108學年度｜微積分模組班｜Class 01

Note 1. Set Theory and Mapping

Date: 108/09/16 (Week 02)

Place: 新生教學樓 301

Part 1. Sets and Operations of Set

Definition 1.1. (Set)

An universal set (宇集) is the set we are working in, for example, when we deal with real valued continuous function, our universal set $$U = C^0(\mathbb{R},\mathbb{R}) = \{f \colon \mathbb{R} \to \mathbb{R} \mid f \text{ is continuous}\}$$; when we are dealing number theory, our universal set $U = \mathbb{Z}$; ... etc. We suppose in the following definition, we work in the universal set $U$.

A subset (子集) $A$ of a set $U$ is a collection of element in $U$, denote by $A \subset U$.

A power set (冪集) of a set $U$ is the collection of all subsets of $U$, denote by $2^U$.

The union (聯集) of subsets $A,B \subset U$ is defined by: $$A \cup B = \{x \in U \mid x \in A \text{ or } x \in B\}.$$

The intersection (交集) of subsets $A,B \subset U$ is defined by: $$A \cap B = \{x \in U \mid x \in A \text{ and } x \in B\}.$$

The complement (補集/餘集) of subset $A \subset U$ is defined by: $$A^\complement = \{x \in U \mid x \not \in A\}.$$

The relative complement of A in B for $A, B \subset U$ is defined by $$B \setminus A:=\{x \in B \mid x \not \in A\}=B \cap A^\complement.$$

The Cartesian product (笛卡爾積) of two set $A$, $B$ is defined by: $$A \times B = \{(x, y) \mid x \in A, y \in B\}.$$

Two set $A,B$ are equivalent if they have same elements, denote by $A = B$.

Example 1.2.

(Numbers) $\mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R} \subset \mathbb{C}$.

(Intervals) We use the following notation for intervals on real number $\mathbb{R}$:

$[a,b]=\{x \in \mathbb{R} \mid a \leqslant x \leqslant b\}$
$(a,b)=\{x \in \mathbb{R} \mid a < x < b\}$
$(a,b]=\{x \in \mathbb{R} \mid a < x \leqslant b\}$
$[a,b)=\{x \in \mathbb{R} \mid a \leqslant x < b\}$

(Functions) This example is better if you know the definition of continuous and differentiable (which will soon be introduced), we usually denote: $$C^0(\Omega,\mathbb{R}) := \{f \colon \Omega \to \mathbb{R} \mid f \text{ is continuous}\}$$ be the set of all continuous function of a subset $\Omega \subset \mathbb{R}$, namely, every $f \in C^0(\Omega,\mathbb{R})$ is a function continuous on $\Omega$. Moreover, we denote for $n \geqslant 1$ that: $$C^n(\Omega,\mathbb{R}) := \{f \colon \Omega \to \mathbb{R} \mid f \text{ is $n-$times differentiable and $f^{(n)}$ is continuous}\}$$, which is called the $n-$times continuously differentiable function. Moreover, for infinitely differentiable functions (such as $f(x) = \sin x$), we denote the collection of such by: $$C^\infty(\Omega,\mathbb{R}) := \{f \colon \Omega \to \mathbb{R} \mid f \text{ is infinitely differentiable on $\Omega$}\}$$

The product $\mathbb{R} \times \mathbb{R} = \{(x,y) \mid x,y \in \mathbb{R}\}$ is the coordinate plane, usually denote by $\mathbb{R}^2$. Similarly, we use the name $n$-dimensional Euclidean space for $\mathbb{R}^n$.

Remark 1.3.

For union of more than two sets, say, $A_1 \cup A_2 \cup \cdots \cup A_n$, we usually denote by: $$\bigcup_{i = 1}^n A_i := A_1 \cup A_2 \cup \cdots \cup A_n$$The notation is similar if the union is infinitely many.

For intersection of more than two sets, say, $A_1 \cap A_2 \cap \cdots \cap A_n$, we usually denote by: $$\bigcap_{i = 1}^n A_i := A_1 \cap A_2 \cap \cdots \cap A_n$$ The notation is similar if the intersection is infinitely many.

Example 1.4.

$\displaystyle \bigcap_{n = 1}^\infty \left( 0, \dfrac{1}{n}\right) = \varnothing$.

$\displaystyle \bigcup_{n = 1}^\infty (- \infty, n) = \mathbb{R}$.

Proposition 1.5. (De Morgan's Laws)

$\displaystyle \left( \bigcup_{i = 1}^n A_i \right)^\complement = \bigcap_{i = 1}^n A_i^\complement$.

$\displaystyle \left( \bigcap_{i = 1}^n A_i \right)^\complement = \bigcup_{i = 1}^n A_i^\complement$.

Proof. (點擊顯示)

This is actually really simple:
\begin{align*} x \in \left( \bigcup_{i = 1}^n A_i\right)^\complement &\Longleftrightarrow x \not \in \bigcup_{i = 1}^n A_i \\
&\Longleftrightarrow x \not \in A_i \text{ for every } 1 \leqslant i \leqslant n \\
&\Longleftrightarrow x \in A_i^\complement \text{ for every } 1 \leqslant i \leqslant n \\
&\Longleftrightarrow x \in \bigcap_{i = 1}^n A_i^\complement
\end{align*}

Exercise

Exercise 1.6.

Given universal set $U=\{1,2,3,4,5,6,7,8,9,10\}$ and $A=\{1,2,3\}, B=\{1,3,4,7\}, C=\{4,8,9\}$, find the following:

$A^\complement$、$B^\complement$、$C^\complement$.
$A \setminus B$、$B \setminus A$、$B \setminus C$、$C \setminus B$.
$A \cup B$、$A \cap B$、$B \cup C$、$B \cap C$.
$(A \cup B)^\complement$、$(A \cap B)^\complement$、$(B \cup C)^\complement$、$(B \cap C)^\complement$.
$A^\complement \cup B^\complement$、$A^\complement \cap B^\complement$、$B^\complement \cup C^\complement$、$B^\complement \cap C^\complement$.

Show the following:

$A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$.
$A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$.

Cardinality of a finite set (a set of finite element) $S$ is the number of elements of $S$, usually denote by $|S|$, $\# S$ or card$(S)$. Show the following (we assume $A$, $B$ are finite sets.):

$|A \cup B| = |A| + |B| - |A \cap B|$.
$|A \times B| = |A||B|$.
$|2^A| = 2^{|A|}$.

Part 2. Mapping / Function

Definition 2.1.

Domain of a function $f$ is the set of the "input" of the function, denote by Dom($f$).
Codomain of a function $f$ is the set where all its "output" are contained. (Note that under this definition, codomain of a function is not uniquely determined)
Range of a function $f$ is the set of all its "output", suppose $f$ has domain $A$, then the range is usually denote by $f(A)$.

Example 2.2.

The mapping $f \colon \mathbb{R} \to \mathbb{R}$ defined by $f(x) = x^2$ has domain $\mathbb{R}$ and codomain $\mathbb{R}$ (In this particular case), with range $f(\mathbb{R}) = \mathbb{R}_{\geqslant 0}$.
The mapping $f \colon \mathbb{C} \to \mathbb{C}$ defined by $f(x) = x \cdot i$ where $i = \sqrt{-1}$ has domain $\mathbb{C}$ and codomain $\mathbb{C}$, with range $f(\mathbb{C}) = \mathbb{C}$. (Why?)
The mapping $f \colon \mathbb{R}^2 \to \mathbb{R}$ defined by $f(x,y) = \sqrt{x^2 + y^2}$ has domain $\mathbb{R}^2$ and codomain $\mathbb{R}$, with range $f(\mathbb{R}) = \mathbb{R}_{\geqslant 0}$.

Definition 2.3.

A mapping / function $f \colon A \to B$ is said to be injective/one-to-one (單射/嵌射) if for all $x_1,x_2 \in A$ with $f(x_1)=f(x_2)$, we have $x_1=x_2$. We also use the term injection for injective map.
A mapping/function $f \colon A \to B$ is said to be surjective/onto (蓋射/滿射) if for all $y \in B$, there exist $x \in A$ such that $f(x)=y$. Equivalently speaking, the codomain and range coicides, $B = f(A)$. We also use the term surjection for surjective map.
A mapping/function $f \colon A \to B$ is said to be bijective (雙射/對射) if it is injective and surjective. We also use the term bijection for bijective map.

Example 2.4.

The mapping $f \colon \mathbb{R} \to \mathbb{R}$ defined by $f(x)=x^2$ is neither injective nor surjective, since $f(-1)=f(1)$, it is not one-to-one, moreover, there does not exist $x \in \mathbb{R}$ such that $f(x)=-1$, hence, not onto.
The identity mapping on a set $A$, $\text{Id}|_A \colon A \to A$ is defined by $\text{Id}|_A(a) = a$ for all $a \in A$, it is clearly a bijective mapping. As an example, the polynomial $f \colon \mathbb{R} \to \mathbb{R}$ defined by $f(x) = x$ is the identity mapping on $\mathbb{R}$.

Definition 2.5.

Suppose we have two functions $f \colon A \to B$ and $g \colon B \to C$, the composition of $f$ and $g$ is a function $h \colon A \to C$ defined by $h(a) = g(f(a))$ for all $a \in A$. In convention, we denote $h$ by $h = g \circ f$.
Suppose $f \colon A \to B$ a function, we say $g \colon B \to A$ is the \textbf{inverse function} of $f$ if $g \circ f = \text{Id}|_A$ and $f \circ g = \text{Id}|_B$. If such inverse function exist, we say $f$ is invertible.

Example 2.6.

Let $f \colon \mathbb{R} \to \mathbb{R}^2$ defined by $f(x) = (x^2,x^3)$ and $g \colon \mathbb{R}^2 \to \mathbb{R}$ defined by $g(x,y) = xy$, then $(g \circ f)(x) = g(f(x)) = g(x^2,x^3) = x^5$.
The function $f \colon \mathbb{R}_{\geqslant 0} \to \mathbb{R}_{\geqslant 0}$ defined by $f(x) = x^2$ is invertible, defined by $g(x) = \sqrt{x}$.

Proposition 2.7.

If $f \colon A \to B$ is invertible, then the inverse function is unique, usually denote by $f^{-1}$.
A bijective function is invertible.

Proof. Exercise.

Exercise 2.8.

Determine the natural domain of the following function, show that weather it is injective or not. (If possible, try to calculate the range of each function.

$f_1(x) = x$
$f_2(x) = x^2 + x$
$f_3(x) = \sqrt{1-x^2}$
$f_4(x) = \tan(x) + \sqrt{x^2 - 2}$
$f_5(x) = \log_3(x) - \dfrac{1}{x - 5}$
$f_6(x) = 2^x + 2^{-x}$
$f_7(x) = 3 \cdot 5^x + 7 \cdot 5^{-x}$

Show that if $f : A \rightarrow B$ is an injection and $g : B \rightarrow C$ is an injection, then $g\circ f : A \rightarrow C$ is an injection.
Show that if $f : A \rightarrow B$ is a surjection and $g : B \rightarrow C$ is a surjection, then $g\circ f : A \rightarrow C$ is a surjection.
Show that if $f : A \rightarrow B$ is a bijection and $g : B \rightarrow C$ is a bijection, then $g\circ f : A \rightarrow C$ is a bijection.
Show that if we have mappings $f : A \rightarrow B$ and $g : B \rightarrow C$, where $g\circ f : A \rightarrow C$ is an injection, then $f$ is a injection. Give an example where $g$ is not an injection but $g \circ f$ is.
Show that if we have mappings $f : A \rightarrow B$ and $g : B \rightarrow C$, where $g\circ f : A \rightarrow C$ is a surjection, then $g$ is a surjection. Give an example where $f$ is not a surjection but $g \circ f$ is.

Part 3. Limit of Sequence

Definition 3.1. (Limit)

A sequence of real numbers is a mapping $a \colon \mathbb{N} \to \mathbb{R}$ defined by $a(n) = a_n$, usually denote by $\{a_n\}_{n=1}^{\infty}$.
Given a sequence $\{a_n\}_{n=1}^{\infty}$, we say $a_n$ is convergent and \textbf{converges to $L$} if that for every $\epsilon > 0$, there exist $n_0 \in \mathbb{N}$ such that: $$|a_n - L|<\epsilon \text{ if }n \geqslant n_0$$ ,usually denote by $\displaystyle \lim_{n \to \infty}a_n = L$. $L$ is called the limit of $\{a_n\}_{n = 1}^\infty$.

Remark 3.2.

We use the term divergent to describe a sequence that is not convergent. (Can you write down the rigorous definition of divergent by the definition of convergent ?)

Example 3.3.

For the sequence $\left\{a_n = \dfrac{1}{n}\right\}_{n = 1}^{\infty}$, we have $\displaystyle \lim_{n \to \infty} a_n = 0$. Since for every $\epsilon > 0$, we may pick $n_0 = \left\lceil \dfrac{1}{\epsilon} \right \rceil + 1$, then for any $n \geqslant n_0$, we have $|a_n-0| = \dfrac{1}{n} < \epsilon$.
For the sequence $\left\{a_n = \dfrac{1}{2^n}\right\}_{n = 1}^{\infty}$, we have $\displaystyle \lim_{n \to \infty} a_n = 0$. Since for every $\epsilon > 0$, we may pick $n_0 = \max \left \{1, \log_2\left( \dfrac{1}{\epsilon}\right) + 1 \right \}$, then for any $n \geqslant n_0$, we have $|a_n-0| = \dfrac{1}{2^n} < \epsilon$.

Definition 3.4. (Bound)

A set $S \subset \mathbb{R}$ is said to be bounded above if there exist $U \in \mathbb{R}$ such that $s \leqslant M$ for all $s \in S$. In particular, a sequence $\{a_n\}_{n =1}^{\infty}$ is bounded above if there exist $U \in M$ such that $a_n \leqslant U$ for all $n \in N$.
A set $S \subset \mathbb{R}$ is said to be bounded below if there exist $L \in \mathbb{R}$ such that $s \geqslant L$ for all $s \in S$. In particular, a sequence $\{a_n\}_{n =1}^{\infty}$ is bounded below if there exist $L \in M$ such that $a_n \geqslant L$ for all $n \in \mathbb{N}$.
A set $S \subset \mathbb{R}$ is said to be bounded if it is both bounded above and below, equivalently speaking, that is that there exist $M \in \mathbb{R}^{+}$ such that $|s| \leqslant M$ for all $s \in S$.

Example 3.5.

A finite set $S \subset \mathbb{R}$ is always bounded, we may pick $M = \max\{|s| \mid s \in S\} + 1$.
A subset of a bounded set is bounded.
$\{(-1)^n\}_{n = 1}^\infty$ is bounded, we may pick $M = 2$.
$\{2^n\}_{n = 1}^{\infty}$ is unbounded.
$\mathbb{N}$, $\mathbb{Z}$, $\mathbb{Q}$ is unbounded.

Proposition 3.6.

A convergent sequence is always bounded.

Proof. (點擊顯示)

Suppose $\displaystyle \lim_{n \to \infty}a_n = L$. Given $\epsilon = 1$, there exist $n_0 \in \mathbb{N}$ such that $|a_n - L| < \epsilon = 1$ when $n \geqslant n_0$. Being precise, that is $|a_n| < N_1 = \max \{|L-1|,|L+1|\}$. Moreover, we have $|a_n| < N_2 = \max\{|a_n| \mid 1 \leqslant n < n_0\} + 1$, when $n < n_0$. Taking $N = \max\{N_1,N_2\}$, we have $|a_n| < N$ for all $n \in \mathbb{N}$, hence bounded.

Remark 3.7.

A bounded sequence is not necessarily a convergent sequence, for example, the sequence $\{a_n = (-1)^n\}_{n = 1}^\infty$ is bounded but not convergent.

Proposition 3.8.

Limit of a convergent sequence is unique.

Proof. (點擊顯示)

We ought to prove by contradiction. Suppose $\{a_n\}_{n = 1}^\infty$ is a convergent sequence, with $\displaystyle \lim_{n \to \infty} a_n = L$, $\displaystyle \lim_{n \to \infty} a_n = M$ and $L < M$. Now by definition of limit, picking $\epsilon = \dfrac{1}{2}(M-L)$, there exist $n_1, n_2 \in \mathbb{N}$ such that: $$|a_n - L|< \epsilon \text{ when }n \geqslant n_1$$ $$|a_n - M|< \epsilon \text{ when } n \geqslant n_2$$ Picking $n \geqslant n_0 = \max\{n_1,n_2\}$, we have both: $$|a_n - L|< \epsilon \Longleftrightarrow L - \epsilon < a_n < L + \epsilon \Longleftrightarrow \dfrac{3L-L}{2} < a_n < \dfrac{M+L}{2}$$ $$|a_n - M|< \epsilon \Longleftrightarrow M - \epsilon < a_n < M + \epsilon \Longleftrightarrow \dfrac{M+L}{2} < a_n < \dfrac{3M-L}{2}$$ which is a contradiction since $\left( \dfrac{3L-L}{2}, \dfrac{M+L}{2}\right) \cup \left( \dfrac{M+L}{2}, \dfrac{3M-L}{2}\right) = \varnothing$, so we must have $L = M$, limit of a convergent sequence is unique.

Exercise 3.9.

Suppose both sequence $\{a_n\}_{n = 1}^\infty$, $\{b_n\}_{n =1}^\infty$ converges, with $\displaystyle \lim_{n \to \infty} a_n = A$, $\displaystyle \lim_{n \to \infty}b_n = B$ and $a_n \leqslant b_n$ for all $n \in \mathbb{N}$, show that $A \leqslant B$.
(Squeeze theorem) Suppose both sequence $\{a_n\}_{n = 1}^\infty$, $\{b_n\}_{n =1}^\infty$ converges, with there limit coincide, that is, $\displaystyle \lim_{n \to \infty} a_n = \lim_{n \to \infty}b_n = L$. Moreover, $a_n \leqslant c_n \leqslant b_n$ for all $n \in \mathbb{N}$, then the sequence $\{c_n\}_{n = 1}^\infty$ converges and $\displaystyle \lim_{n \to \infty}c_n = L$.

Remark 3.10.

Try to show the two statement in Exercise 3.9 still holds if the condition "for all $n \in \mathbb{N}$" is weaken into "for all $n \geqslant n_0$ for some $n_0 \in \mathbb{N}$".

Proposition 3.11.

Suppose $\displaystyle \lim_{n \to \infty}a_n = A$ and $\displaystyle \lim_{n \to \infty}b_n = B$, then:

$\{c \cdot a_n\}_{n = 1}^{\infty}$ is convergent for any $c \in \mathbb{R}$, and $\displaystyle \lim_{n \to \infty}(c \cdot a_n) = c \cdot A$

$\{a_n + b_n\}_{n = 1}^{\infty}$ is convergent, and $\displaystyle \lim_{n \to \infty}(a_n + b_n) = A + B$

$\{a_n - b_n\}_{n = 1}^{\infty}$ is convergent, and $\displaystyle \lim_{n \to \infty}(a_n - b_n) = A - B$

$\{a_n \cdot b_n\}_{n = 1}^{\infty}$ is convergent, and $\displaystyle \lim_{n \to \infty}(a_n \cdot b_n) = A \cdot B$

Suppose $B \neq 0$ and $b_n \neq 0$ for all $n \in \mathbb{N}$, then $\displaystyle \left\{\dfrac{a_n}{b_n}\right\}_{n = 1}^{\infty}$ is convergent, and $\displaystyle \lim_{n \to \infty} \dfrac{a_n}{b_n} = \dfrac{A}{B}$

Proof. (點擊顯示)

Exercise.

Since $\displaystyle \lim_{n \to \infty}a_n = A$, we have that for any $\epsilon > 0$, there exist $n_1 \in \mathbb{N}$ such that: $$|a_n - A|<\dfrac{\epsilon}{2(|B|+1)} \text{ for all } n \geqslant n_1.$$ Moreover, since convergent sequence is bounded (Proposition 3.6), there exist $M>0$ such that $|a_n|<M$ for all $n \in \mathbb{N}$. Again, by the convergent $\displaystyle \lim_{n \to \infty}b_n = B$, we have that for any $\epsilon > 0$, there exist $n_2 \in \mathbb{N}$ such that: $$|b_n - B|<\dfrac{\epsilon}{2M} \text{ for all } n \geqslant n_2.$$ Let $n_0 = \max\{n_1,n_2\}$, then if $n \geqslant n_0$, we have: \begin{align*} |a_nb_n-AB| &= |a_nb_n - a_nB +a_nB -AB| \\ &=|a_n(b_n-B)+B(a_n-A)| \\ &\leqslant |a_n||b_n-B|+|B||a_n-A| \\ &<M \cdot |b_n-B|+|B||a_n-A| \\ &<M \cdot \dfrac{\epsilon}{2M} + |B| \cdot \dfrac{\epsilon}{2(|B|+1)} \\ &< \dfrac{\epsilon}{2} + \dfrac{\epsilon}{2} = \epsilon \end{align*} Hence the result $\displaystyle \lim_{n \to \infty}a_nb_n = AB$.

Exercise. (Hint: try to show that $\displaystyle \lim_{n \to \infty}\dfrac{1}{b_n} = \dfrac{1}{B}$ first and use 4.)]

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