In this project you will provide a proof different from that

In this project, you will provide a proof, different from that in the text or Project 1.5, of the fact that every Cauchy sequence is convergent. In the process, you will give another proof that every bounded sequence has a convergent subsequence. 1. Let {a_n}^infinity_n=1nm be a bounded sequence. For each n J, define y_n = sup{a_k : k n}. i. Prove that {y_n}^infinity_n=1 is decreasing and bounded from below, hence convergent. ii. Let A be the limit of the sequence {y_n}^infinity_n=1. Prove that {a_n}^infinity_n=1 has a subsequence that converges to A. 2. Use item 1 to prove that every Cauchy sequence is convergent.

Solution

Ans(1.i):

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: kn} x{a_n} x < U. Then for

each n, {a_k: kn} is a nonempty set bounded above and so indeed has a supremum.
Next, since {a_n} is bounded below, let L be a lower bound. Then for any n, y_n = sup {a_k: kn} a_k > L, so {y_n} is bounded below by L. To show that it is decreasing, note that if m>n, then x{a_k: km} x{a_k: kn} xsup{a_k: kn} = y_n. Thus y_n is an upper bound for {a_k: km}, so y_m = sup {a_k: km} y_n. Thus m>n y_my_n, so y is decreasing.

Ans(1.ii):

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) < A. However, A is the limit of a decreasing sequence, so A = inf {y_k} y_(k_n + 1), so this is a contradiction. Therefore, such a natural number must exist, and {a_(k_n)} is well-defined. It remains only to show that it converges to A.

Let >0. Then choose N_1 such that 1/N_1 < and N_2 such that n>N_2, y_n < A+ (it is possible since y_n A). Then let N = max (N_1, N_2). Then suppose 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-. 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 < A+. So we have A- < a_(k_n) < A+, so |A - a_(k_n)| < . Thus n>N |A - a_(k_n)| < , and since we can find such n for any >0, it follows that a_(k_n) A.

Ans(2):

Let {a_n} be a Cauchy sequence. By the definition of cauchiness, N s.t. n, m>N |a_n - a_m| < 1. So for m>N, a_m < a_(N+1) + 1. Thus {a_n} is bounded above by max (a_1, a_2... a_N, a_(N+1) + 1) (there are only finitely many terms, so the maximum exists). For similar reasons, it is bounded below by min (a_1, a_2... a_N, a_(N+1) - 1). Thus {a_n} is bounded, thus by #1 it has a convergent subsequence {a_(k_n)}, which approaches some limit L. Then we need merely show that a_n L.

Let >0. Choose N such that |a_n - a_m| < /2. Now, since {a_(k_n)} L, M such that m>M |a_(k_m) - L| < /2. So choose m>max (M, N). Then we have indeed m>M, so |a_(k_m) - L| < /2. Also, we have that k_m m > N. Thus, if n>N, we have that |a_n - a_(k_m)| < /2. So from this and the triangle inequality we have |a_n - L| = |a_n - a_(k_m) + a_(k_m) - L| |a_n - a_(k_m)| + |a_(k_m) - L| < /2 + /2 = . Thus n>N |a_n - L| < , and since we can find such an N for any >0, it follows that {a_n} L, as claimed.