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Math Help - Boundedness and Convergence

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    Boundedness and Convergence

    Suppose that {x(n)} 1=n=infinity, is a sequence of real numbers with limit x, and suppose that a≤ x(n) ≤b, all n. Prove that a≤ x ≤b.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Snooks02 View Post
    Suppose that {x(n)} 1=n=infinity, is a sequence of real numbers with limit x, and suppose that a≤ x(n) ≤b, all n. Prove that a≤ x ≤b.
    start with a \le x_n \le b. now take the limit as n \to \infty of that system
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    Quote Originally Posted by Snooks02 View Post
    Suppose that {x(n)} 1=n=infinity, is a sequence of real numbers with limit x, and suppose that a≤ x(n) ≤b, all n. Prove that a≤ x ≤b.
    If x_n is convergent and x_n \leq b then x\leq b - where x=\lim x_n.
    This is what we will prove. Say we prove that if x_n \leq 0 then x\leq 0. This is sufficient to complete the proof. Why? Because if x_n \leq b with x = \lim x_n then define y_n = x_n - b so y_n \leq 0 and by above y \leq 0 \implies x \leq b (here y = \lim y_n).

    Now let x_n\leq 0. And assume, by contradiction, x > 0. Then there is N so that |x_N - x| < \frac{x}{2} and so x_N > \frac{x}{2} > 0 a contradiction.

    Similary do it for x_n \geq a \implies x \geq a by repeating the argument.
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