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A monotone decreasing sequence ${x_{n}}$ converges if and only if is bounded from below

Could they please help me with this exercise?

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Anyone agrees Miguel's autogenerated gravatar is borderline inappropriate? =D –  Pedro Tamaroff May 15 '12 at 22:03
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Hint for the reverse implication: Let $\alpha$ be the greatest lower bound of the sequence. Show that in fact the sequence converges to $\alpha$. –  David Mitra May 15 '12 at 22:06
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@PeterTamaroff Why you you consider it inappropriate? –  David Mitra May 15 '12 at 22:06
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@DavidMitra Please trust me that if you don't know, then you will regret finding out. –  MJD May 15 '12 at 22:09
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Hint for the forward implication: contrapositive. –  Cameron Buie May 16 '12 at 0:51

1 Answer 1

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If $x_n$ is bounded from below, there exist $I = \inf x_n $. Given $\epsilon > 0$. Chose $n_0$ such that \begin{equation} I \le x_{n_0} < I+ \epsilon. \end{equation} Hence as $(x_n)$ is decreasing \begin{equation} n \ge n_0 \Rightarrow I \le x_n \le x_{n_0} < I + \epsilon. \end{equation} Then $(x_n)$ converges. Reciprocally if $(x_n)$ converges to $x$, for $n > > 1$, $x_n > x -1$ and $x_n$ is bounded below.

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