MCT and BWT

Monotone Convergence Theorem

Monotone increasing and bounded above \(\Rightarrow\) converges to supremum.

Def'n. Monotonic

A sequence \((a_n)\) is (strictly) monotone increasing if

\[a_n\leq a_{n+1}(a_n<a_{n+1}), \forall n\geq 1\]

Thrm 1.

\[\lim_{n\rightarrow\infty}a_n = \inf(a_n)\]

proof. We'd like to show that

\[\forall \epsilon. \exists N. \forall n \geq N. |a_n - L| \leq \epsilon\]

Let \(L=\inf(a_n)\). Let \(\epsilon > 0\), take \(a_N \in (a_n)\) s.t. \(L-\epsilon < a_N\).
Then \(L-\epsilon < a_N \leq a_n \leq L\) for all \(n\geq N. \Rightarrow \forall n\geq N. |a_n-L|<\epsilon\)

Corollary 2.

Consider closed, non-empty nested intervals \(I_n\supseteq I_{n+1}\supseteq I_{n+2}...\), then then \(\cap^\infty_n I_n\neq \emptyset\)

proof. Let \(I_n=[a_n,b_n]\), then we have two monotone sequence \((a_n)\) is increasing, \((b_n)\) is decreasing. Then, \(\forall k\geq 1. a_k \leq \lim_\infty a_n\leq \lim_\infty b_n \leq b_k\)

Then, we can show \(\Rightarrow [lim_\infty a_n, \lim_\infty b_n]=[\sup(a_n),\inf(b_n)]\subseteq \cap_{n\geq 1} I_n\)

Example 1

(D&D 2.6.B) Let \(a_1 = 0, a_{n+1}=\sqrt{5 + a_n}\), find whether it's convergent and if convergent what's the limit.

Monotone

\[\begin{align*} a_2 &= \sqrt 5 > 0 = a_1\\ a_{n+1} &= \sqrt{5 + a_n} > \sqrt{5 + a_{n-1}}=a_n \end{align*}\]

Bounded above
\(a^2_{n+1} =5 + a_n < 5 + a_{n+1}\). Let \(x = a_{n+1}\Rightarrow x^2 < 5 + x\Rightarrow x^2 -x -5 < 0\).
\(x\in [\frac{1-\sqrt{21}}{2}, \frac{1+\sqrt{21}}{2}]\Rightarrow a_{n+1}=x\) is bounded above.

In fact, \(L = \lim\sqrt{5 + a_{n-1}}=\sqrt{5+L} \Rightarrow L^2 = L - 5\). solve and take \(L > 0\) since monotone increasing, then \(L=\frac{1+\sqrt{21}}{2}\)

Example 2

(D&D 2.6.I) Let \((a_n)\) be bounded, define \(\lim\sup a_n = b_n = \sup\{a_k: k\geq n\}\) for \(n\geq 1\). Prove that \((b_n)\) converges.

Monotone \(b_n\leq b_{n+1}\)
since \(b_n = \max(a_n, \sup\{a_k: k\geq n+1\})=\max(a_n,b_{n+1})\geq b_{n+1}\)

Bounded below \(\forall n \geq 1. a_n\geq M, b_n = \sup(a_k:k\geq n )\geq M\)

\[\exists L<\infty. \lim b_n = \lim_{n\rightarrow\infty}\sup_{k\geq n} a_nL\]

Remark \(L_u = \lim\sup a_k \geq \lim\inf a_k = L_l\), if \(L_u = L_l = L\Rightarrow \lim a_k = L\)

Bolzano-Weierstrass Theorem

Every bounded sequence of real numbers has a convergent subsequence

Def'n Subsequence

A subsequence of \((a_n)\) is a sequence \((a_{n_k})\) where \(n_1<n_2<...\)

proof. Let \((a_n)\subseteq [-M, M]\).
Construct the subsequence by - picking \(I_1\subset [-M,M]\) that contains infinitely many \(a_n\) s.t. \(|I_1|\leq M/2\) - ... - picking \(I_n\subset I_{n-1}\) that contains infinitely many \(a_n\) s.t. \(|I_n| \leq |I_{n-1}|/2 \leq M/2^n\).

By Nested interval lemma, \(\cap_{n\geq 1}I_n \neq\emptyset\), take \(L\in \cap I_n\).

Pick \(a_n\in I_n, \forall n \geq 1\) Since \(L\in I_n\Rightarrow |a_n - L|\leq |I_n|<M/2^n\).

Then \(\forall \epsilon > 0\), take \(N_\epsilon\) s.t. \(\epsilon > M/2^{N_\epsilon} > |a_{N_\epsilon} - L|\)
but \(|a_n - L | \leq |I_n|\leq |I_{N_\epsilon}| / 2^{n-N_\epsilon}, \forall n\geq N_\epsilon\)

\[\epsilon > M / 2^{N_\epsilon} > |a_n - L|. \forall n\geq N_\epsilon\Rightarrow \lim{a_n} = L\]