# Why does $(a_n)$ bounded imply that $(b_n)$ is decreasing?

Why does $(a_n)$ bounded imply that $(b_n)$ is decreasing?

$$(a_n)=a_1,a_2,\dots\tag{1}$$

$$b_n=\sup (a_n,a_{n+1},\dots), c_n=\inf (a_n,a_{n+1},\dots)$$

If $\left(a_n\right)$ is bounded, then $\left(b_n\right)$ exists and $(b_n)$ is decreasing, $(c_n)$ is increasing.

Why?

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$(a_n)$ bounded implies that each $b_n$ exists. That $(b_n)$ is decreasing is obvious (look at the sets you're supping over). – David Mitra Jul 25 '13 at 12:16
@DavidMitra Yes, the original words is that obvious--,me silly to see that fact obviously. :( – User19912312 Jul 25 '13 at 12:18
@Did Typo, fixed – User19912312 Jul 25 '13 at 12:25

That $(a_n)$ is bounded insures that each $b_n$ and $c_n$ is defined.

To see that $(b_n)$ is decreasing:

Fix an $n$.

Any upper bound of $\{a_n, a_{n+1}, \cdots\}$ is also an upper bound of $\{ a_{n+1}, a_{n+2}, \cdots\}$. In particular, $b_n$ is an upper bound of $\{ a_{n+1}, a_{n+2}, \cdots\}$. As $b_{n+1}$ is the least upper bound of $\{ a_{n+1}, a_{n+2}, \cdots\}$, we have $b_{n+1}\le b_n$.

A similar argument will establish that $(c_n)$ is increasing.

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I understand--! haha:) – User19912312 Jul 25 '13 at 12:48
• The hypothesis "$(a_n)$ is bounded" is to make sens to the definition of $b_n$ and $c_n$ i.e. they have a finite value.
• We have this general result: if $A\subset B$ then $$\sup(A)\leq \sup (B)\quad \text{and}\quad\inf(A)\geq \inf(B)$$ so take $A_n=\{a_n,a_{n+1},\cdots\}$ and we have $A_{n+1}\subset A_n$ and apply the previous result.
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Hi, A=[1,2], sup(A)=2? B=[1,3], sup(A)=3? – User19912312 Jul 25 '13 at 12:44
I edited my answer. – user63181 Jul 25 '13 at 12:47
you are welcome, and thanks, I understand. – User19912312 Jul 25 '13 at 12:50

$$b_n=\sup(a_n,b_{n+1})\implies b_n\geqslant b_{n+1}$$

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sorry, I donot know what's the meaning of $sup(a_n,b_n+1)$ – User19912312 Jul 25 '13 at 12:45
You do not know what $\sup(x,y)$ is, when $x$ and $y$ are real numbers? – Did Jul 25 '13 at 12:50
I know sup(A set) if (x,y) means set {x,y} then I understand, if x,y are arbitrary real numbers, that is complex number system – User19912312 Jul 25 '13 at 12:51
"complex number system". What are you talking about? // Yes, $\sup(x,y)$ is $\sup\{x,y\}$. – Did Jul 25 '13 at 12:56
Did, do you think you can help me with a little something about sequences? I have put my foot in the mud here but I haven't obtained any solution really. – Pedro Tamaroff Jul 25 '13 at 14:09