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Suppose $x_n$ is bounded with $x_n \in [a,b]$, let $b_n = \sup\{x_n, x_{n+1}, ...\}$. Prove that $b_n$ has a limit $b_\infty$ and that $b_\infty \leq b$.

I've been having a strange time wrapping my head around this. It makes sense to me that this should be true and such but I am having a hard time proving it. I don't think it should be hard, I feel like I am just complicating it. Here is my best attempt, could use some advice.

Note that $a \leq x_n \leq b_n$ so $b_n$ is bounded below. Also, note that $b_n = \sup\{x_n, x_{n+1}, ...\} = \sup\{x_n, \sup\{x_{n+1}, x_{n+2},...\}\} = \sup\{x_n, b_{n+1}\}$. So $b_n$ is decreasing.

Since $b_n$ is decreasing and bounded below then it has a limit $b_\infty$.

I really can't get the conclusion that $b_\infty \leq b$, it makes sense to be true but I can't decided how to show it.

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    Each $b_n\le b$ because again, $b$ is an upper bound for $x_n$ and hence being a least upper bound it must be at most $b$ ($b_n\le b_1 =\sup_{n\ge 1}x_n \le b$) – 2017-02-23
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    Oh yes of course, I feel like I am way complicating the idea of a limsup. Thank you! – 2017-02-23

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Hint: show that $b_1\leq b$, and then use that $b_n$ is decreasing.

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    I think that the comment above helps me answer this. $b_1$ is the least upper bound of $x_n$ and since $b$ is an upper bound it must be true that $b_1 \leq b$ and then since $b_n$ is decreasing then $b_\infty \leq b_1 \leq b$. Thank you! – 2017-02-23
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    @danny Yup, that's the one. – 2017-02-23