Cauchy Sequence Proof?

Question

1. Let {a_n} from n = 1 to infinity be a bounded sequence. For each n that is an element of the natural number set N, define
y_n = sup{a_k: k>=n}.

(i) Prove that {y_n} from n = 1 to infinity is decreasing and bounded from below, hence convergent.
(ii) Let A be the limit of the sequence {y_n} from n = 1 to infinity. Prove that {a_n} from n = 1 to infinity has a subsequence that converges to A.

2. Use item 1 to prove that every Cauchy sequence is convergent.

Answer 1

Hi Gabe.

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 < U. Then for each n, {a_k: k≥n} 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: k≥n} ≥ 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: 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) < 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 ε>0. Then choose N_1 such that 1/N_1 < ε and N_2 such that ∀n>N_2, y_n < A+ε (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-ε. 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+ε. So we have A-ε < a_(k_n) < A+ε, so |A - a_(k_n)| < ε. Thus n>N ⇒ |A - a_(k_n)| < ε, and since we can find such n for any ε>0, it follows that a_(k_n) → A.

2: Let {a_n} be a Cauchy sequence. By the definition of cauchiness, ∃N∈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 ε>0. Choose N such that |a_n - a_m| < ε/2. Now, since {a_(k_n)} → L, ∃M such that m>M ⇒ |a_(k_m) - L| < ε/2. So choose m>max (M, N). Then we have indeed m>M, so |a_(k_m) - L| < ε/2. Also, we have that k_m ≥ m > N. Thus, if n>N, we have that |a_n - a_(k_m)| < ε/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| < ε/2 + ε/2 = ε. Thus n>N ⇒ |a_n - L| < ε, and since we can find such an N for any ε>0, it follows that {a_n} → L, as claimed.