Prove absolutely convergent double series $\sum u_{\mu,\nu}$ (which tends to S) when rearranged has the same limit.
(1) First prove it is true for absolutely convergent series whose items are all non-negative.
Say the rearranged series is $\sum u'_{\mu,\nu}$ , $\sum_{\mu\leq p,\ \nu\leq q} u'_{\mu,\nu}=S'_{p,q}$. All items in $S'_{p,q}$ are contained in $S_{m,n}$ where $m\geq p, n\geq q$, so $\forall p, q, S'_{p,q}\leq S_{m,n}\leq S$, so the limit of $\sum u'_{\mu,\nu}$ exist, (a 1-dim bounded monotonous sequence has an upper limit, is that true for 2-dim bounded 'monotonous' sequence has an upper limit?) say $S'$, and $S'\leq S$. Similarly $S\leq S'$, and so $S'= S$.
(2) Then we prove it is true for any absolutely convergent series. Its non-negative items are denoted as $v_{\mu,\nu}$, negative items are denoted as $-w_{\mu,\nu}$. Decompose the original series to two series:
$\sum v_{\mu,\nu}$ where $v_{\mu,\nu} = u_{\mu,\nu}$ if $u_{\mu,\nu}\geq 0; 0$ if $u_{\mu,\nu}< 0$,
$\sum -w_{\mu,\nu}$ where $-w_{\mu,\nu} = u_{\mu,\nu}$ if $u_{\mu,\nu} < 0; 0$ if $u_{\mu,\nu} \geq 0$.
Since $\sum |v_{\mu,\nu}| \leq \sum |u_{\mu,\nu}|$ which has a limit, $\sum v_{\mu,\nu}$ has a limit, similarly, $\sum -w_{\mu,\nu}$ has a limit, and both series when rearrganed tend to the same limits.
My questions:
From this how to go on to prove that $\sum u_{\mu,\nu}$ when rearranged tend to the same limit?
Besides, does the proposition, that an absolutely convergent double series $\sum u_{\mu,\nu}$ when rearranged has the same limit, imply that we can arrange all items in the series to the same row and the limit is still the same? (But then the double series becomes linear, and the items in other rows are simply 'nothing', even not 0, which is quite weird in my eyes).