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title: Sequences by Bryce

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# Sequences by Bryce

**Definition 1.** $\{x_n\}$ is constant if $x_n=x_{n+1}$ $\forall n\in {\mathbb N}$.

**Example 1.** 
* $\{17\}$ is constant since $x_n=17=x_{n+1}$ $\forall n\in {\mathbb N}$.

**Counterdefinition 1.** $\{x_n\}$ is not constant if $\exists n\in{\mathbb N}$ such that $x_n≠x_{n+1}$ 

**Example 1'.** $\{(-1)^n\}$ 
* It is not constant since $\{(-1)^n\}=\{1,-1\}$ Let n=2, then $\exists  n\in{\mathbb N}$ such that $x_n≠x_{n+1}$ where $1≠-1$.

**Definition 2.** $\{x_n\}$ is bounded above if $\exists K\in{\mathbb R}$ and
$\forall n\in {\mathbb N}$.   $x_n\leq K$,  

**Example 2.** Prove that $\{(-1)^n\}$ is bounded above. 
* Let $K=2$.  Since the range of $\{(-1)^n\}$ is{1,-1\}. $K\geq x_n$ $\forall n\in {\mathbb N}$  Therefore K is an upper bound.

**Counterdefinition 2.** 
* $\{x_n\}$ is not bounded above if  $\forall K\in{\mathbb R}$, $\exists n\in{\mathbb N}$ such that $x_n\geq K$.

**Example 2'.**
$\{(2)^n\}$ has no upper bound. 
**Proof** 
* Suppose $\{(2)^n\}$ has an upper bound. Let $K=2^j$ $j\in\{2^n\}$ for $j\in{\mathbb N}$ be an upper bound. Now consider$2^{j+1}$. Since 2 is a positive base then, $2^j < 2^{j+1}$. Then $K< 2^{j+1}$, so K is not an upper bound. Contradiction.  $\{(2)^n\}$ has no upper bound.


**HW**

**Defintion 3 Lower Bound:**
* $\{x_n\}$ is bounded bellow if $\exists K\in{\mathbb R}$,$\forall n\in {\mathbb N}$  $x_n\geq K$.

**Example 3 Lower bound**:Prove that $\{(0.5)^n\}$ is bounded bellow. 
* Let $K=-2$.  Since the range of $\{(0.5)^n\}$ is $\{0<k:k\in{\mathbb R}\}$. Then $K\leq x_n$ $\forall n\in {\mathbb N}$  Therefore K is a lower bound.

**Counter Definition  3 Lower bound:**
* $\{x_n\}$ is not bound bellow if $\forall K\in{\mathbb R}$, $\exists n\in{\mathbb N}$ such that $x_n\leq K$ .


**Example 3'Lower bound:**

* $\{(-3)^n\}$ has no lower bound. 
**Proof** 
* Suppose $\{(-3)^n\}$ has a lower bound.  Let $j\in\{(-3)^n\}$ for $j\in{\mathbb N}$ such that $K=2^j$ be a lower bound. Now consider$2^{j+2}$.  Then $(-3)^{j+2} < (-3)^{j}$.Therefore, $\exists m\in{\mathbb N}$, such that $-3^m<K$.  $K$ is not a lower bound. Hence $\{(-3)^n\}$ has no lower bound.

**Definition 4 Bounded:** $\{x^n\}$ is bounded if  $\exists K\in{\mathbb R}$, ${|K|\geq |x^n|}$ $\forall n\in{\mathbb N}$ 

* Needs both an upper and lower bound

**Example 4 Bounded 4:** $\{(sin(x_n)\}$ Is bounded.

**Proof**
*  Let $K_1,K_2\in{\mathbb R}$ such that $K_1=2$ and $K_2=-2$ Therefore $|K_1|=|K_2|=2$. WLOG consider $K_1$. since the codomain of $sin(x_n)$ is $\{1\geq|x|:x\in{\mathbb R}\}$ Therefore $|sin(x)|<|K_1|$, since $|1|<|2|. \{(sin(x_n)\}$ is bounded.

**Counter Defintion Bounded 4:**
* $\{x^n\}$  is not bounded $K\in{\mathbb R}$, $\exists n\in{\mathbb n}$ such that $|x^n|\geq|K|$ .

* Note it means that is does not have a lower bound or a upper bound. Therefore you can combine the converses of both of those to get definition

**Example 4' Bounded:**

* Note: all you need to show is that is does not have a lower bound or upper bound to show that is it unbounded. Therefore Example 2',3',4' all work.

**Additional example: $\{(1.5)^n\}$ is bounded**
    * Proof by contradiction
* WLOG suppose that$\{(1.5)^n\}$ has an upperbound. Then $\{(1.5)^n\}$ has a least upper bound.  $j\in\{1.5^n\}$ for $j\in{\mathbb N}$ such that $K=1.5^j$ be an upper bound. Now consider$1.5^{j+1}$. Since 1.5 is a positive base greater than 0 then, $1.5^j < 1.5^{j+1}$. Then $K< 1.5^{j+1}$, so K is not an upper bound. Contradiction. So $\{(1.5)^n\}$ is not bounded

Continue similarly with  bounded below and bounded: definition, example, counterdefinition, example. Show all your work.


**Week 2**


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1. Prove that if $\{x_n\}$ is infinitely large then $\{x_n\}$ is unbounded
    * (Contradiction) Suppose $\{x_n\}$ is infintely large. Then $\forall K\in{\mathbb R}$  $\exists m\in\{x_n\}$ such that $\forall n\in{\mathbb N}$  $n\geq m$, ${|x_n|>K}$. Let $\{x_n\}$ be bounded.  Then $\exists K_1\in{\mathbb R}$ such that $x_n\leq K_1$,$\forall n\in {\mathbb N}$. However if $x_n\leq K_1$ then there $\exists K\in{\mathbb R}$ such that $K\geq x_n$ $\forall n{\mathbb N}$. This is a contradiction on the assumption that $\forall K\in{\mathbb R}$ there exists $m\in\{x_n\}$ such that $\forall n\in{\mathbb N}$  $n\geq m$ and ${|x_n|>K}$. Therefore $\{x_n\}$ is unbounded.
3. Prove that $\{(-1)^nn\}$ is inf. large
    *  (Unboundedness)Suppose $\{(-1)^nn\}$ is bounded. Let $K_1\in{\mathbb R}$ such that $|K_1|=|{(-1)^jj}|$,  $j\in{\mathbb N}$ be a bound. Then $|K_1|\geq|(-1)^nn|$ $\forall n\in{\mathbb N}$.
        *  Case 1: Let  J be even. Consider $J+2 \in{\mathbb N}$  Then $|(-1)^{j+2}(j+2)|= |j+2| > |j|=|(-1)^jj|=|K_1|$ This is a contradiction. 
        *  Case 2: Let J be odd. Consider $J+1 \in{\mathbb N}$$|(-1)^{j+1}(j+1)|= |j+1| > |j|=|(-1)^jj|=|K_1|$ This is a contradiction. $\{(-1)^nn\}$ is not bounded.  
    * (Inf. Large) Now Consider $j+1$. Then $\forall n\in{\mathbb N}$ such that  $n\geq j+1$ ,${|(-1)^nn|>|K_1|=|{(-1)^jj}|\geq K_1}$. Therefore it is inf. large


5. Counter the definition of infinitely large.
    * $\{x_n\}$ is not infinietly large if   $\exists K\in{\mathbb R}$ and $\forall m\in{\mathbb N}$   $\exists n\geq m$,  $|x_n|<K.$
7. Prove that  $$x_n=\begin{cases} n,&\text{if $n$ is even}\\ 0,& \text{if $n$ is odd}\end{cases}$$is not infinitely large, but unbounded
    * Let $$x_n=\begin{cases} n,&\text{if $n$ is even}\\ 0,& \text{if $n$ is odd}\end{cases}$$
    * (Proving unboundedness) Suppose $x_n$ is bounded. Then $x_n$ is bounded above. Then $\exists K\in{\mathbb R}$ such that $K$ is an upper bound. Then $x_n\leq K$, $\forall n\in {\mathbb N}$.  However consider  $j\in{\mathbb N}$ and even, such that $x_j= 2|K|$.  Then $x_j> K$. Contradiction. Therefore it is unbounded.
    * (Not Infinitely large) Suppose $x_n$ is infinitley large. Then $\forall K\in{\mathbb R}$ there exists $n^*\in{\mathbb N}$ such that $\forall n\in{\mathbb N}$  $n\geq N*$ and ${|x_n|>K}$. Let $m\in\{x_n\}$. 
        *  Case (1): Suppose $m$ is odd. Then $x_{m+2}$ is also odd. Then $x_{m+2}=0$. However, $x_{m+1}$  is even, so $x_{m+1} =m+1>0$. Then $\exists K\in{\mathbb R}$ such that $K\geq x_{m+1}$. However, $x_{m+2}=0<|X_{m+1}|\leq K$ This is a contradiction on the definition of inf large. 
        *  Case (2):Suppose $m$ is even. Then $x_{m+2}$ is also even. Then $\exists K\in{\mathbb R}$ such that $K\geq x_{m+2}$.  However, $x_{m+3}=0$ since it is odd. Then $x_{m+3}=0<x_{m+2}\leq K$ This Contradiction on the definition of inf large

    * Therefore $$x_n=\begin{cases} n,&\text{if $n$ is even}\\ 0,& \text{if $n$ is odd}\end{cases}$$is not infinitely large, but unbounded 
    
9. Counter the definition of $\lim x_n=L$.
    * $\lim x_n\neq L$ if $\exists\epsilon>0$ such that $\forall m\in{\mathbb N}$, $\exists n\geq m$ such that  $|x_n-L|>\epsilon$
11. Prove that $\lim(-1)^n\neq 0$
    * Suppose $\lim(-1)^n= 0$. Given any $\epsilon >0$, let $m=1$. Then $\forall n\geq m$ for $n,m\in{\mathbb N}$ be odd or even.
        * Case 1 Let $n$ be odd. Then $|(-1)^n-L|=|-1-0|=1<\epsilon$
        * Case 2: Let $n$ be even. Then  $|(-1)^n-L|=|1-0|=1<\epsilon$  
    *    Consider $\epsilon =\frac{1}{2}$. Then $|(-1)^n-L|=|1-0|=|-1-0|=1<\frac{1}{2}$ Which a contradiction since $\epsilon>0$. Then $\lim(-1)^n\neq 0$.
13. Prove that $\lim(-1)^n\neq -1$
    * Suppose $\lim(-1)^n= -1$. Given any $\epsilon >0$, let $m=1$. Then $\forall n\geq m$ for $n,m\in{\mathbb N}$ be odd or even.
        * Let $n$ be even $|(-1)^n-L|=|1-(-1)|=2<\epsilon$  Consider. $\epsilon =\frac{1}{2}$. Then $|(-1)^n-L|=|1-(-1)|=2<\frac{1}{2}$ Which a contradiction  since. 
    * Then $\lim(-1)^n\neq -1$.

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**Week 3**


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1. Negate and justify how we negate the following statement. In particular, why we do not change the inequality $\epsilon>0$.

$\forall \epsilon>0, ~~~~\epsilon$ is a bad number. 
* Orginial: $\forall\epsilon>0$ $\epsilon$ is a bad number.  
* Negations: $\exists \epsilon >0$ that  $\epsilon$ is good number
* In any Negatation you suppose A  implies not B. Since this is a $\forall$ statement, then it is a biconditional. Therefore the original and its converse are true.  Since that is the case, we need to negate A and B. Since this can be written as 2 statements For example:
    * $A\implies B$ 
        * If $\epsilon>0$, then$\epsilon$ is a bad number
    * $B\implies A$
        * If $\epsilon$ is a bad number, then$\epsilon>0$.

*   In order for us to negate this logically it makes sense to break down the $\forall$ statement, we only need one to counter so we need an $\exists$ .This handles one direction. Now in order to do the other direction the only quantifier is bad. Therefore we switch that as good.  Therefore we get the negation above. $\exists \epsilon >0$ that  $\epsilon$ is good number. We do not change the inequality, because it defines the values for $\epsilon$ so negating it would change the statement all together. All the inequality does is define what values we are looking at. 


2. Let $A\subset \mathbb R$. State the definition of $m=\inf~A.$
    *  Let $A\subset \mathbb R$ $m=\inf~A.$ if $\forall a\in A$ $a\geq m$
    *  $\forall \epsilon >0$, $\exists b\in A$ such that $m\leq b < m+ \epsilon$
3. Prove that $\sup\{1-\frac{1}{n}\}=1$
    *  Note: $\forall x\in \{1-\frac{1}{n}\}$, $x\leq 1$. Now let $\epsilon >0$, then $x^*=\max\{\frac{2}{3},\frac{1}{\lceil{\epsilon\rceil}}\}$. Then $1 -\epsilon<\max\{\frac{2}{3},\frac{1}{\lceil{\epsilon\rceil}}\}\leq 1$ Therefore $\sup\{1-\frac{1}{n}\}=1$

4. Prove that $\inf\{1-\frac{1}{n}\}=0$
    *  Note: $\forall x\in \{1-\frac{1}{n}\}$, Now$0\leq x<1$.Let $\epsilon >0$, then $x^*=\min\{\frac{2}{3},1-\frac{1}{\lfloor{{\epsilon\rfloor}+1}}\}$. Then $0\leq\min\{\frac{2}{3},1-\frac{1}{\lfloor{{\epsilon\rfloor}+1}}\}<\epsilon$. Therefore $\inf\{1-\frac{1}{n}\}=0$

5. Prove that $\inf\{(0,1)\}=0$
    * Note: $\forall x\in\{(0,1)\}$ $0<x<1$.Now let $\epsilon >0$, then $x^*=\min\{\frac{1}{2},\frac{\epsilon}{2}\}$. Then $0\leq\min\{\frac{1}{2},\frac{\epsilon}{2}\}<\epsilon$. Therefore $\inf\{(0,1)\}=0$.


6. Let $A=\{0\}\cup \{(1,2)\}$. Prove that $\inf A=0$.
    * $\forall x\in A$ $x=0$ or $1<x<2$.Now let $\epsilon >0$, then $x^*=\min\{0,\frac{1+\epsilon}{2}\}=0$. Then $0\leq\min\{0,\frac{1+\epsilon}{2}\}=0<\epsilon$. Therefore $\inf A=0$.
    
7. Ex. 2.1.3 and 2.1.4 in our text. You must prove your answers, not just provide them. 
    * Is the sequence $\{\frac{(-1)^n}{2n}\}$ Convergent?
        *  The $\lim(\frac{(-1)^n}{2n})=0$
            *  Finding $n^*$:
            *   $\forall n\in{\mathbb N}$ $\lim(\frac{(-1)^n}{2n})=|X_n-L|=|\frac{(-1)^n}{2n}-0|=\frac{1}{2n}< \epsilon$. This Implies $2n \epsilon>1$. Therefore $n>\frac{1}{2 \epsilon}$ 
        *$\forall \epsilon >0.$ Let $n^*=\lfloor{ \frac{1}{2\epsilon}\rfloor}+1$. Then $\forall n\geq n^*$,  $|\frac{(-1)^n}{2n}-L|=|\frac{(-1)^n}{2n}-0|=\frac{1}{2n}\leq \frac{1}{2{n^*}}=\frac{1}{2\lfloor{ \frac{1}{2\epsilon}\rfloor}+2}< \epsilon$   Since the limit exists at an exact value, it is convergent
    * Is the sequence $\{2^{-n}\}%$ convergent?
        *  The $\lim(2^{-n})=0$  
            * Finding $n^*$:
            *  $\forall n\in{\mathbb N}$, $\lim(2^{-n})=|x_n -L|=|2^{-n}-0|=|\frac{1}{2^n}|=\frac{1}{2^n}<\epsilon.$
            * This implies $1<2^n\epsilon$. Then $\log_2(1)<\log(2^n\epsilon)=\log_2(2^n)+\log_2(\epsilon)=n+\log_2(\epsilon)$. This implies $0-\log_2(\epsilon)=\log_2(\epsilon)=\log_2(\frac{1}{\epsilon})<n$ 

        * Now $\forall \epsilon >0$, Let $n^*=\lfloor{\log_2(\frac{1}{\epsilon})\rfloor}+1$. Then $\forall n>n^*, |x_n-L|=|2^{-n}-0|=|2^{-n}|=\frac{1}{2^n}\leq \frac{1}{2^{n^*}}=\frac{1}{2^{\lfloor{\log_2(\frac{1}{\epsilon})\rfloor}+1}} <\epsilon$. Since the limit exists to an exact value, it is convergent.
8. Redo Week 1 HW if you want to (for half of the remaning points)
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**Week 4**


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1. Prove that any subsequence of an infinitely large sequence is infinitely large.
    2. Suppose $\{x_n\}$ is inf. Large. Then $\forall k\in{\mathbb R}$, $\exists n^* \in{\mathbb N}$ s.t. $\forall n\geq n^*$ $|x_n|>K$ Since a subsequence is infinite then $\exists n^{**} \in\{x_{n_i}\}$ s.t. $n^{**}\geq n^*$  Therefore $\{x_{n_i}\}$ is inf large

2. p. 50 Ex. 2.1.7
    *  Show that $\lim(x_n)=0$ iff $\lim(|x_n|)=0$    
        *  Suppose $\lim(x_n)=0$. Then $\forall \epsilon>0$ $\exists n^*$ s.t $\forall n\geq n^*$  $|x_n-0|=|x_n|=x^n<\epsilon$. However,$||x_n||= |x_n|=x_n$. This means $||x_n|-0|=|x_n-0|=|x_n|=x^n<\epsilon$ This implies that $\lim(|x_n|)=0$.
        *  Suppose $\lim(|x_n|)=0$. Then $\forall \epsilon>0$ $\exists n^*$ s.t $\forall n\geq n^*$ $||x_n||=|x_n|=x^n<\epsilon$ This implies that $\lim(x_n)=0$.
        *    Therefore $\lim(x_n)=0$ iff $\lim(|x_n|)=0$
    *  Find and example s.t $|x_n|$ converges, but $x_n$ diverges
        *  Consider $\{(-1)^n\}$. 
        *  The $\lim(|(-1)^n|)=1$ since $\forall \epsilon>0$ $\exists n^*$ s.t $\forall n\geq n^*$; $|x_n-L|=|{|(-1)^n|}-1|=|1-1|=0<\epsilon$. However, the $\lim((-1)^n)$ does not exists.
            *  Proof: Let $\lim(-1)^n$=a.
                *  Case 1: Let $a\neq 1$ and $a\neq -1$. Then there $\exists\epsilon$ s.t $\forall n^*$ $\exists n\geq n^*$  $|x_n-a|\geq \epsilon$
                    *  Note: $\epsilon$ can be $|a-1|$ or $|a+1|$. Let $\epsilon'=|a-1|$ and $\epsilon''=|a+1|$.  
                * Let $\epsilon=\frac{\min(\epsilon',\epsilon'')}{2}$. Then $\forall n^*$ $\exists n=n^*+1$ s.t. $|x_n-a|\geq.\epsilon=\frac{\min(\epsilon',\epsilon'')}{2}$. WLOG let $\min(\epsilon',\epsilon'')=\epsilon'$ Then $|x_n-a|\geq \epsilon=\frac{\min(\epsilon',\epsilon'')}{2}=\epsilon'=\frac{|a+1|}{2}$.  Therefore $\lim(-1)^n\neq a$.  
            * Case 2: Let $a=-1$.$\lim(-1)^n= -1$. Given any $\epsilon >0$, let $m=1$. Then $\forall n\geq m$ for $n,m\in{\mathbb N}$ be odd or even.
                * Let $n$ be even. Then $|(-1)^n-L|=|1-(-1)|=2<\epsilon$.  Consider $\epsilon =\frac{1}{2}$. Then $|(-1)^n-L|=|1-(-1)|=2<\frac{1}{2}$ Which is a contradiction.
            *   Case 3: Let $a=1$.$\lim(-1)^n= -1$. Given any $\epsilon >0$, let $m=1$. Then $\forall n\geq m$ for $n,m\in{\mathbb N}$ be odd or even.
                * Let $n$ be odd $|(-1)^n-L|=|1-(1)|=2<\epsilon$  Consider. $\epsilon =\frac{1}{2}$. Then $|(-1)^n-L|=|1-(-1)|=2<\frac{1}{2}$ Which is a contradiction.
     

3. p. 50 Ex. 2.1.10
    * Show that the sequence $\{\frac{n+1}{n}\}$ is monotone, bounded, and use thm 2.1.10 to find the limit.
        * (1) Monotone:  Let $x_{n^*}\in\{\frac{n+1}{n}\}$. Consider $n^{**}=n^*+k$. Then $\frac{(n^*+k)+1}{n^*+k}<\frac{n^*+1}{n}$ Then $\forall n>n^*$,   $\frac{n^*+1}{n^*}>\frac{n+1}{n}$. Therefore it is monotone decreasing.
        * (2) Bounded: Ler $K=2$ for $K\in{\mathbb R}$. Since $\sup(\{\frac{n+1}{n}\})=2$ and $\inf(\{\frac{n+1}{n}\})=1$, then $\forall n\in{\mathbb N}$ ,$|x_n|\leq |K|$.
            * Proof of Sup:  $\epsilon > 0$ let$n^*= {\lceil\epsilon\rceil}$. Then $2-\epsilon<\frac{n^*+1}{n^*}=\frac{\lceil\epsilon\rceil+1}{\lceil\epsilon\rceil}\leq 2$
            * Proof of Inf:$\epsilon > 0$ let $n^*=\frac{1}{\lfloor\epsilon\rfloor}+1.$ Then $1\leq\frac{n+1}{n}\leq \frac{n^*+1}{n^*}= \frac{\frac{1}{\lceil\epsilon\rceil}+2}{{\frac{1}{\lceil\epsilon\rceil}+1}}<1+\epsilon$
        * $\lim(\{\frac{n+1}{n}\})=1$. From previous proofs we know that $\{\frac{n+1}{n}\}$ is monotone decreasing and bounded. Therefore $\lim(\{\frac{n+1}{n}\})=\inf(\{\frac{n+1}{n}\})=1$ by thm 2.1.10.

4. p. 50 Ex. 2.1.13
    *  Let $\{x_n\}$ be a monotone convergent sequence. Supose there is $k\in{\mathbb N}$ s.t. $\lim(x_n)=x_k$ Show that $x_n=x_k ~\forall n\geq k$
        *  Contradiction: Suppose $\exists n^*\geq k$ s.t. $x_{n^*}\neq x_k$. Then $x_{n^*}>x_k$ or $x_{n^*}<x_k$.
            *  Case(1):$x_{n^*}>x_k$. Then$\{x_n\}\nearrow$, Then $\forall n\geq n^*$ $x_n>x_{n^*}$. However, by thm 2.1.10, $\lim({x_n})=\sup({x_n})=x_k$, but , $x_{n^*}>x_k$.Therefore, $\lim({x_n})=\sup({x_n})\neq x_k$. This is a contradiction on that $\lim(x_n)=x_k$.  
            * Case(2):$x_{n^*}<x_k$. Then$\{x_n\}\searrow$, Then $\forall n\geq n^*$ $x_n<x_{n^*}$. However, by thm 2.1.10, $\lim({x_n})=\inf({x_n})=x_k$, but , $x_{n^*}<x_k$.Therefore, $\lim({x_n})=\inf({x_n})\neq x_k$. This is a contradiction on that $\lim(x_n)=x_k$.
        * Therefore$x_n=x_k$

5. p. 50 Ex. 2.1.16
    *    Let $\{x_n\}$ be a sequence. Suppose there are 2 convergent subsequences $\{x_{n_i}\}$ and $\{x_{m_i}\}$ Suppose $\lim(x_{n_i})=a$ and $\lim(x_{m_i})=b$ Where $a\neq b$, show that $\{x_n\}$ is divergent without prop 2.1.17
            *  Let $\lim(\{x_n\})=L.$ Then $\forall \epsilon >0$ $\exists n^*\in{\mathbb N}$ s.t $\forall n\geq n^*$  $|X_n-L|<\epsilon$. However, $\{{x_{n_i}}\}$ is a subsequence, so $\exists n^{**}\in\{{x_{n_i}}\}$ s.t $\forall n\geq n^{**}$ $|x_{n_i}-L|<a$. Similiarly, since $\{{x_{m_i}}\}$ is a subsequence, so $\exists n^{***}\in\{{x_{m_i}}\}$ s.t $\forall n\geq n^{***}$ $|x_{m_i}-L|<b$ This implies $\lim(\{x_{m_i}\})=b=\lim(\{x_{n_i}\})=a=\lim(\{x_n\})=L.$ Therefore $a=b$. This is a contradiction. Therefore $\{x_n\}$ is divergent.
6. p. 50 Ex. 2.1.17
    * Find a sequence $\{x_n\}$, such that $\forall y\in{\mathbb R}$, $\exists \{x_{n_i}\}$ converging to $y$
    * Consider the decimal reprisenation of a number. Let $d=[0,9].$ Then $D_n={\frac{d_1}{10^1},\frac{d_2}{10^2},...,\frac{d_{n-1}}{10^{n-1}},{\frac{d_n}{10^n}}}$. Consider $\{D_n\}$. This contains every decimal expansion between $[0,1]$ Since we know that ${\mathbb Q}$ is denumerable and the decimal repricentation is not unique. $\exists \{D_{x_i}\} \in \{D_n\}$ s.t $\lim(\{D_{x_i}\})=y$ $\forall y\in [0,1]$ This can be expanded for and sequence with a whole number in front to include $y \forall y\in{\mathbb R}$.

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     **Week 7**


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1. Without using Lemma on nested Intervals prove that
$$\bigcap_{n=1}^\infty[-1/n, 1/n]=\{0\}.$$
    *    Let $\{a_n\}=\frac{-1}{n}$ and $\{b_n\}=\frac{1}{n}$ $\implies a_n\leq b_n~~~\forall n\in{\mathbb N}.$ Then $\lim a_n=\lim b_n=0\implies~~~\forall \epsilon>0~~\exists n^*$ s.t $\forall n \geq n^*~~~|a_n-0|=|b_n-0|<\epsilon$. 
Let  $n^*=\lfloor\frac{1}{\epsilon} \rfloor +1$ $\implies |a_n-0|=|b_n-0|=|\frac{-1}{n}|=|\frac{1}{n}|=\frac{1}{n}\leq \frac{1}{n^*}<\epsilon.$ 
Let $x_n\in\bigcap_{n=1}^\infty[-1/n, 1/n]$. $\implies x_n\in [a_n,b_n]~~~\forall n~~~\implies a_n\leq x_n\leq b_n\implies \lim x_n=0$ by the squeeze thm.  Now let $\{x_{n_k}\}\in[a_{n_k},b_{n_k}]$ s.t  $\sup a_{n_k}=\inf b_{n_k}=\lim a_n=\lim b_n=0$ Since $a_n$ snd $b_n$ converge to $0$.Then $\sup a_{n_k}\leq x_{n_k}\leq \inf a_{n_k} \implies$ $0\leq a_{n_k}\leq 0 \implies a_{n_k}=0 \implies  x_n\in\bigcap_{n=1}^\infty[-1/n, 1/n]={0}$.
2. Exercise 2.3.5
    a) let $x_n:= \frac{(-1)^n}{n}$ find $\lim\sup x_n$ and $\lim\inf x_n$ 
    Proof: $\lim x_n=0:$ $\forall \epsilon>0$ Let $n^*=\lfloor\frac{1}{\epsilon}\rfloor+1$ $\implies \forall n\geq n^*\in{\mathbb N}.$ $|\frac{(-1)^n}{n}-0|=|\frac{(-1)^n}  {n}|=\frac{1}{n}\leq \frac{1}{n^*}<\epsilon$ By thm. 2.3.5 $\lim\sup x_n=\lim\inf x_n=\lim x_n=0$.
    b) $x_n:= \frac{(n-1)(-1)^n}{n}$ find $\lim\sup x_n$ and $\lim\inf x_n$.
    * $\{x_n\}=\{0,\frac{-1}{2},.\frac{2}{3}\}\implies$ divergent but is bounded. Consider the two subsequenses:
    $$x_n=\begin{cases} x_{n_k},&\text{if $n$ is even}\\ x_{n_j},& \text{if $n$ is odd}\end{cases}$$
    Then $\{x_{n_k}\}=\{\frac{1}{2},\frac{3}{4},\frac{4}{6},...,\frac{2n_k-1}{2n_k}\}$ and $\{x_{n_j}\}=\{0,\frac{-2}{3},\frac{-4}{5},\frac{-6}{7},...,\frac{-2n_j+2}{2n_j-1}\}$ Since $\{x_{n_k}\}>0 \forall k$ and $\{x_{n_j}\}\leq0 \forall j$ and $\{x_n\}=\{x_{n_k}\}\cup\{x_{n_j}\}\implies$ $\lim x_{n_k}=\lim\sup x_n=1$ and $\lim x_{n_j}=\lim\inf x_n=-1$
    * $\lim\inf {x_n}=\lim x_{n_j}=-1$ Let $n_j^*=\lfloor\frac{1}{2\epsilon-1}\rfloor +1$ then $\forall \epsilon > 0$ and $\forall n_j\geq n_j^*$ $\implies|x_{n_j}+1|=|\frac{-2n_j+2}{2n_j-1}+1|=|\frac{1}{2n_j-1}|=\frac{1}{2n_j-1}\leq\frac{1}{2n_j^*-1}<\epsilon$
    *  $\lim\sup {x_n}=limx_{n_k}=1$ Let $n_k^*=\lfloor\frac{1}{2\epsilon}\rfloor +1$ then $\forall \epsilon > 0$ and $\forall n_k\geq n_j^*$ $\implies|x_{n_k}-1|=|\frac{2n_k-1}{2n_k}|=|\frac{1}{2n}|=\frac{1}{2n_k}\leq\frac{1}{2n_k^*}<\epsilon$  

3. Pove that every unbounded sequence, contains an infinitely large subsequesnce. The proof goes by construction.
    * Let $\{X_n\}$ be unbounded. $\implies\forall k\in{\mathbb R}~~\exists n^*$ s.t $|x_n^*|>|K|$. WLOG suppose $\{x_n\}$ is unbounded above.$\implies \forall K~~~\exists n^{**}$ s.t $x_{n^{**}}>K$
    Consider the $\{x_{n_k}\}=\{x_{n_1},x_{n_2},...,x_{n_{k-1}},x_{n_k}\}$
        1. $x_{n_1}={x_1}$
        1. $n_2$
        Note: ${n_2}>n_1 +1$ 
        Consider $K_1\in{\mathbb R}$ s.t. $K_1\geq x_{n_1}$ However $\{x_n\}$ is unbounded, $\implies\exists n_2^*$ s.t $x_{n^*_2}>K_1> x_{n_1}\implies x_{n^*_2}>x_{n_1}$.Let $x_{n_2}=x_{n^*_2}$
        1.   $n_3$
        Note: ${n_3}>n_2 +1$ 
        Consider $K_2\in{\mathbb R}$ s.t. $K_2\geq x_{n_2}$ However $\{x_n\}$ is unbounded, $\implies\exists n_3^*$ s.t $x_{n^*_3}>K_2> x_{n_2}\implies x_{n^*_3}>x_{n_2}$.Let $x_{n_3}=x_{n^*_3}$
        1. .$n_k$  
        Note: ${n_k}>n_{k-1} +1$ 
        Consider $K_{k-1}\in{\mathbb R}$ s.t. $K_{k-1}\geq x_{n_{k-1}}$ However $\{x_n\}$ is unbounded, $\implies\exists n_k^*$ s.t $x_{n^*_k}>K_{k-1}> x_{n{k-1}}\implies x_{n^*_k}>x_{n_{k-1}}$.Let $x_{n_k}=x_{n^*_k}$
    
    *  Now $\{x_{n_k}\}\nearrow$ by construction. 
    *    (proof of unboundness). Let $\{x_{n_k}\}$ be bounded.$\implies \forall j\in{\mathbb N}$ $\exists K_j\in{\mathbb R}$ s.t. $|K_j|\geq |x_{n_j}|$. However, by construction and $\{x_n\}$ is being unbounded $\exists {j^*}\in\{x_n\}\in\{x_{n_k}\}$ s.t $|{x_{n_j}}|\geq|K_j|$. Contradiction. $K_j$ is not a bound. Therefore $\{x_{n_k}\}$ is unbounded.
    *    $\{x_{n_k}\}$  is unbounded and $\{x_{n_k}\}\nearrow\implies$$\{x_{n_k}\}$ inf larg 

        
