# Random math: 03

Source of most content: Gunhee Cho

# Density in $\mathbb{R}$

## Density of $\mathbb{Q}$

$\text{For any }x, y \in \mathbb{R} \text{ with }x < y,\text{ there exists }r \in \mathbb{Q}\text{ s.t. }x < r < y. \nonumber$

### Proof

It is suffice to consider the case $0 < x < y$. Since $y-x > 0$, by A.P. with 1, $\exists n \in \mathbb{N}$ s.t. $n(y-x)>1$. Define $A := \{ k \in \mathbb{N} \mid nx < k \} \ne \emptyset \leadsto \emptyset \ne A \subseteq \mathbb{R}$. By W.O., there exists $\min A = m_0 \in A$, i.e., $nx < m_0$ but $nx \geq m_0 -1$. So $m_0 -1 \leq nx < m_0 \leadsto nx < m_0 \leq nx+1 < ny$, where the last inequality is from A.P. $\Leftrightarrow x < \frac{m_0}{n} < y$, which concludes the proof.

NB; $m_0$ was coming from W.O. which is an axiom about natural numbers and $n$ was coming from A.P. which is an axiom about real numbers.

## Density of $\mathbb{I}$

$\text{By density of }\mathbb{Q},\text{ it is clear that there exists } q \text{ s.t. }\frac{a}{\epsilon} < q < \frac{b}{\epsilon}\text{ where }a,b \in \mathbb{R}, \epsilon \in \mathbb{I}, q \in \mathbb{Q} \leadsto a < \epsilon q < b. \nonumber$

# Convergence of seq. in $\mathbb{R}$

Def. $(a_n)^\infty_{n=1} (\subseteq X)$, where $X$ is topological space, is a sequence if it is a function $(a_n)^\infty_{n=1}: \mathbb{N} \rightarrow X \ (n \mapsto a_n)$

Def. We say a sequence $( a_n)^\infty_{n=1} (\subseteq \mathbb{R})$ converges to $\alpha \in \mathbb{R}$ if for each $\epsilon > 0, \exists N_\epsilon \in \mathbb{N}$ s.t. for any $n \geq N_\epsilon$, $|a_n - \alpha |< \epsilon$. And denote $\lim_{n\rightarrow \infty}a_n=\alpha$, otherwise we say $(a_n)$ diverges.

NB; To check convergence, we first fix $\epsilon >0$. Then there exists corressponding $N_\epsilon \in \mathbb{N}$, i.e., depends on fixed $\epsilon$. And finally, we choose arbitrary fixed $n$ s.t. $n \geq N_\epsilon$.

Def. Sequence $(a_n)^\infty_{n=1} (\subseteq \mathbb{R})$ is bounded if there exists $M > 0$ s.t. for any $n \in \mathbb{N}, | a_n | \leq M$, where M does not depend on $n$.

NB1; Following statement is wrong: seq. $(a_n)$ is bounded if for any $n\in\mathbb{N}$, there exists $M>0$ s.t. $| a_n | \leq M$. Since $M$ depends on $n$, we can set $M=|a_n|$ for all $n$.
NB2; Boundeness is not a topological property.

## Proposition

$\text{If }(a_n)^\infty_{n=1} (\subseteq \mathbb{R})\text{ is convergent, then } (a_n)^\infty_{n=1}\text{ is bounded}. \nonumber$

### Proof

Let $\alpha = \lim a_n \in \mathbb{R}$, i.e., $a_n$ is convergent. Take $\epsilon = 1$, then $\exists N_1 \in \mathbb{N}$ s.t. for any $n \geq N_1, n\in \mathbb{N}, | a_n - \alpha | < \epsilon = 1$ $\Leftrightarrow -1 < a_n - \alpha < 1 \Leftrightarrow \alpha -1 < a_n < \alpha +1$. We can choose $M=\max \{ |\alpha+1 |, | a_1 |, \cdots, | a_{N_1 -1}| \}$.
Then, for any $n \in \mathbb{N}, | a_n| \leq M$.
NB; In general, boundedness does not mean convergence. (oscillation)

# Limit Theorem

Let $\lim_{n \rightarrow \infty}a_n = \alpha, \lim_{n \rightarrow \infty}b_n = \beta$. Then, following holds:

1. $\lim_n (a_n + b_n) = \alpha+ \beta$
2. $\lim_n(ka_n) = k\alpha$ for $k\in\mathbb{R}$
3. $\lim_n(a_n-b_n)= \alpha - \beta$
4. $\lim_n(a_n b_n)=\alpha\beta$
5. $\lim_n \frac{a_n}{b_n} = \frac{\alpha}{\beta}$, where $\forall n \in \mathbb{N}, \ b_n,\beta \ne 0$

## Proof

1. Let $c_n = a_n + b_n, \ \epsilon >0$ be fixed. Since $\lim_n a_n =\alpha, \ \lim_n b_n = \beta$, there exists $N_\epsilon, N_\epsilon’ \in \mathbb{N}$ s.t. for any $n_1 \geq N_\epsilon, \ n_2 \geq N_\epsilon’, \ |a_{n_1} - \alpha | < \frac{\epsilon}{2}, \ |b_{n_2} - \beta | < \frac{\epsilon}{2}$. Take $N = \max (N_\epsilon, N_\epsilon’)$, then $n \geq N, \ |c_n - (\alpha+\beta) | = | a_n - \alpha + b_n - \beta | \leq | a_n - \alpha | + | b_n - \beta | < \epsilon$.
2. Since $\lim_n a_n =\alpha$, there exists $N_\epsilon\in \mathbb{N}$ s.t. for any $n \geq N_\epsilon, \ |a_{n} - \alpha | < \frac{\epsilon}{|k|}$. $\leadsto | k a_n - k \alpha | \leq |k| |a_n - \alpha | \leq \epsilon$.
3. Similar to 1.
4. Let $c_n = a_n b_n$. Then, $| a_n b_n - \alpha\beta | = | (a_n-\alpha)b_n + \alpha b_n - \alpha \beta | = | (a_n - \alpha)b_n + \alpha (b_n - \beta) | \leq |a_n - \alpha | | b_n | + | b_n -\beta | |\alpha|$. Since $b_n$ is bounded, $\exists M \in \mathbb{R}, |b_n| \leq M \leadsto |a_n - \alpha | | b_n | + | b_n -\beta | |\alpha| \leq M |a_n - \alpha | + | b_n -\beta | |\alpha|$. By convergence of $a_n, b_n$, there exists $N_\epsilon, N_\epsilon’ \in \mathbb{N}$ s.t. for any $n_1 \geq N_\epsilon, \ n_2 \geq N_\epsilon’, \ |a_{n_1} - \alpha | < \frac{\epsilon}{2M}, \ |b_{n_2} - \beta | < \frac{\epsilon}{2|\alpha|}$, which concludes the proof by taking $N = \max (N_\epsilon, N_\epsilon’)$.
5. Fix $\epsilon > 0$ and take $\epsilon_1 = \frac{| \beta |}{2} >0$. Since $\lim_n b_n = \beta, \ \exists N_1 \in \mathbb{N}$ s.t. $\forall n \geq N_1, \ |b_n - \beta | < \epsilon_1$ $\leadsto \forall n \geq N_1, \ |\beta|- |b_n| \leq | b_n - \beta| < \epsilon_1 =\frac{| \beta |}{2} \Leftrightarrow \forall n \geq N_1, \ |b_n| > \frac{| \beta |}{2} \Leftrightarrow \forall n \geq N_1,\ \frac{1}{|b_n|} < \frac{2}{|\beta|}$. Next, take $\epsilon_2 = \frac{\epsilon}{2}\frac{|\beta|^2}{2|\beta|}, \ \epsilon_3 = \frac{\epsilon}{2}\frac{|\alpha|^2}{2|\alpha|}$. By convergence of $a_n, \ b_n \ \exists N_2,N_3 \in \mathbb{N}$ s.t. $\forall n \geq N_2, |a_n -\alpha | < \epsilon_2,$ $\forall n \geq N_3, |b_n -\beta | < \epsilon_3$. Take $N=\max (N_1,N_2, N_3)$. Then, $\forall n \geq N, | \frac{a_n}{b_n} - \frac{\alpha}{\beta} | = | \frac{\beta a_n - \alpha b_n}{\beta b_n} | = | \frac{\beta (a_n-\alpha) + \alpha (\beta - b_n)}{\beta b_n} |$$\leq \frac{2}{|\beta|^2} | \beta (a_n-\alpha) + \alpha (\beta - b_n) | \leq \frac{2}{|\beta|^2} | \beta | | a_n-\alpha | + \frac{2}{|\beta|^2} | \alpha| | b_n - \beta | \leq \frac{2}{|\beta|^2} | |\beta|\epsilon_2 + |\alpha|\epsilon_3 | = \epsilon$.