# MATH 409 Lecture 6

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Lecture Slides

## New Challenges

### Challenge 5

Prove that the set ${\displaystyle \mathbb {R} \times \mathbb {R} }$ is of the same cardinality as ${\displaystyle \mathbb {R} }$.

## Limits of Sequences

A sequence of elements of a set ${\displaystyle X}$ is a function ${\displaystyle f:\mathbb {N} \to X}$.

Notation: ${\displaystyle x_{1},x_{2},\ldots }$, where ${\displaystyle x_{n}=f(n)}$. Alternatively, ${\displaystyle \left\{x_{n}\right\}_{n\in \mathbb {N} }}$ or simply ${\displaystyle \left\{x_{n}\right\}}$.

A sequence ${\displaystyle \left\{x_{n}\right\}}$ of real numbers is said to converge to a real number ${\displaystyle a}$ if for any ${\displaystyle \epsilon >0}$ there exists ${\displaystyle N\in \mathbb {N} }$ such that ${\displaystyle |x_{n}-a|<\epsilon }$ for all ${\displaystyle n\geq N}$.

The number ${\displaystyle a}$ is called the limit of ${\displaystyle \left\{x_{n}\right\}}$.

Notation: ${\displaystyle \lim _{n\to \infty }x_{n}=a}$, or ${\displaystyle x_{n}\to a}$ as ${\displaystyle n\to \infty }$.

A sequence is called convergent if it has a limit and divergent otherwise.

The condition ${\displaystyle |x_{n}-a|<\epsilon }$ is equivalent to ${\displaystyle a-\epsilon or to ${\displaystyle x\in (a-\epsilon ,a+\epsilon )}$. This interval is called the ε-neighborhood of the point ${\displaystyle a}$. The convergence ${\displaystyle x_{n}\to a}$ means that any ε-neighborhood of ${\displaystyle a}$ contains all but finitely many elements of the sequence ${\displaystyle \left\{x_{n}\right\}}$.

### Examples

Theorem. The sequence ${\displaystyle \left\{{\frac {1}{n}}\right\}_{n\in \mathbb {N} }}$ converges to 0.

Proof. By the Archimedean Principle, for any ${\displaystyle \epsilon >0}$ there exists a natural number ${\displaystyle N}$ such that ${\displaystyle N\,\epsilon >1}$. Then for any ${\displaystyle n\geq N}$, we have ${\displaystyle n\,\epsilon \geq n\,\epsilon >1}$ so that (dividing by ${\displaystyle n}$) ${\displaystyle {\frac {1}{n}}<\epsilon }$. Since ${\displaystyle {\frac {1}{n}}>0}$, we obtain ${\displaystyle \left|{\frac {1}{n}}\right|<\epsilon }$ for ${\displaystyle n\geq N}$.

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Theorem. The constant sequence ${\displaystyle \left\{x_{n}\right\}}$, where ${\displaystyle x_{n}=a}$ for some ${\displaystyle a\in \mathbb {R} }$ and all ${\displaystyle n\in \mathbb {N} }$.

Proof. Since ${\displaystyle |x_{n}-a|=0}$ for all ${\displaystyle n\in \mathbb {N} }$, the sequence converges to ${\displaystyle a}$.

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Theorem. The sequence ${\displaystyle \left\{\left(-1\right)^{n}\right\}_{n\in \mathbb {N} }}$ is divergent.

(to be shown later)

Theorem. The sequence ${\displaystyle \left\{n\right\}_{n\in \mathbb {N} }}$ is divergent.

Proof. By the Archimedean Principle above, ${\displaystyle {\frac {1}{n}}}$ gets small, so its inverse must get large.

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### Properties

Theorem. The limit is unique.

Proof. Suppose ${\displaystyle a}$ and ${\displaystyle b}$ are distinct real numbers. Let ${\displaystyle \epsilon ={\frac {\left|b-a\right|}{2}}}$. Then the ε-neighborhoods of ${\displaystyle a}$ and ${\displaystyle b}$ are disjoint. Hence they cannot both contain all but finitely many elements of the same sequence.

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Theorem. Any convergent sequence ${\displaystyle \left\{x_{n}\right\}}$ is bounded, which means that the set of its elements is bounded.

Proof. This follows from three facts:

1. Any ε-neighborhood is bounded,
2. any finite set is bounded, and
3. the union of two bounded sets is also bounded.

Take the first set to be the one that is bounded in the ε-neighborhood, and the remaining elements outside the ε-neighborhood is finite. Thus the union of these two sets is the sequence, which must also be bounded.

The sequence ${\displaystyle \left\{n\right\}_{n\in \mathbb {N} }}$ discussed above breaks this property.

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Theorem. Any subsequence converges to the same limit.

Proof. Here a subsequence of a sequence ${\displaystyle \left\{x_{n}\right\}}$ is any sequence of the form ${\displaystyle \left\{x_{n_{k}}\right\}}$, where ${\displaystyle \left\{n_{k}\right\}}$ is an increasing sequence of natural numbers. Note that ${\displaystyle n_{k}\geq k}$. The proof follows easily by induction.

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The sequence ${\displaystyle \left\{\left(-1\right)^{n}\right\}}$ discussed above breaks this property.

### Divergence to Infinity

A sequence ${\displaystyle \left\{x_{n}\right\}}$ is said to diverge to infinity if for any ${\displaystyle C>0}$ there exists ${\displaystyle N\in \mathbb {N} }$ such that ${\displaystyle \left|x_{n}\right|>C}$ for all ${\displaystyle n\geq N}$.

Observe that such a sequence is indeed divergent as it is not bounded: given an interval, we can always get out of the interval and never turn back.

A sequence ${\displaystyle \left\{x_{n}\right\}}$ is said to diverge to positive infinity (${\displaystyle +\infty }$) if for any ${\displaystyle C\in \mathbb {R} }$ there exists ${\displaystyle N\in \mathbb {N} }$ such that ${\displaystyle x_{n}>C}$ for all ${\displaystyle n\geq N}$.

Likewise ${\displaystyle \left\{x_{n}\right\}}$ is said to diverge to negative infinity (${\displaystyle -\infty }$) if for any ${\displaystyle C\in \mathbb {R} }$ there exists ${\displaystyle N\in \mathbb {N} }$ such that ${\displaystyle x_{n} for all ${\displaystyle n\geq N}$.

For example, the sequence ${\displaystyle \left\{n\right\}_{n\in \mathbb {N} }}$ diverges to positive infinity.

## Limit Theorems

### Squeeze Theorem

Theorem. [Squeeze Theorem]. Suppose ${\displaystyle \left\{x-n\right\}}$, ${\displaystyle \left\{y_{n}\right\}}$, and ${\displaystyle \left\{w_{n}\right\}}$ are three sequences of real numbers such that ${\displaystyle x_{n}\leq w_{n}\leq y_{n}}$ for all sufficiently large ${\displaystyle n}$.

If the sequences ${\displaystyle \left\{x_{n}\right\}}$ and ${\displaystyle \left\{y_{n}\right\}}$ both converge to the same limit ${\displaystyle a}$, then ${\displaystyle \left\{w_{n}\right\}}$ converges to ${\displaystyle a}$ as well.

Proof. Since ${\displaystyle \lim _{n\to \infty }x_{n}=\lim _{n\to \infty }y_{n}=a}$, for any ${\displaystyle \epsilon >0}$ there exist natural numbers ${\displaystyle N_{1}}$ and ${\displaystyle N_{2}}$ such that ${\displaystyle a-\epsilon for all ${\displaystyle n\geq N_{1}}$ and ${\displaystyle a-\epsilon for all ${\displaystyle n\geq N_{2}}$. Besides, there exists ${\displaystyle N_{0}\in \mathbb {N} }$ such that ${\displaystyle x_{n}\leq w_{n}\leq y_{n}}$ for all ${\displaystyle n\geq N_{0}}$. We can choose ${\displaystyle N}$ to be the max of ${\displaystyle N_{0}}$, ${\displaystyle N_{1}}$, and ${\displaystyle N_{2}}$. Then for any natural number ${\displaystyle n\geq N}$, we have ${\displaystyle a-\epsilon , which implies that ${\displaystyle a-\epsilon . Thus ${\displaystyle \lim _{n\to \infty }w_{n}=a}$.

quod erat demonstrandum
Note: alternate name: theorem of two police men and a drunk guy

### Comparison Theorem

Theorem. [Comparison Theorem]. Suppose ${\displaystyle \left\{x_{n}\right\}}$ and ${\displaystyle \left\{y_{n}\right\}}$ are convergent sequences. If ${\displaystyle x_{n}\leq y_{n}}$ for all sufficiently large ${\displaystyle n}$, then ${\displaystyle \lim _{n\to \infty }x_{n}\leq \lim _{n\to \infty }y_{n}}$.

Proof. Let ${\displaystyle a=\lim _{n\to \infty }x_{n}}$ and ${\displaystyle b=\lim _{n\to \infty }y_{n}}$. Assume the contrary that ${\displaystyle a>b}$. Then ${\displaystyle \epsilon ={\frac {a-b}{2}}}$ is a positive number. Hence there exists a natural number ${\displaystyle N}$ such that ${\displaystyle |x_{n}-a|<\epsilon }$ and ${\displaystyle |y_{n}-b|<\epsilon }$ for all ${\displaystyle n\geq N}$. In particular, ${\displaystyle y_{n} and ${\displaystyle a-\epsilon for ${\displaystyle n\geq N}$. However, ${\displaystyle x+\epsilon =a-\epsilon ={\frac {a+b}{2}}}$, which implies that ${\displaystyle y_{n} for all ${\displaystyle n\geq N}$, a contradiction.

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Corollary. If all elements of a convergent sequence ${\displaystyle \left\{x_{n}\right\}}$ belong to a closed interval ${\displaystyle \left[a,b\right]}$, then the limit belongs to ${\displaystyle \left[a,b\right]}$ as well.

Proof. This follows from two applecations of the previous theorem. let ${\displaystyle \left\{y_{n}\right\}}$ be the constant sequence ${\displaystyle \left\{b\right\}}$ and compare with ${\displaystyle \left\{x_{n}\right\}}$. If all elements satisfy ${\displaystyle x_{n}, then the limits must follow the same pattern. Reverse this step with ${\displaystyle \left\{y_{n}\right\}=\left\{x_{n}\right\}}$ and ${\displaystyle \left\{x_{n}\right\}=\left\{a\right\}}$.

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### Convergence and Arithmetic Operations

Theorem. [Addition]. Suppose ${\displaystyle \left\{x_{n}\right\}}$ and ${\displaystyle \left\{y_{n}\right\}}$ are convergent sequences of real numbers. Then the sequences ${\displaystyle \left\{x_{n}+y_{n}\right\}}$ and ${\displaystyle \left\{x_{n}-y_{n}\right\}}$ are also convergent.

Moreover, if ${\displaystyle a=\lim _{n\to \infty }x_{n}}$ and ${\displaystyle b=\lim _{n\to \infty }y_{n}}$, then ${\displaystyle \lim _{n\to \infty }\left(x_{n}+y_{n}\right)=a+b}$ and ${\displaystyle \lim _{n\to \infty }\left(x_{n}-y_{n}\right)=a-b}$

Proof. Since ${\displaystyle \lim _{n\to \infty }x_{n}=a}$ and ${\displaystyle \lim _{n\to \infty }y_{n}=b}$, for any ${\displaystyle \epsilon >0}$ there exists a natural number ${\displaystyle N}$ such that ${\displaystyle \left|x_{n}a\right|<{\frac {\epsilon }{2}}}$ and ${\displaystyle \left|y_{n}-b\right|<{\frac {\epsilon }{2}}}$ for all ${\displaystyle n\geq N}$. Then for any ${\displaystyle n\geq N}$ we obtain

{\displaystyle {\begin{aligned}\left|\left(x_{n}+y_{n}\right)-\left(a+b\right)\right|&=\left|\left(x_{n}-a\right)+\left(y_{n}-b\right)\right|\\&\leq \left|x_{n}-a\right|+\left|y_{n}-b\right|\\&<{\frac {\epsilon }{2}}+{\frac {\epsilon }{2}}=\epsilon \\\left|\left(x_{n}-y_{n}\right)-\left(a-b\right)\right|&=\left|\left(x_{n}-a\right)+\left(b-y_{n}\right)\right|\\&\leq \left|x_{n}-a\right|+\left|b-y_{n}\right|=\left|x_{n}-a\right|+\left|y_{n}-b\right|\\&<{\frac {\epsilon }{2}}+{\frac {\epsilon }{2}}=\epsilon \end{aligned}}}.

Thus ${\displaystyle x_{n}+y_{n}\to a+b}$ and ${\displaystyle x_{n}-y_{n}\to a-b}$ as ${\displaystyle n\to \infty }$.

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Theorem. [Products]. Suppose ${\displaystyle \left\{x_{n}\right\}}$ and ${\displaystyle \left\{y_{n}\right\}}$are convergent sequences of real numbers. Then the sequence ${\displaystyle \left\{x_{n}\,y_{n}\right\}}$ is also convergent.

Moreover, if ${\displaystyle a=\lim _{n\to \infty }x_{n}}$ and ${\displaystyle b=\lim _{n\to \infty }y_{n}}$, then ${\displaystyle \lim _{n\to \infty }x_{n}\,y_{n}=a\,b}$.

Proof. Since ${\displaystyle x_{n}\to a}$ and ${\displaystyle y_{n}\to b}$ for ${\displaystyle n\to \infty }$, for any ${\displaystyle \delta >0}$ there exists ${\displaystyle N(\delta )\in \mathbb {N} }$ such that ${\displaystyle \left|x_{n}-a\right|<\delta }$ and ${\displaystyle \left|y_{n}-b\right|<\delta }$ for all ${\displaystyle n\geq \mathbb {N} (\delta )}$. Then for any ${\displaystyle n\geq N(\delta )}$ we obtain

{\displaystyle {\begin{aligned}\left|x_{n}\,y_{n}-a\,b\right|&=\left|x_{n}\,y_{n}-a\,y_{n}+a\,y_{n}-a\,b\right|\\&=\left|\left(x_{n}-a\right)\,y_{n}+a\,\left(y_{n}-b\right)\right|\\&=\left|\left(x_{n}-a\right)\,y_{n}-\left(x_{n}-a\right)\,b+\left(x_{n}-a\right)\,b+a\,\left(y_{n}-b\right)\right|\\&=\left|\left(x_{n}-a\right)\,\left(y_{n}-b\right)+\left(x_{n}-a\right)\,b+a\,\left(y_{n}-b\right)\right|\\&\leq \left|\left(x_{n}-a\right)\left(y_{n}-b\right)\right|+\left|\left(x_{n}-a\right)\,b\right|+\left|a\,\left(y_{n}-b\right)\right|\\&=\left|x_{n}-a\right|\,\left|y_{n}-b\right|+|b|\,|x_{n}-a|+|a|\,|y_{n}-b|\\&<\delta ^{2}+(|a|+|b|)\,\delta \end{aligned}}}

Now, given ${\displaystyle \epsilon >0}$, we set ${\displaystyle \delta =\mathrm {min} \left(1,\left(1+|a|+|b|\right)^{-1}\,\epsilon \right)}$. Then ${\displaystyle \delta >0}$ and ${\displaystyle \delta ^{2}+(|a|+|b|)\,\delta \leq (1+|a|+|b|)\,\delta \leq \epsilon }$. By the above, ${\displaystyle |x_{n}\,y_{n}-a\,b|<\epsilon }$ for all ${\displaystyle n\geq N(\delta )}$.

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Theorem. [Quotients]. Suppose that a sequence ${\displaystyle \left\{x_{n}\right\}}$ converges to some ${\displaystyle a\in \mathbb {R} }$. If ${\displaystyle a\neq 0}$ and ${\displaystyle x_{n}\neq 0}$ for all ${\displaystyle n\in \mathbb {N} }$, then the sequence ${\displaystyle \left\{{x_{n}}^{-1}\right\}}$ convergest to ${\displaystyle a^{-1}}$.

Proof. Since ${\displaystyle x_{n}\to a}$ as ${\displaystyle n\to \infty }$, for any ${\displaystyle \delta >0}$ there exists ${\displaystyle N(\delta )\in \mathbb {N} }$ such that ${\displaystyle |x_{n}-a|<\delta }$ for all ${\displaystyle n\geq N(\delta )}$.

Given ${\displaystyle \epsilon >0}$, we set ${\displaystyle \delta =\mathrm {min} \left({\frac {|a|}{2}},{\frac {|a|^{2}\,\epsilon }{2}}\right)}$. Then for any ${\displaystyle n\geq N(\delta )}$ we have ${\displaystyle |x_{n}-a|<{\frac {|a|}{2}}}$. Since

${\displaystyle |a|\leq |a_{n}-x_{n}|+|x_{n}|=|x_{n}-a|+|x_{n}|}$

it follows that ${\displaystyle |x_{n}|\geq |a|-|x_{n}-a|>|a|-{\frac {|a|}{2}}={\frac {|a|}{2}}}$

Furthermore, for any ${\displaystyle n\geq N(\delta )}$, we obtain

${\displaystyle \left|{\frac {1}{x_{n}}}-{\frac {1}{a}}\right|=\left|{\frac {a-x_{n}}{a\,x_{n}}}\right|={\frac {|x_{n}-a|}{|a|\,|x_{n}|}}\leq {\frac {2|x_{n}-a|}{|a|^{2}}}<{\frac {2\delta }{|a|^{2}}}\leq \epsilon }$

We conclude that ${\displaystyle {\frac {1}{x_{n}}}\to {\frac {1}{a}}}$ as ${\displaystyle n\to \infty }$.

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Corollary 1. If ${\displaystyle \lim _{n\to \infty }x_{n}=a}$ then ${\displaystyle \lim _{n\to \infty }c\,x_{n}=c\,a}$ for any ${\displaystyle c\in \mathbb {R} }$.

Proof. Use the multiplication theorem with constant sequence ${\displaystyle \left\{c\right\}}$.

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Corollary 2. If ${\displaystyle \lim _{n\to \infty }x_{n}=a}$ then ${\displaystyle \lim _{n\to \infty }(-x_{n})=-a}$.

Proof. There are two ways: take above ${\displaystyle c=-1}$, or use difference of sequnces with constant sequence ${\displaystyle \left\{0\right\}}$.

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Corollary 3. If ${\displaystyle \lim _{n\to \infty }x_{n}=a}$, ${\displaystyle \lim _{n\to \infty }y_{n}=b}$, and, moreover, ${\displaystyle b\neq 0}$ and ${\displaystyle y_{n}\neq 0}$ for all ${\displaystyle n\in \mathbb {N} }$, then ${\displaystyle \lim _{n\to \infty }{\frac {x_{n}}{y_{n}}}={\frac {a}{b}}}$.

Proof. Since ${\displaystyle b\neq 0}$ and ${\displaystyle y_{n}\neq 0}$ for all ${\displaystyle n\in \mathbb {N} }$, it follows that ${\displaystyle y_{n}^{-1}\to b^{-1}}$ as ${\displaystyle n\to \infty }$. Since ${\displaystyle {\frac {x_{n}}{y_{n}}}=x_{n}\,y_{n}^{-1}}$ for all ${\displaystyle n\in \mathbb {N} }$, the proposition follows from the theorem regarding multiplication.

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