MATH 409 Lecture 6

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Lecture Slides

New Challenges

Challenge 5

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

Challenge 6

…

Limits of Sequences

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

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

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

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

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

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

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

Examples

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

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

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

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

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

(to be shown later)

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

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

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Properties

Theorem. The limit is unique.

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

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Theorem. Any convergent sequence $\left\{x_{n}\right\}$ 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 $\left\{n\right\}_{n\in \mathbb {N} }$ discussed above breaks this property.

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Theorem. Any subsequence converges to the same limit.

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

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

Divergence to Infinity

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

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

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

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

Limit Theorems

Squeeze Theorem

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

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

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

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

Comparison Theorem

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

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

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

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

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Convergence and Arithmetic Operations

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

quod erat demonstrandum