$\{x_n\}_{n=1}^{\infty}$ denotes the set of real numbers $\{x_1,x_2,x_3...x_n,..\}$. Suppose that $\{x_n\}_{n=1}^{\infty}$ is a decreasing and bounded sequence and there exists an $M>0$ such that $|x_n| \leq M$ for all natural $n$. How can I use the axiom of completeness to show that $\lim_{n\to +\infty} x_n = a$ for some $a \in \mathbb{R}$ ?

So we know that for all $x$ in our set, $-M < x < M$ and that the infimum is the limit of our decreasing & bounded sequence. We can't try and use the least-upper-bound property since we are not dealing with a supremum. How else can we go about solving this?

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    $\begingroup$ If you know $\inf$ is the limit, what are you trying to prove? $\endgroup$ – John Oct 23 '14 at 18:26
  • $\begingroup$ If you know how to deal with the supremum, just consider $-x_n.$ $\endgroup$ – mfl Oct 23 '14 at 18:26
  • $\begingroup$ @JohnZHANG I am trying to prove that the inf of the sequence is $a$ for some $a \in \mathbb{R}$. I am asked to used the axiom of completeness to do that though and i'm not sure how. $\endgroup$ – lynom Oct 23 '14 at 18:59
  • $\begingroup$ @lynom Axiom of completeness guarantee the set of all sequence elements has an inf. Do you want to prove the sequence converges to the inf? $\endgroup$ – John Oct 23 '14 at 19:11

Let $\varepsilon>0$ be given.

Since the set of real numbers $\{x_n\}_{n=1}^{\infty}$ is bounded below it admits an infimum in $\mathbb{R}$. Let $a=\underset{n \in \mathbb{N}}\inf \{x_n\}$. We know that there exists $x_N \in \{x_1, x_2, \ldots\}$ so that $x_N<a+\varepsilon$. Since $x_1 \geq x_2 \geq x_3 \geq \cdots$, it follows that $x_n < a + \varepsilon$ whenever $n \geq N$. Clearly $a-\varepsilon < x_n$ for $n=1,2,\ldots$ . Thus there is a positive integer $N$ so that $|x_n-a|<\varepsilon$ whenever $n \geq N$. Therefore $\lim_{n\to +\infty} x_n = a$ by definition.

  • $\begingroup$ Why is this using the completeness axiom? Isn't this just using information of what inf means? That is $x_n < a + \epsilon$. Maybe I don't understand the axiom well enough. $\endgroup$ – lynom Oct 23 '14 at 19:07
  • $\begingroup$ Since the set $\{x_1, x_2, \ldots\}$ is bounded below by $-M$, it follows that the set $\{-x_1, -x_2, \ldots\}$ is bounded above by $M$. Let $-a=\underset{n \in \mathbb{N}}{\sup} \{-x_n\}$. We know that there exists $-x_N \in \{-x_1, -x_2, \ldots\}$ so that $-a-\varepsilon<-x_N$. Does that work better? $\endgroup$ – Matt A Pelto Oct 23 '14 at 19:21
  • $\begingroup$ ohhh. I understand. Thanks. $\endgroup$ – lynom Oct 24 '14 at 2:43

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