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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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    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

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Hint: for each $n$ and $l$, $a_{n,l}\leqslant \sup_j\sup_ka_{j,k}.$ Taking the supremum over $n$, then over $k$, we get the first inequality. A similar argument for the other gives what we want.

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    @UH1 It would be better if you write an answer.2012-12-13
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Let $C=\sup_{(n,l)}a_{n,l}$ and $A$ be your number above. Then obviously $C\ge A$. It is equally easy to see that $C\le A$. So $C=A$. Same proof for the other one. (Of course, you need to show a bit more details to get full credit. $\epsilon$-proof is not necessary.)

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    No problem. Unrelated: I like your nickname!2012-12-13