Let {a_n} n=1 to infinity, be a bounded sequence of real numbers. Prove that {a_

Question

Let {a_n} n=1 to infinity, be a bounded sequence of real numbers. Prove that {a_n} n=1 to infinity, has a convergent sequence. (Hint: You may want to use the Bolzano-Weierstrass Thereom)

Explanation / Answer

1i: First let us prove that {y_n} is well defined -- since {a_n} is bounded above, let U be an upper bound. Then for any n, x?{a_k: k=n} ? x?{a_n} ? x L, so {y_n} is bounded below by L. To show that it is decreasing, note that if m>n, then x?{a_k: k=m} ? x?{a_k: k=n} ? x=sup{a_k: k=n} = y_n. Thus y_n is an upper bound for {a_k: k=m}, so y_m = sup {a_k: k=m} = y_n. Thus m>n ? y_m=y_n, so y is decreasing. 1ii: Define a subsequence {a_(k_n)} of {a_n} as follows: Let k_1 = 1, and then k_(n+1) be the smallest natural number greater than k_n such that a_(k_(n+1)) > A - 1/(n+1). We must show that such a natural number exists. Suppose it does not -- then A - 1/(n+1) is an upper bound for the set {a_k: k = k_n +1}, so y_(k_n + 1) = sup {a_k: k = k_n +1} = A - 1/(n+1) 0. Then choose N_1 such that 1/N_1 N_2, y_n N. By the definition of a_(k_n), we have that {a_(k_n)} > A - 1/n > A - 1/N = A - 1/N_1 > A-e. Also, we have that k_n = n, so since y is decreasing and by the definition of y, we have that a_(k_n) = y_(k_n) = y_n M, so |a_(k_m) - L| N. Thus, if n>N, we have that |a_n - a_(k_m)|

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