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Math Help - intro topology proof linking continuity and convergence

  1. #1
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    intro topology proof linking continuity and convergence

    I'm trying to figure out how to prove this theorem straightforward, i have figured out the proof by contradiction letting a < 0 and epsilon being -a/2, but can't seem to get it straightforward, here is the proof:

    if a sequence of numbers (x_n) approaches "a", and the sequence of numbers (x_n) is larger or equal to 0, then "a" is larger than or equal to 0.

    Thank you.
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    Quote Originally Posted by ml692787 View Post
    I'm trying to figure out how to prove this theorem straightforward, i have figured out the proof by contradiction letting a < 0 and epsilon being -a/2, but can't seem to get it straightforward, here is the proof:

    if a sequence of numbers (x_n) approaches "a", and the sequence of numbers (x_n) is larger or equal to 0, then "a" is larger than or equal to 0.

    Thank you.
    For what it's worth, I cannot see a straightforward proof either. At some point you must say, "If a < 0", and show that leads to x_n < 0, contradicting a hypothesis.
    Last edited by JakeD; September 2nd 2007 at 06:37 PM.
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    Quote Originally Posted by ml692787 View Post
    I'm trying to figure out how to prove this theorem straightforward, i have figured out the proof by contradiction letting a < 0 and epsilon being -a/2, but can't seem to get it straightforward, here is the proof:

    if a sequence of numbers (x_n) approaches "a", and the sequence of numbers (x_n) is larger or equal to 0, then "a" is larger than or equal to 0.

    Thank you.
    Theorem: Let (x_n) be convergent sequence of real numbers. And x_n \geq M for sufficiently large n then \lim \ x_n \geq M.

    Proof: Let L = \lim x_n. That means |x_n - L| < \epsilon  \mbox{ for }n\geq N\in \mathbb{N}. Now in sake of a contradiction say L<M. That means \frac{M-L}{2}>0, and so choese \epsilon = \frac{M-L}{2}. This means, |x_n - L | < \frac{M-L}{2} \mbox{ for }n\geq N. Thus, x_n -L < \frac{M-L}{2} \implies x_n < \frac{M-L}{2}+L = \frac{M+L}{2} < \frac{M+M}{2} = M. We have shown that x_n < M \mbox{ for }n\geq N. Which is a contradiction because we assumed that x_n \geq M for sufficiently large n which fails heir. Thus, \lim \ x_n = L \geq M. Q.E.D.
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