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I was reading Rudin's Principles of Mathematical Analysis, and I came across the definition 7.19, where it says that a sequence of functions $f_n(x)$ is pointwise bounded on E if there exists a finite-valued function $\phi$ defined on E such that $ |f_n(x)| < \phi(x) $ for x element of E, n = 1, 2 ,3 ... While $f_n$ is uniformly bounded on E if there exists a number M s.t. $|f_n(x)| < M$ for x element of E, n = 1, 2 , 3 ... But if we define the set U as the values of $\phi(x)$ from our first definition and define the sup of the set as R, then don't we get the second definition. Wouldn't that mean that pointwise bounded implies uniform bounded?

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    Pointwise boundedness means that for EACH $x_0 \in E$, the sequence $\{f_n(x_0)\}$ is a bounded sequence of real numbers. So, if all of the $f_n$'s are the same thing (for example), then for each $x_0$, the sequence $\{f_n(x_0)\}$ will be a constant sequence, hence bounded. But again, if we let the $x_0$'s vary, then what this constant sequence is might be arbitrarily large.2010-11-17

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The short answer is: there need not be a real number that is the supremum of the values of $\phi(x)$. You may have $\sup\{\phi(x)\mid x \in E\} = \infty$. If that is the case, you're out of luck.