Most of my previous math courses have had some graphs or pictures to help explain the ideas and concepts. I have tried looking for visual representation of the concepts of Real analysis; however, I have not found much. Is there a good resource like "Proof without words" for real analysis?
The topics I have seen so far:
- The open internal $(0,1)$ is uncountable
- A nonempty set $E$ is at most countable if and only if there is a function $g$ from $\mathbb{N}$ onto $E$
Suppose that $A$ and $B$ are sets
i. If $A \subseteq B$ and $B$ is at most countable, then $A$ is at most countable
ii. If $A \subseteq B$ and $A$ is uncountable, then $B$ is uncountable.
iii. $\mathbb{R}$ is uncountable
Let $A_1, A_2, \ldots$ be at most countable sets.
i. Then $A_1 \times A_2$ is at most countable
ii. if $E = \bigcup\limits_{j=1}^{\infty} A_j := \bigcup\limits_{j\in\mathbb{N}} A_j := \{x:x\in A_j$ for some $j \in \mathbb{N} \}$ then $E$ is at most countable
The sets $\mathbb{Z}$ and $\mathbb{Q}$ are countable, but the set of irrationals is uncountable.
A sequence of real numbers $\{x_n\}$ is said to converge to a real number $a \in \mathbb{R}$ if and only if for every $\epsilon > 0 $ there is an $N \in \mathbb{N}$ such that $n \geq N$ implies $\vert x_n - a \vert < \epsilon$
A sequence can have at most one limit.
By a subsequence of a sequence $\{x_n\}_{N\in\mathbb{N}}$, we shall mean a sequence of the form $\{x_{n_k}\}_{k\in\mathbb{N}}$, where each $n_k \in \mathbb{N}$ and $n_1 < n_2 < \cdots$
If $\{x_n\}_{n\in\mathbb{N}}$ converges to $a$ and $\{x_{n_k}\}_{k\in\mathbb{N}}$ is any subsequence of $\{x_n\}_{n\in\mathbb{N}}$, then $x_{n_k}$ converges to $a$ as $k \to \infty$
Let $\{x_n\}$ be a sequence of real numbers
i. The sequence $\{x_n\}$ is said to be bounded above if and only if the set $\{x_n : n\in \mathbb{N}\}$ is bounded above.
ii. The sequence $\{x_n\}$ is said to be bounded below if and only if the set $\{x_n : n\in \mathbb{N}\}$ is bounded below
iii. $\{x_n\}$ is said to be bounded if and only if it is bounded both above and below.
Every convergent sequence is bounded
Suppose that $\{x_n\}, \{y_n\}$, and $\{w_n\}$ are real sequences.
i. If $x_n \to a$ and $y_n \to a$ (the SAME a) as $n \to \infty$, and if there is any $N_0 \in \mathbb{N}$ such that $x_n\leq w_n \leq y_n$ for $n \geq N_0$, then $w_n \to a$ as $n \to infty$
ii. If $x_n \to 0$ as $n\to \infty$ and $\{y_n\}$ is bounded, then $x_n y_n \to 0$ as $n\to \infty$
Let $E \subset \mathbb{R}$. If $E$ has a finite supremeum (respectively, a finite infimum), then there is a sequence $x_n \in E$ such that $x_n \to \sup E$ (respectively, a sequence $y_n \in E$ such that $y_n \to \inf E$) as $n \to \infty$
Suppose that $\{x_n\}$ and $\{y_n\}$ are real sequences and that $a \in \mathbb{R}$. if $\{x_n\}$ and $\{y_n\}$ are convergent, then
i. $\lim\limits_{n\to\infty} (x_n + y_n) = \lim\limits_{n\to\infty} x_n + \lim\limits_{n\to\infty} y_n$
ii. $\lim\limits_{n\to\infty} (\alpha x_n) = \alpha \lim\limits_{n\to\infty} x_n$
iii. $\lim\limits_{n\to\infty} (x_n y_n) = \left(\lim\limits_{n\to\infty} x_n \right)\left(\lim\limits_{n\to\infty} y_n \right)$
iv: If, in addition, $y_n \neq 0$ and $\lim\limits_{n\to\infty} y_n \neq 0$ then $\lim\limits_{n\to\infty} \frac{x_n}{y_n} = \frac{\lim\limits_{n\to\infty} x_n}{\lim\limits_{n\to\infty} y_n} $
Let $\{x_n\}$ be a sequence of real numbers
i. $\{x_n\}$ is said to diverge to $+\infty$ (notation: $x_n \to + \infty$ or $\lim\limits_{n\to\infty} x_n = + \infty$) if and only if for each $M \in \mathbb{R}$ there is an $N \in \mathbb{N}$ such that $n \geq N$ implies $x_n > M$
ii. $\{x_n\}$ is said to diverge to $-\infty$ (notation : $x_n \to -\infty$ or $\lim\limits_{n\to\infty} x_n =- \infty$) if and only if for each $M \in \mathbb{R}$ there is an $N \in \mathbb{N}$ such that $n \geq N$ implies $x_n < M$
Suppose that $\{x_n\}$ and $\{y_n\}$ are real sequences such that $x_n \to + \infty$ (respectively, $x_n \to -\infty$) as $n \to \infty$
i. If $y_n$ is bounded below (respectively, $y_n$ is bounded above), then $\lim\limits_{n\to\infty} (x_n + y_n) = +\infty$ (respectively, $\lim\limits_{n\to\infty} (x_n + y_n) = -\infty$ ).
ii. If $\alpha > 0$, then $\lim\limits_{n\to\infty} (\alpha x_n)= + \infty$ (respectively \lim\limits_{n\to\infty} (\alpha x_n) = - \infty$).
iii. If $y_n > M_0$ for some $M_0 > 0$ and all $n \in \mathbb{N}$, then $\lim\limits_{n\to\infty} (x_n y_n) = +\infty$ (respectively, $\lim\limits_{n\to\infty} (x_n y_n) = -\infty$).
iv. If $\{y_n\}$ is bounded and $x_n \neq 0$, then $\lim\limits_{n\to\infty} \frac{y_n}{x_n} = 0$
Let $\{x_n\}, \{y_n\}$ be real sequences and $\alpha, x , y $ be extended real numbers. If $x_n \to x$ and $y_n \to y$, as $n\to \infty$, then $\lim\limits_{n\to\infty}(x_n + y_n ) = x+y$ provided that the right side is not of the form $\infty - \infty$, and $\lim\limits_{n\to\infty} (\alpha x_n ) = \alpha x$, $\lim\limits_{n\to\infty} (x_n y_n) = xy$ provided that none of these products is of the form $0 \cdot \pm \infty$
Suppose that $\{x_n\}$ and $\{y_n\}$ are convergent sequences. If there is an $N_0 \in\mathbb{N}$ such that $x_n \leq y_n$ for $n \geq N_0$, then $\lim\limits_{n\to\infty} x_n \leq \lim\limits_{n\to\infty} y_n$ In particular, if $x_n \in [a,b]$ converges to some point $c$ then $c$ must belong to $[a, b]$
Let $\{x_n\}_{n\in \mathbb{N}}$ be a sequence of real numbers.
i. $\{x_n\}$ is said to be increasing (respectively, strictly increasing) if and only if $x_1 \leq x_2 \leq \cdots$ (respectively, $x_1 < x_2< \cdots$).
ii. $\{x_n\}$ is said to be decreasing (respectively, strictly decreasing), if and only if $x_1 \geq x_2 \geq \cdots$ (respectively, $x_1 > x_2> \cdots$).
iii. $\{x_n\}$ is said to be monotone if and only if it is either increasing or decreasing.
If $\{x_n\}$ is increasing and bounded above, or if $\{x_n\}$ is decreasing and bounded below, then $\{x_n\}$ converges to a finite limit.
A sequence of sets $\{I_n\}_{n\in\mathbb{N}}$ is said to be nested if and only if $I_1 \subseteq I_2 \subseteq \cdots$
If $\{I_n\}_{n\in\mathbb{N}}$ is a nested sequence of nonempty closed bounded intervals, then $E:=\bigcap_{n=1}^{\infty} I_n$ is nonempty. Moreover, if the lengths of these intervals satisfy $\vert I_n \vert \to 0$ as $n \to \infty$ then $E$ is a single point.
The Nested Interval Property might not hold if "closed" is omitted.
The Nested Interval Property might not hold if "bounded" is omitted.
Every bounded sequence of real numbers has a convergent subsequence.
A sequence of points $x_n \in \mathbb{R}$ is said to be Cauchy (in $\mathbb{R}$) if and only if for every $\epsilon >0$ there is an $N \in \mathbb{N}$ such that $n,m \geq N$ implies $\vert x_n - x_m \vert < \epsilon$
If $\{x_n\}$ is convergent, then $\{x_n\}$ is Cauchy
Let $\{x_n\} be a sequence of real numbers. Then ${x_n}$ is Cauchy if and only if ${x_n}$ converges (to some point in $\mathbb{R}$).
A sequence that satisfies $x_{n+1} - x_n \to 0$ is not necessarily Cauchy.
