

    Since the sequence \{ x_n \} approaches a, we have \lim_{n \to \infty} x_n = a. Similarly, since \{ y_n \} approaches b, we have \lim_{n \to \infty} y_n = b

    Now, we have x_n \geq y_n for all n

    \Leftrightarrow \lim_{n \to \infty}x_n \geq \lim_{n \to \infty}y_n

    \Leftrightarrow a \geq b

    as desired
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  4. #4
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    Quote Originally Posted by Jhevon View Post
    Hey, TPH, could we have said the following?

    Since the sequence \{ x_n \} approaches a, we have \lim_{n \to \infty} x_n = a. Similarly, since \{ y_n \} approaches b, we have \lim_{n \to \infty} y_n = b

    Now, we have x_n \geq y_n for all n

    \Leftrightarrow \lim_{n \to \infty}x_n \geq \lim_{n \to \infty}y_n

    \Leftrightarrow a \geq b

    as desired
    But you are accepting a theorem that we are trying to prove!

    For example, this is what you are doing. Say we want to be that the sum of the triangles add up 180 degrees (let us called this Euclid's Principle). We prove Euclid's principle using the Parallel Postulate. Then you want to justify the Parallel Postulate and you use Euclid's principle. How is that a proof?
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