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This is what I am trying to prove:

Suppose $\{x_n\}$ is a bounded sequence ($a \leq x_n \leq b$). Let $b_n = \sup\{x_n, x_{n+1}, ...\}$ Prove $b_n$ has a limit $b_\infty$ as $n \rightarrow \infty$ and $b_\infty \leq b$.

Not really sure what the correct way of doing this is, I made an attempt but could use some guidance.

Let $b_\infty$ be the least upper bound of $x_n$. Then we know $x_n \leq b_\infty \leq b.$ Let $\epsilon > 0$, there exists some $N_1$ such that for all $n \geq N_1$, $x_n \leq b_n < b_\infty + \epsilon$.

There also exists some $N_2$ such that for all $n \geq N_2$ we have $b_\infty - \epsilon < x_n \leq b_n \leq b$.

Let $N = \max\{N_1, N_2\}$, then for all $n \geq N$ we have $b_\infty - \epsilon < b_n < b_\infty + \epsilon$ and we have $|b_n - b_\infty| < \epsilon$ and $b_\infty \leq b$ otherwise $b_\infty$ would not be a least upper bound.

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HINT: the limit superior of a sequence $(x_n)_{n\in\Bbb N}$ can be defined by the limit of the sequence

$$\left(\sup_{k\ge n}x_n\right)_{n\in\Bbb N}$$

Now prove, using a $\epsilon{-}N$ proof if you want, that the above sequence is bounded and decreasing, hence convergent.

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  • $\begingroup$ I see where this is going, but how would I show that it is convergent to this $b_\infty$ $\endgroup$ – student_t Jan 29 '17 at 19:59
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    $\begingroup$ Ok: suppose that a monotone and bounded sequence is not convergent. Show, by a $\epsilon-N$ argument, that this is a contradiction for a real-valued sequence. $\endgroup$ – Masacroso Jan 29 '17 at 20:03
  • $\begingroup$ Ah yes ok! Thank you. $\endgroup$ – student_t Jan 29 '17 at 20:04
  • $\begingroup$ Would choosing $\epsilon = b - b_\infty$ be a good choice, since it would lead to $b_n > b$, which would contradict $b_n$ being a least upper bound. $\endgroup$ – student_t Jan 29 '17 at 20:13
  • $\begingroup$ Idk now... In theory any $\epsilon$ you choose must work. It is enough to show that $(b_n)$ is a Cauchy sequence. $\endgroup$ – Masacroso Jan 29 '17 at 20:24
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For each $n$, $a\leq b_{n}\leq b$, and we have that $b_{n+1}\leq b_{n}$ for all $n\geq 1$. Thus, the sequence of $(b_{n})_{n\geq1}$ is bounded and monotonically decreasing, so it has a limit $b_{\infty}=\inf\{b_{n}:n\geq1\}(=\lim_{n\rightarrow\infty}b_{n}).$

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  • $\begingroup$ showing that it is bounded and monotonically decreasing is much nicer! But I should have clarified I need to do this using $\epsilon$ proof. And how did you get to $b_\infty$ is the inf of the sequence $b_n$ $\endgroup$ – student_t Jan 29 '17 at 19:54
  • $\begingroup$ Since the $b_{n}$ are decreasing, their limit is their greatest lower bound. In any case, it looks like you've found your answer. $\endgroup$ – RideTheWavelet Jan 29 '17 at 20:14
  • $\begingroup$ That makes sense, thank you. $\endgroup$ – student_t Jan 29 '17 at 20:17

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