MthT 430 Notes Chap7c Bounded Monotone Sequences Have Limits

BISHL: Bounded Increasing Sequences Have Limits

Theorem. Let {x_n}_n=1^∞ be a bounded monotone increasing sequence; i.e.

x1 ≤ x2 ≤ …,

and there is a number M such that for n = 1,2, …,

xn ≤ M.

Then there is a number L such that

lim n → ∞  xn = L.

Proof using (P13-BIN): Without loss of generality, we assume that

0 ≤ x1 ≤ x2 ≤ … ≤ xn ≤ … < 1.

We will construct a binary expansion for L.

A picture is helpful!

Divide the interval [0,1) into two halves.

Is there an n₁ such that x_n1 ≥ m₀ = [1/2] = 0._bin1?

If NO, let c₁ = 0, a₁ = 0 = 0._binc₁, b₁ = m₀ = a₁ + [1/2]. If YES, let c₁ = 1, a₁ = m₀ = [1/2] = 0._binc₁, b₁ = a₁ +[1/2] = 1.

In both cases, for n ≥ n₁, a₁ = 0._binc₁ ≤ x_n ≤ b₁ = a₁ + [1/(2¹)] and b₁ − a₁ = [1/(2¹)].

Next Divide the interval [a₁,b₁) = [0._binc₁, a₁ + [1/(2¹)]) into two halves.

Is there an n₂ > n₁ such that x_n2 ≥ m₁ = a₁ + [1/(2²)]?

If NO, let c₂ = 0, a₂ = a₁ = 0._binc₁c₂, b₂ = m₁ = a₂ + [1/(2²)]. If YES, let c₂ = 1, a₂ = m₁ = 0._binc₁c₂, b₂ = b₁ = a₂ + [1/(2²)].

In both cases, for n ≥ n₂, a₂ = 0._binc₁c₂ ≤ x_n < b₂ = a₂ + [1/(2²)] and b₂ − a₂ = [1/(2²)].

By recursion (on k), if n_k > n_k−1, c₁, …, c_k, a_k = 0._binc₁…c_k, b_k = a_k + [1/(2^k)] have been defined so that for n ≥ n_k,

ak = 0.binc1…ck ≤ xn < bk = ak + 1 2k ,

divide the interval [a_k,b_k) into two halves.

Is there an n_k+1 > n_k such that x_nk+1 ≥ m_k = a_k + [1/(2^k+1)]?

If NO, let c_k+1 = 0, a_k+1 = a_k = 0._binc₁c₂…c_k+1, b_k+1 = m_k = a_k+1 + [1/(2^k+1)]. If YES, let c_k+1 = 1, a_k+1 = m_k = 0._bin c₁c₂…c_k+1, b_k+1 = b_k = a_k+1 + [1/(2^k+1)].

In both cases, n_k+1 > n_k, c₁, …, c_k, c_k+1, a_k+1 = 0._binc₁…c_kc_k+1, b_k+1 = a_k+1 + [1/(2^k+1)] have been defined so that for n ≥ n_k+1,

ak+1 = 0.binc1…ck+1 ≤ xn < bk+1 = ak+1 + 1 2k+1 ,

Let

L = 0.binc1 …ck … = lim k → ∞  ak = lim k → ∞  bk

We have that L = lim_{n → ∞}x_n since for all k, and n > n_k,

0 ≤ L − xn ≤ bk − xn ≤ bk − ak ≤ 1 2k .

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On 18 Sep 2014, 21:22.