[Lecture Summary] 03 Real Anaysis : Sequence

This contents is based on Lecture of 18.100A of MIT Open Course. Coverage will be Lecture from 4~. link

Overview

Analysis is the study of limits

Definition and Symbols

Sequence of Reals

• function $x: \mathbb{N}\to\mathbb{R}$
• $x(n)=x_n$
• $\lbrace x_n \rbrace_{n=1}^{\infty}, \lbrace x_n\rbrace$

Bounded

• sequence is bounded if $\exists B\ge 0 s.t. \forall n, \lvert x_n\rvert\le B$
• to prove that a sequence is unbounded, then derive other values which is less than any value of sequence

Convergence

• if $\forall \epsilon >0, \exists M\in \mathbb{N} s.t. \forall n\ge M \lvert x_n - x \rvert < \epsilon$
• As a sequence progress, sequence and convergent value are getting closer!

Limits

• $x=\lim_{n \to \infty}x_n$ or $x_n\to x$

Monotone increasing(decreasing): monotonic

• $\forall n\in \mathbb{N}, x_n \le(\ge) x_{n+1}$

Subsequence

• sequence with entries coming from another given sequence.
• $x_n$ be a parent sequence and $n_k$ is strictly increasing sequence of $\mathbb{N}$. Then the subsequence is $\lbrace x_{n_{k}}\rbrace_{k=1}^{\infty}$

Limsup(limit superior)/Liminf(limit inferior)

• For bounded sequence $x_n$
• $\limsup x_n = \lim (\sup\lbrace x_k \mid k\ge n\rbrace)$
• $\liminf x_n = \lim (\inf\lbrace x_k \mid k\ge n\rbrace)$
• Of course, supremum and infimum might be different with each other, they can be same for limsup and liminf(maybe convergence?!)

Cauchy

• A sequence $x_n$ is Cauchy if $\forall \epsilon >0 \exists s.t. \forall n,k\ge M, \lvert x_n-x_k\rvert<\epsilon$
• any 2 entries get closer to each other.

Not Cauchy

• A sequence $x_n$ is not Cauchy if $\exists \epsilon_0 s.t. \forall M\in\mathbb{N}, \exists n,k\ge M s.t. \lvert x_n - x_k\rvert \ge \epsilon_0$

Questions to be solved

1. How do limits interact with ordering, algebric operations?
2. Does a bounded sequene have a convergent subsequence?
   • Motivation for $\limsup, \liminf$

Theorem

$x,y \in \mathbb{R}. \mbox{ If } \forall \epsilon > 0, \lvert x-y \rvert <\epsilon, \mbox{ then } x=y$

Approach

• Contradiction
  • Suppose that $x\ne y$
  • Prove that absolute value is negative!

(Uniqueness) If $\lbrace x_n\rbrace$ converges for x and y, then x=y.

Approach

• Use just the above theorem
  • To do so, we need to derive $\lvert x-y \rvert <\epsilon$
  • Inequality comes from the definition of convergence

If $x_n$ is convergent then $x_n$ is bounded

Let $x_n$ be a monotone increasing sequence. $x_n$ is convergent $\iff x_n$ is bounded. Also $\lim_{n\to \infty}x_n = \sup\lbrace x_n \mid n\in \mathbb{N}\rbrace$

Approach

• iff relation is proved with previous theorem.
• Definition of limit.
• Then define $x$ as the supremum of $x_n$.

Assumption

• AP and WOP of $\mathbb{R}$ to show it is bounde above and below.

Want to show

• limit of $x_n$ is $x$.

If $c\ge 1$ then $c^n$ is unbounded

Approach

• for $x\ge -1$, $\forall n\in\mathbb{N}, (1+x)^n \ge 1+nx$

Want to show

• $\forall B\ge0, \exists n\in \mathbb{N} s.t. c^n > B$

If $c\in(0,1)$, then $\lim c^n = 0$

Approach

• for $x\ge -1$, $\forall n\in\mathbb{N}, (1+x)^n \ge 1+nx$
• Induction!

Want to show

• $\forall n\in\mathbb{N}, 0<c^{n+1}<c^n$ : monotonic decreasing and bounded
• Then it converges and show that limit of it is 0!

If $x_n$ converges to $x$, then any subsequence of $x_n$ will converge to $x$ either.

(Squeeze theorem) $a_n, b_n, x_n$ are sequences s.t. $a_n \le x_n \le b_n$. Both $a_n, b_n$ converges to $x$, then $x_n$ converges to x either.

Approach

• Using definition of convergence, define the number that each sequence $a_n, b_n$ converges.
• Then choose larger value for $x_n$!

$\lim x_n = x \iff \lim \lvert x_n - x \rvert = 0$

Approach

• Use definition
• $\lvert x_n - x \rvert = \lvert \lvert x_n - x\rvert - 0\rvert$

Let $x_n, y_n$ be sequences of $\mathbb{N}$. If both sequences are convergent and $\forall n\in\mathbb{N}, x_n \le y_n$ then $\lim x_n \le \lim y_n$

Remark

• limit can lose strict inequality relation.
  • limit of two sequences can be equal to each other!

Let $x_n, y_n$ be sequences of $\mathbb{N}$. If $x_n$ is convergent and $\forall n \in \mathbb{N}, a_n \le x_n \le b_n$ then $a\le \lim x_n \le b$

$x_n$ is convergent s.t. $\forall n\in \mathbb{N}, x_n\ge 0$ then $\lbrace \sqrt{x_n}\rbrace$ is convergent and $\lim \sqrt{x_n} = \sqrt{\lim x_n}$

If $x_n$ is convergent and $\lim x_n = x$ then $lim \lvert x_n\rvert = \lvert x\rvert$

Approach

• Reverse inequality

Remark

• converse statement is not true!

(Special sequences) If $p>0$ then $\lim n^{-p} = 0$

Approach

• Use definition of limit

(Special sequences) If $p>0$ then $\lim p^ = 1$

Approach

• For $x\ge -1$ then $(1+x)^n\ge 1+nx$
• p = (1+(p^1-1))^n
• Squeeze Theorem!

(Special sequences) $\lim n^ = 1$

Approach

• Binomial theorem!
• $n = (1+x_n)^n$
  • $x_n = n^-1$

For bounded seuqence $x_n, a_n = \sup\lbrace x_k \mid k\ge n\rbrace, b_n = \inf\lbrace x_k \mid k\ge n\rbrace$, then $a_n$ is monotone decreasing and bounded and $b_n$ is monotone increasing and bounded

Approach

• To show that $x_n$ is bounded, AP of $\mathbb{R}$!
• To show that $a_n, b_n$ are monotonic, use definition of monotone.
• Finally, we can show that $a_n, b_n$ are bounded use $x_n$

Remark

• $a_n , b_n$ is subsequence!

For bounded seuqence $x_n, a_n = \sup\lbrace x_k \mid k\ge n\rbrace, b_n = \inf\lbrace x_k \mid k\ge n\rbrace$, then $\liminf x_n \le \limsup x_n$

Approach

• Use previous theorem
• Use definition of infimum and supremum of $x_n$ and derive inequality!

For bounded sequence $x_n$, there exists subsequences $x_{n_k}, x_{m_k} s.t. \lim x_{n_k} = \limsup x_n , \lim x_{m_k} = \liminf x_n$

Approach

• definie supremum of $x_k$.
• derive ineqaulity relation repeatedly and use Squeeze Theorem!
  • We can find an element which satisfies supremum inequality relation
  • Gather them! they are subsequence!

(Bolzano-Weierstrass) Every bounded sequence has a convergent subsequence!

For bounded seuqence $x_n$, $x_n$ converges $\iff \liminf x_n = \limsup x_n (=\lim x_n)$

Approach

• Sequeze theorem for converse statement
• define limsup and liminf for each and use the fact that $x_n$ is convergent!

If $x_n$ is Cauchy, then $x_n$ is bounded

Approach

• Once we define $\epsilon_0$ then we can calculate corresponding interval for Cauchy assumption and then prove that it is bounded!

If $x_n$ is Caucyand subsequence $x_{n_k}$ converges then $x_n$ converges

Approach

• Definition
• Convergence of subsequence does not guarantee that of full sequence!
• So we need to “concatenate” inequality from Cauchy assumption and subsequence convergence

A sequence of real numbers $x_n$ is Cauchy if and only if $x_n$is convergent

Remark

• If we only work in natural numbers, convergence implies Cauchy but Cauch does not imply convergence.
• For $\mathbb{R}, l.u.b.p is the reason why converges $\iff$ Cauchy
• To use BW theorem, we need to derive limit of sup and inf which says we need sup and inf literally. $\mathbb{R}$ has l.u.b.p. but $\mathbb{Q}$ does not.
• If we can derive Cauchy then we can prove convergence without calculating the exact x