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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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    "if and only if" means there are two things to prove. What are they? Can you do one of them?2012-05-15
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    You want to prove that statement, right?2012-05-15
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    Anyone agrees Miguel's autogenerated gravatar is borderline inappropriate? =D2012-05-15
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    Hint: If the sequence $x_n$ converges to $L$, then $L$ is a lower bound for the sequence (why)? The other direction is a direct consequence of the Montone Convergence Theorem, proved in most introductory analysis/calculus textbooks.2012-05-15
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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 $\alpha$.2012-05-15
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    @PeterTamaroff Why you you consider it inappropriate?2012-05-15
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    @DavidMitra Please trust me that if you don't know, then you will regret finding out.2012-05-15
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    Hint for the forward implication: contrapositive.2012-05-16
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    @David: At a guess, he thinks that it resembles a swastika. This seems to me a very considerable stretch.2012-05-23

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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.