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!