STATEMENT: Every bounded sequence of real numbers has a convergent subsequence.
PROOF: Let {un} be a bounded sequence, Then there is a closed and bounded interval, say I [a, b], such that Un E for every n EN. e = ~ and I' [a, e), I" = [e, bJ, Then at least one of the intervals I' and I" contains infinitely many elements of {un}. Let It = [al, bl) be such an interval. Then It c I and I It I = the length of the interval = ~ (b - a). Let el = and Ii =. [al, el], If = [ell bd, Then at least one of the intervals Ii and If' contains infinitely many elements of {Un}. Let 12 [a2, b2) be such an interval. Then 12 chand I 12 ~ lIt I· Continuing thus, we obt~in a sequence of closed and bounded intervals {In} such that (i) In+l C In for all n E,N; (ii) I In \= 2I", (b - a) and therefore lim 1 In OJ. and n_oo (iii) each In contains infinitely many elements of {un}· By Cantor's theorem on nested intervals, there exists a unique point 00 a: such that o:E n In· . n=l We prove that 0: is a subsequential limit of the sequence {un}· Let us choose E > O. There exists a natural number k such that o < b:}ka. < E. That is, \ h 1< E. . Since 0: E h and I h 1< E,1;e is entirely contained in the neighbourhood (0: - €, 0: + E) and consequently, the E-neighbourhood of 0: contains infinitely many elements of {un}. . Since E is arbitrary, each neighbourhood of 0: contains infinitely many elements of {un}' Therefore 0: is a subsequential limit of {un}' Therefore there exi.sts a subsequence of {Un} that converges to 0:, In other words, {Un} has a convergent subsequence, This completes the proof
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