11.25 Total Order
Proposition
If \(B\) has a supremum (resp. infimum), then it is unique.
Proof
Suppose that \(a\) and \(a'\) are both supremum for \(B\).
- \(a \leq a'\) because \(a\) is a supremum and \(a'\) is an upper bound,
- \(a' \leq a\) because \(a'\) is a supremum and \(a\) is an upper bound.
Therefore, \(a = a'\).
From this proposition, we denote \(\sup B = \text{supremum of } B\), \(\inf B = \text{infimum of } B\).
Proposition
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If \(a\) is a maximum for \(B\), then \(a = \sup B\).
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If \(a\) is a minimum for \(B\), then \(a = \inf B\).
Exercise.
Some Definitions
Total order
Let \((A, \leq)\) be a poset. We say that \(\leq\) is a total order if \(\forall x, y \in A\), we have \(x \leq y\) or \(y \leq x\).
In other words, a relation \(\leq\) on \(A\) is a total order if:
- \(\leq\) is reflexive.
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\(\leq\) satisfies trichotomy: \(\forall x, y \in A\), one and only one of the following conditions holds:
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\(x < y\),
- \(x = y\),
- \(x > y\).
- \(\leq\) is transitive.
Example
The usual order on \(\mathbb{R}\) is a total order.
Non-example
\(\mid\) on \(\mathbb{N}\) is not a total order because \(2 \nmid 3\), \(3 \nmid 2\), \(3 \neq 2\).
Notation
A set \(A\) with a total order is called an ordered set.
Example
-
Let \(A = \mathbb{R}^2\) and \(\leq\) is defined as:
\((x,y)\leq(x^{\prime},y^{\prime})\text{ if }(x<x^{\prime})\text{ or }(x=x^{\prime}\text{ and }y\leq y^{\prime}).\)
This is known as the lexicographical order.
-
\(\leq\) is a total order.
Exercise
Well-Ordered
An ordered set \((A, \leq)\) is called well-ordered if any non-empty subset \(B \subseteq A\) has a minimum (or first element).
Example: \(\mathbb{N}\) with the usual order is a well-ordered set.
Theorem
Any set \(A\) admits a total order \(\leq\) such that \((A, \leq)\) is well-ordered.
Counterexamples:
- \((0, 1)\) does not have a first element.
- \(\mathbb{R}\) with the usual order does not have a first element.
Function
Definition
Given two sets \(A, B\), a function from \(A\) to \(B\) is a relation \(f \subseteq A \times B\) such that:
- \(\text{Dom } f = A\),
- \(x f y \land x f y' \implies y = y'\).
In other words, a function is a relation such that any \(x \in A\) is related to one and only one element of \(B\).
Non-example
The diagram shows a relation that is not a function from \(A\) to \(B\) because some elements of \(A\) are related to multiple elements of \(B\).
Notation
If \(f\) is a function from \(A\) to \(B\), then we write \(f: A \to B\) or \(A \xrightarrow{f} B\).
If \(x \in A\), \(y \in B\), and \(f(x) = y\), then \(y\) is called the value of \(f\) at \(x\).
More notation
- \(A =\) domain of \(f = \text{Dom } f\).
- \(B =\) codomain of \(f = \text{Codom } f\).
- \(\text{Rg } f =\) image of \(f = \text{Im } f = \{ y \in B : \exists x \text{ s.t. } f(x) = y \}\).
Given \(y \in B\), the preimage of \(y\) is the set: \(f^{-1}(y)=\{x\in A:f(x)=y\}.\)
Example
Let \(A = \{1, 2, 3, 4\}\) and \(B = \mathbb{R}\).
-
This is a function, where:
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\(\text{Dom } f = A\),
- \(f(1) = 2\), \(f(2) = 1\), \(f(3) = \pi\), \(f(4) = 1\),
- \(\text{Codom } f = \mathbb{R}\),
- \(\text{Im } f = \{2, 1, \pi\}\).
For preimages:
- \(f^{-1}(1) = \{2, 4\} \subseteq A\),
- \(f^{-1}(2) = \{1\}\),
- \(f^{-1}(\pi) = \{3\}\),
- \(f^{-1}(y)=\emptyset\ \forall y\notin\{1,2,\pi\}\).
Remark
Some relations are "close" to being functions.
Example 1: \(\{(x, \frac{1}{x}) : x \neq 0\} \subseteq \mathbb{R} \times \mathbb{R}\).
- This is not a function from \(\mathbb{R}\) to \(\mathbb{R}\) because the domain \(\neq \mathbb{R}\).
- However, it is a function from \(\mathbb{R} \setminus \{0\} \to \mathbb{R}\).
Example 2: \(\{(x^2, x) : x \in \mathbb{R} \geq 0\}\).
- This is a relation from \(\mathbb{R}_{\geq 0}\) to \(\mathbb{R}\), but it is not a function from \(\mathbb{R}_{\geq 0}\) to \(\mathbb{R}\) because \(1R1\) and \(1R-1\).
- However, it is a function from \(\mathbb{R}_{\geq 0} \to \mathbb{R}_{\geq 0}\), defined as \(f(x) = \sqrt{x}\).
Remark
To define a function, we need to tell:
- \(A\) and \(B\)
- What is \(f(x)\) for any \(x \in A\).
Example: \(f(n) = n^2\) is a function from \(\mathbb{N} \to \mathbb{N}\).
Remark:
Sometimes, we only give \(f(x)\) and deduce \(A\) and \(B\) from the context.
Example: \(f(x) = \frac{1}{x}\)
Remark:
When are two functions \(A \to B\) and \(A' \to B'\) the same?
- When \(A = A'\), \(B = B'\) and \(f(x) = f'(x)\) for all \(x \in A\).
Example: \(f: \mathbb{N} \to \mathbb{N}\), \(f(n) = n^2\). \(g: \mathbb{N} \to \mathbb{N}\), \(g(n) = \sum_{k=1}^{n} (2k-1)\). By induction, we can prove that \(g(n) = g(n)\) for all \(k \in \mathbb{N}\), so \(f = g\).
SOME SPECIAL FUNCTIONS.
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The identity function of a set \(A\): \(\text{Id}_{A}:A\to A.\quad\text{Id}_{A}(x)=x\quad\forall x\in A\)
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Let \(A \subseteq U\) be a subset. The inclusion is the function \(i: A \to U\), where \(i(x) = x \quad \forall x \in A\).
- Characteristic function of a subset.
Let \(A \subseteq U\) be a subset. We define \(\chi_{A}:U\to\{0,1\},\quad\chi_{A}(x)=\begin{cases}1 & \text{if }x\in A\\ 0 & \text{if }x\in U-A\end{cases}\)
* Let \(U = \mathbb{R}\), \(A = [0,1] \cup [2,3]\).
Given a partition \(\mathcal{P} = \{A_i: i \in I\}\) of a set \(A\), and given \(b_i \in B\) for each \(i\), we define a function \(f: A \to B\) as \(f(x) = b_i \quad \text{if} \quad x \in A_i\)
* Let \(\mathbb{R} = \bigcup_{n \in \mathbb{Z}} [n, n+1)\) and define \(f: \mathbb{R} \to \mathbb{Z}\) as \(f(x) = n \quad \text{if} \quad x \in [n, n+1)\)
This function is also called the floor function, and we write \(f(x) = \left\lfloor x \right\rfloor\)
* Given an equivalence relation \(\mathcal{R}\) on \(X\), we define the canonical function \(\pi: X \to X / \mathcal{R}\), where \(\pi(x)=x/\mathcal{R}\in X/\mathcal{R}\)
Example Let \(X = \mathbb{Z}\), \(\mathcal{R}=\equiv _m\) and \(X/\mathcal{R}=\mathbb{Z}_{m}=\{\overline0,\overline1,\dots,\overline{m-1}\}\).
We define \(\pi(a) = \bar{a} = \{a + km : k \in \mathbb{Z}\}\) * A function \(f: \mathbb{N} \to B\) is called a sequence.
We denote \(f(1) = f_1\), \(f(2) = f_2\), ..., \(f(n) = f_n\).
Where:
- The first element is \(f_1\)
- The \(n\)-th element is \(f_n\).
Example: \(f_n = 1 + \frac{1}{n}\)