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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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    @Cristian You're welcome!2012-09-10