00:01
Hello students we are given proof that every bound sequence has a convergent sub sequence right so basically if i write for this we can write proof as so basically we need to let something wn so basically be a bounded wn be a bounded if i write here wn be a bounded bounded sequence right so if i write for this and then there exist an interval then there exist an interval and interval if i write a one comma b one such that such that a one is less than w n and w n and it is less than b and for all n now now if i write for this either a1 comma a1 plus v1 divided by 2 or a1 plus b1 divided by 2 comma b1 contains infinitely contains infinitely many terms infinitely many terms many terms many terms of wn right so if i write for this so that is there exist infinitely many n in j such that a n is in such that a n is in a 1 to a 1 plus b1 divided by 2 or there it exists infinitely many n in j such that an is in so again a n is in so again a n if i again an is in a1 plus b1 divided by 2 up to up to this will goes to b1 so if i write if a 1 and this will contain if if i write for this if i write for this if a 1 these are if a 1 comma a 1 plus b1 divided by 2 if i write for then contain contains infinitely infinitely many terms of many terms of w n then definitely what can we suggest let let a2 comma b2 is equal to a1 comma a1 plus b1 divided by 2 otherwise let otherwise otherwise we need to let a2 comma b2 a2 is equal to a1 plus v1 divided by 2 comma b1 right so if i correct this all the brackets here must be larger brackets if i write for them all the brackets must be converted to larger brackets because these are the values for the interval and this must be converted something like this and this must be converted to and these are larger brackets now so by using mathematical induction by using mathematical induction mathematical induction we can continue this by using mathematical induction we can continue this construction we can continue this construction, this construction, write this construction and obtain a sequence of interval and obtain sequence of interval such that, such that if i write for this, such that, c.
04:30
B n, a n comma b n if i write such that if i write for this so if i write for the first case it will be encountered as it will be encountered as for each n one for each and our a n coma b n contains infinitely infinitely many terms of w n many terms of w n if i write for this many terms of w n and second if i write is for each and for each a n plus one comma b n plus one is the subset of a n comma b n plus one is the subset of a n comma b n and if i write for this in the third case for each end for each and bn plus 1 minus a n plus 1 equals to half of half of bn minus a n minus a n right in the end we can conclude the nested interval nested interval theorem nested interval interval theorem implies that intersection of all the intervals, intersection of all the intervals, of all the intervals, a, and comma, v, is a single point, w, is a single point, w, is a single point, w, is a single point, is a single point w if i write so we will now construct a subsequence of w and we will now construct construct construct a subsequence of w and a subsequence we will now construct a subsequence of a subsequence of wn if i write for this which will converge to which will converse to w right? hence using mathematical instruction we can state after continuing each and every subsequent we will find our subsequence as our conclusion which will converge to w infinitely so basically we can prove hence proved that every bound sequence between a and b has a convergent subsequent hence, we can say, we can say every, every bound sequence, every bound sequence has a convergent, has a convergent, has a convergent subsequence...