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In the process of constructing a proof, I obtain the following set. Let C be the set of continuous and nondecreasing functions defined on $[0,1]$. Let the set $B$ be given as follows:

$$ B=\{b_i\in[0,1]: b_i=\sup_{t\in[0,1]}|h_i(t)|,\,h_i \in C\} $$

My question is whether the set $B$ is bounded. In particular, si $B$ bounded above? I don't know how to approach this question so any help/suggestion/comment on how to proceed is really appreciated it!

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    It seems that $B$ is bounded above by 1 as it is a subset of $[0,1]$. A set $S$ is bounded above if there exists a $k$ such that $s \leq k$ for all $s \in S$. Therefore, 1 seems to work here.2012-08-31

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In your definition, $B$ is clearly a subset of $[0, 1]$. Every element $b \in B$ must also be in $[0, 1]$. $[0, 1]$ is bounded. Therefore, $B$ is also bounded.

If, on the other hand, you don't restrict the elements of $B$ to $[0, 1]$:

$$ B = \{b \in \mathbb{R} : b = \sup_{t \in [0,1]} \left|h(t)\right|, \, h \in C\} $$

Then $B$ is not bounded. No matter what $M \in \mathbb{R}$ you choose, you can construct the following function which belongs to $C$: $$ f : [0, 1] \to \mathbb{R} \\ f(t) = t + |M| $$

$f$ is continuous and non-decreasing. However: $$ \sup_{t \in [0,1]} \left|f(t)\right| = |M| + 1 > M $$

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    Then, according to Ayman's answer, the set of continuous and nondecreasing functions defined on $[0,1]$ is unbounded, isn't it?2012-09-10
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    @Cristian - Every continuous function on a closed interval must be bounded. However, the set of upper bounds for those functions is not bounded.2012-09-10
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    Yup..but I was wondering if there exists a bound on the set of continuous and nondecreasing functions...that is, if there is a number $B$ such that $|f(x)|\leq B$ for all $x\in[0,1]$ and all $f$ continuous and nondecreasing. According to your pretty nice answer above, it seems that the answer to my question is no, there is no such $B$...2012-09-10
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    @Cristian - Indeed. There is no such $B$.2012-09-10
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    Thanks for your help @ayman!2012-09-10
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    @Cristian You're welcome!2012-09-10