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# Preliminaries

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### Sets

• Sets are denoted by capital letters.

• Elements are denoted by lower case letters.

• Subsets:

• Every element of B is an element of A, written as $$B\subset A.$$

• $$B=\{a \in A:$$ …..condition on okay!.

• Trivial subsets of $$A$$ are $$A$$ and $$\emptyset$$ (empty set).

• How do we test if an element belongs to a set?

Example $$\PageIndex{1}$$

$$M_{22}(\mathbb{R}):=$$ set of all 2 x 2 matrices with real numbers, where $$\mathbb{R}:=$$ set of all real numbers.

Let  $$A=\big{\{}\begin{bmatrix}x\end{bmatrix} \in M_{22}(\mathbb{R})|\begin{bmatrix}x\end{bmatrix}\begin{bmatrix}1 &1\\ 0 &1\end{bmatrix}=\begin{bmatrix}1 &1\\ 0 &1\end{bmatrix}\begin{bmatrix}x\end{bmatrix}\big{\}}$$.

Determine whether the following entries belong to $$A$$.

1. $$\begin{bmatrix}1 & 0\\ 0 & 0\end{bmatrix}$$ does not belong to $$A$$.

$$\begin{bmatrix}1 & 0\\ 0 & 0\end{bmatrix} \cdot \begin{bmatrix}1 &1\\ 0 &1\end{bmatrix}=\begin{bmatrix}1 &1\\ 0 &0\end{bmatrix}$$ and

$$\begin{bmatrix}1 &1\\ 0 &1\end{bmatrix} \cdot \begin{bmatrix}1 & 0\\ 0 & 0\end{bmatrix}=\begin{bmatrix}1 &0\\ 0 &0\end{bmatrix}$$

Since $$\begin{bmatrix}1 &1\\ 0 &0\end{bmatrix} \ne \begin{bmatrix}1 &0\\ 0 &0\end{bmatrix}$$, $$\begin{bmatrix}1 & 0\\ 0 & 0\end{bmatrix} \notin A$$.◻

1. $$\begin{bmatrix}0 & 1\\ 0 & 0\end{bmatrix}$$ does belong to $$A$$.

$$\begin{bmatrix}0 & 1\\ 0 & 0\end{bmatrix} \cdot \begin{bmatrix}1 &1\\ 0 &1\end{bmatrix}=\begin{bmatrix}0 &1 0 &0\end{bmatrix}$$

$$\begin{bmatrix}1 &1\\ 0 &1\end{bmatrix} \cdot \begin{bmatrix}0 & 1\\ 0 & 0\end{bmatrix}=\begin{bmatrix}0 &1\\ 0 &0\end{bmatrix}$$

Since $$\begin{bmatrix}0 &1\\ 0 &0\end{bmatrix} = \begin{bmatrix}0 &1\\ 0 &0\end{bmatrix}$$, $$\begin{bmatrix}1 & 0\\ 0 & 0\end{bmatrix} \in A$$.◻

1. $$\begin{bmatrix}0 & 0\\ 1 & 0\end{bmatrix}$$ does not belong to $$A$$.

$$\begin{bmatrix}0 & 0\\ 1 & 0\end{bmatrix} \cdot \begin{bmatrix}1 &1\\ 0 &1\end{bmatrix}=\begin{bmatrix}0 &0\\ 1 &1\end{bmatrix}$$

$$\begin{bmatrix}1 &1\\ 0 &1\end{bmatrix} \cdot \begin{bmatrix}0 & 0\\ 1 & 0\end{bmatrix}=\begin{bmatrix}1 &0\\ 1 &0\end{bmatrix}$$

Since $$\begin{bmatrix}0 &0\\ 1 &1\end{bmatrix} \ne \begin{bmatrix}1 &0\\ 1 &0\end{bmatrix}$$, $$\begin{bmatrix}0 & 0\\ 1 & 0\end{bmatrix} \notin A$$.◻

1. $$\begin{bmatrix}0 & 0\\ 0 & 1\end{bmatrix}$$ does not belong to $$A$$.

$$\begin{bmatrix}0 & 0\\ 0 & 1\end{bmatrix} \cdot \begin{bmatrix}1 &1\\ 0 &1\end{bmatrix}=\begin{bmatrix}0 &0\\ 0 &1\end{bmatrix}$$

$$\begin{bmatrix}1 &1\\ 0 &1\end{bmatrix} \cdot \begin{bmatrix}0 & 0\\ 0 & 1\end{bmatrix}=\begin{bmatrix}0 &1\\ 0 &1\end{bmatrix}$$

Since $$\begin{bmatrix}0 &0\\ 0 &1\end{bmatrix} \ne \begin{bmatrix}0 &1\\ 0 &1\end{bmatrix}$$, $$\begin{bmatrix}0 & 0\\ 0 & 1\end{bmatrix} \notin A$$.◻

1. $$\begin{bmatrix}0 & 0\\ 0 & 0\end{bmatrix}$$.  does belong to $$A$$.

Solution

This one is obviously true, thus $$\begin{bmatrix}0 & 0\\ 0 & 0\end{bmatrix} \in A$$.

• $$\cup$$ or / union

• $$\cap$$ and / intersection

• $$\{\}^c$$ complement of a set (aka $$\{\}^{'}$$).

• $$A \backslash B$$ is equivalent to $$A \cap B^{'}$$.

• $$\mathbb{Z}=$$ set of all integers $$=\{\ldots,-2,-1,0,1,2,\ldots\}$$.  A countable set.

• $$\mathbb{Q}=$$ set of all rational numbers $$=\{\frac{a}{b}: a,b \in \mathbb{Z}, b \ne 0 \}$$.   A countable set.

• $$\mathbb{Q}^c=$$  set of all irrational numbers $$=\{\pm e, \pm \pi, \pm \sqrt{2}, \ldots\}$$.   Not a countable set.

• $$\mathbb{R}=$$ set of all real numbers $$=\mathbb{Q} \cup \mathbb{Q}^c$$.  Note $$\mathbb{Q} \cap \mathbb{Q}^c=\emptyset$$.  Not a countable set.

• $$\mathbb{N} \subseteq \mathbb{Z} \subseteq \mathbb{Q} \subseteq \mathbb{R} \subseteq \mathbb{C}$$.

• The cardinality of a set $$A$$ is written as $$|A|$$.

Example $$\PageIndex{2}$$

Let $$A=\{1,2,3,5\}$$.  Therefore $$|A|=4$$. In this case $$A$$ is called a finite set.

### Functions

Mapping

Let $$A$$ be a set.

Then $$A \times A$$ is the Cartesian product of $$A$$.

Example:  If $$A = \mathbb{R}$$ then this is the Cartesian plane -  Like $$\{x,y: x,y \in A\}$$.

A function $$f$$ is a subset (symbol is $$\subseteq$$) of $$A \times B$$ such that for every $$a \in A$$ there exists a unique $$b \in B$$ such that $$f(a)=b$$.

Denoted by $$f: A \rightarrow B$$ or $$A \xrightarrow{f} B$$.

$$A$$ is the domain of the function $$f$$.

$$B$$ is the co-domain of the function $$f$$.

The range of $$f$$ is $$f(A)=\big{\{}f(x): x \in A \big{\}} \subseteq B$$

A well-defined function is one that for every value of $$A$$, there is one and only one value of $$B$$.

Definition:

Let $$f$$ be a function from $$A \rightarrow B$$.

1. $$f$$ is called onto (surjective) if $$f(A)=B$$.

2. $$f$$ is called one to one (injective) if $$f(x_1)=f(x_2), x_1,x_2 \in A$$ then $$x_1=x_2$$.

3. $$f$$ is bijective if it is surjective and injective.

4. A function can be injective, surjective, bijective (both injective & surjective), or neither.

Example $$\PageIndex{1}$$

Let

$$\mathbb{Q}=\big{\{}\frac{a}{b}: b\ne 0,\text{and } gcd(a,b)=1 \big{\}}$$

and $$f: \mathbb{Q} \rightarrow \mathbb{Z}$$ that is defined by

$$f(\frac{a}{b})=a.$$

a) Is $$f$$ a well-defined function? b)  Is it surjective?  c) Is it injective?  d) Is it bijective?

Solution

1. The function is well-defined.

Let $$A$$ and $$B$$ be set.

Let $$a \in A$$.

We need to show that if $$f(a)=b$$ then $$b\in B$$ and if $$f(a)=b_1$$ that $$b=b_1$$.

We are given $$f: \mathbb{Q} \rightarrow \mathbb{Z}$$, which means that this is a function thus $$f(a)$$ will map to one and only one element of $$B$$.  Thus if $$f(a)=b$$ and if $$f(a)=b_1$$ then $$b=b_1$$.

In b) below, we have shown that the function is surjective, thus \)f(a)=b\),

1. The function is surjective

Proof:

We shall show that $$f(\mathbb{Q})=\mathbb{Z}$$ by showing that  $$f(\mathbb{Q}) \subseteq \mathbb{Z}$$ and  $$\mathbb{Z} \subseteq f(\mathbb{Q})$$.

Since we are given $$f: \mathbb{Q} \rightarrow \mathbb{Z}$$,  $$f(\mathbb{Q}) \subseteq \mathbb{Z}$$.

Next, we shall show that $$\mathbb{Z} \subseteq f(\mathbb{Q})$$.

Let $$n \in \mathbb{Z}$$, then $$\frac{n}{1} \in \mathbb{Q}$$ and $$f\big(\frac{n}{1}\big)=n$$.

Hence $$n \in f(\mathbb{Q})$$.

Hence $$\mathbb{Z} \subseteq f(\mathbb{Q})$$.

Since $$\mathbb{Z} \subseteq f(\mathbb{Q})$$ and $$f(\mathbb{Q}) \subseteq \mathbb{Z}, f(\mathbb{Q})=\mathbb{Z}$$.◻

1. The function is not injective.

Counterexample:

Let $$A=\big{\{}\frac{a}{b}: b\ne 0,\text{and } gcd(a,b)=1 \big{\}}$$ and $$B=\mathbb{Z}$$.

Note $$\frac{1}{4}, \frac{1}{3} \in A$$ and that $$f(\frac{1}{4}), f(\frac{1}{3}) \in B$$ and that $$f(\frac{1}{4})=f(\frac{1}{3})=1$$.

Since $$\frac{1}{4} \ne \frac{1}{3}$$ , the function is not injective.

1. The function is not bijective since it has been shown not to be both injective and surjective.

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