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Is $\sup_n\sup_l a_{n,l} = \sup_l \sup_n a_{n,l}$? Prove or disprove. I am preparing for my analysis final and this is one of the practice problems. Any help would be really appreciated! My try:

Let $A = \sup_n\sup_l a_{n,l} \ and\\x_n = \sup_l a_{n,l}\\$

$For\ all\ \epsilon > 0 \ \exists \ an \ N \ such\ that\ if \ n, \ l_* > N then\\$ $x_n - \epsilon < a_{n,l_*} <= x_n\\$

$\sup_n\ (x_n - \epsilon) < \sup_n\ a_{n,l_*} <= \sup_n\ x_n\\$

$ A - \epsilon < \sup_n\ a_{n,l_*} <= A \\$

$ A - \epsilon < \sup_l \sup_n\ a_{n,l_*} <= A \\$

$\ Let\ \epsilon\ go\ to\ zero$

$\ Therefore,\ \sup_l \sup_n a_{n,l} = A$

Thanks!

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    It will prepare you better if you try something.2012-12-13
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    I have already, i am just too lazy to write it down.2012-12-13
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    ok i wrote it down.2012-12-13
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    The $N$ you selected depends on $n$, as far as I can tell. So, when you take the supremum over all $n$, the inequality $x_n - \epsilon < a_{n,l^\ast} \leq x_n$ will no longer hold.2012-12-13

2 Answers 2