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Thread: 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 $\displaystyle a \le x_n \le b$. now take the limit as $\displaystyle n \to \infty$ 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 $\displaystyle x_n$ is convergent and $\displaystyle x_n \leq b$ then $\displaystyle x\leq b$ - where $\displaystyle x=\lim x_n$.
    This is what we will prove. Say we prove that if $\displaystyle x_n \leq 0$ then $\displaystyle x\leq 0$. This is sufficient to complete the proof. Why? Because if $\displaystyle x_n \leq b$ with $\displaystyle x = \lim x_n$ then define $\displaystyle y_n = x_n - b$ so $\displaystyle y_n \leq 0$ and by above $\displaystyle y \leq 0 \implies x \leq b$ (here $\displaystyle y = \lim y_n$).

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

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