WM wrote: > On 27 Okt., 11:27, "Dirk Van de moortel" > <dirkvandemoor...@nospAm.hotmail.com> wrote: > >>> For my third sequence I did not calculate the limit but ask you and >>> others: What is it? > >> >> Yes, that's the first sequence. I got the same result with >> S_n = { 1 , ... , n }: >> LimInf (S_n) >> = V(n = 1 ... oo)[/\(k = n ... oo) S_k] >> = V(n = 1 ... oo)[ { 1, 2, ... , n } ] >> = |N >> LimSup (S_n) >> = /\(n = 1 ... oo)[V(k = n ... oo) S_k] >> = /\(n = 1 ... oo)[ { 1, 2, ... } ] >> = |N >> Therefore LimInf (S_n) = LimSup (S_n) = |N >> OK >> >> With the second sequence >> T_n = { 1/2 (n-1) n + 1 , ... , 1/2 (n-1) (n+2) + 1 } >> I'm not so sure: >> >> LimInf (T_n) >> = V(n = 1 ... oo)[/\(k = n ... oo) T_k] >> = V(n = 1 ... oo)[ { } ] >> = { } >> LimSup (T_n) >> = /\(n = 1 ... oo)[V(k = n ... oo) T_k] >> = /\(n = 1 ... oo)[ { 1/2 (n-1) n + 1, 1/2 (n-1) n + 2, ... } ] >> = { } >> Therefore LimInf (T_n) = LimSup (T_n) = { } >> OK. >> >> Now for the final sequence: >> TS_n = { ( 1/2 (n-1) n + 1 , 1 ) , ... , ( 1/2 (n-1) (n+2) + 1 , n ) >> } >> LimInf (TS_n) >> = V(n = 1 ... oo)[/\(k = n ... oo) TS_k] >> = V(n = 1 ... oo)[ { } ] >> = { } >> LimSup (TS_n) >> = /\(n = 1 ... oo)[V(k = n ... oo) TS_k] >> = /\(n = 1 ... oo)[ { ( 1/2 (n-1) n + 1 , 1 ) , ( 1/2 (n-1) n + 2 , >> 2 ), .... } ] = { } >> Therefore LimInf (TS_n) = LimSup (TS_n) = { } >> Seems OK? > > I disagree. You can consider the first or the second index as > negligible or as not written. In my opinion the final result must > depend in a continuous way on each of the indices. Note, we are > calculating the number of elements a_ij. This number does not change > when one of the indices becomes smaller and smaller and finally fades > away or reappeares.
I don' see any error in the proof. Furthermore, there is an obvious bijection between every set T_n of the second sequence and its corresponding set TS_n of the third sequence. You can easily think away the right part of each tuple in TS_n and you end up with the element of T_n. So if T_k converges to { } (which you agree with), then so must TS_k. Pretty straighforward i.m.o.
