given a set S which is not bounded above, prove there exists a sequence (Sn) s.t. : Sn exists in S for all n?

Answer 1

not really sure wat bounded means but even if i did i couldnt help

Answer 2

No. S_n=a million if n is fundamental, in any different case =0 would not converge. to work out if a chain converges, style 2 auxiliary sequences: A_1=the biggest huge style it rather is below each and all of the S_1,S_2,S_3,... A_2=the biggest huge style it rather is below each and all of the S_2,S_3,S_4,... A_3=the biggest huge style it rather is below each and all of the S_3,S_4,S_5,... etc. B_1=the smallest huge style it rather is extra advantageous than each and all of the S_1,S_2,S_3,... B_2=the smallest huge style it rather is extra advantageous than each and all of the S_2,S_3,S_4,... B_3=the smallest huge style it rather is extra advantageous than each and all of the S_3,S_4,S_5,... etc. The a collection would be an nondecreasing sequence and the B sequence would be nonincreasing. in the event that they converge to a similar fee, the unique S sequence converges, in any different case it would not. as an occasion, assume the sequence S={-a million,a million,-a million,a million,-a million,...} then A={-a million,-a million,-a million,...}-->-a million and B={a million,a million,a million,...}--> a million So S would not converge. on the different hand, assume the sequence S={a million,-a million/2,a million/3,-a million/4,...} then A ={-a million/2,-a million/2,-a million/4,-a million/4,...}-->0 and B ={a million,a million/3,a million/3,a million/5,a million/5...}-->0 So S converges to 0.