Pointwise vs. uniform convergence

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SUMMARY

The discussion centers on the concepts of pointwise and uniform convergence in the context of bounded functions. A function is pointwise bounded on a set E if for every x in E, there exists a finite-valued function φ such that |f_n(x)| < φ(x) for n = 1, 2, .... In contrast, a function is uniformly bounded on E if there exists a constant M such that |f_n(x)| < M for all x in E and all n. The participants clarify that while pointwise boundedness can imply uniform boundedness in finite domains, this equivalence does not hold in countable or uncountable domains due to the potential for supremum to be unbounded despite pointwise boundedness.

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ForMyThunder
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A function is pointwise bounded on a set E if for every x\in E there is a finite-valued function \phi such that |f_n(x)|&lt;\phi(x) for n=1,2,....

A function is uniformly bounded on E if there is a number M such that |f_n(x)|&lt;M for all x\in E, n=1,2,....

I understand that in uniform boundedness, the bound is independent of x and in pointwise convergence it is dependent. My question is this: if we take M=\max\phi(x), then since \phi is finite-valued, wouldn't this make every pointwise bounded function a uniformly bounded function? I don't understand.
 
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If E is the set (0,1) and \phi=\frac{1}{x}...

I think you can figure out the rest
 
Oh, thanks. It said finite-valued which I took to mean bounded.
 
I think for domain being a finite set, both notions coincide because maximum is indeed supremem. However, when it is either countable or uncountable domain, it is not necessary to have the equivalence between maximum and supremum. And it can turn out that supremum is unbounded depsite boundedness at each x. Office_Shredder showed a nice example.
 
ForMyThunder said:
Oh, thanks. It said finite-valued which I took to mean bounded.

Finite-valued must have merely meant that |\phi(x)| &lt; \infty for all x.
 

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