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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 \\$

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

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


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It will prepare you better if you try something. – Davide Giraudo Dec 13 '12 at 14:53
I have already, i am just too lazy to write it down. – UH1 Dec 13 '12 at 15:00
ok i wrote it down. – UH1 Dec 13 '12 at 15:14
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. – Zach L. Dec 13 '12 at 15:38

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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which 'first' inequality are you talking about? also are you sure your subscripts are consistent with the question? – UH1 Dec 13 '12 at 15:46
I don't know whether they are consistent as nothing is specified about them. By first inequality, I mean, proving that $A=B$ is the same as proving that $A\leqslant B$ and $B\leqslant A$. – Davide Giraudo Dec 13 '12 at 15:48
i wrote something but i think the reasoning is not sound, can you help me with proving at least one inequality? – UH1 Dec 13 '12 at 18:37
@UH1 It would be better if you write an answer. – Davide Giraudo Dec 13 '12 at 19:04

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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@Giraudo, I didn't see your answer when I wrote mine. – TCL Dec 13 '12 at 15:56
No problem. Unrelated: I like your nickname! – Davide Giraudo Dec 13 '12 at 16:05

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