0.1: Sets

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Sets $Screen Shot 2023-06-27 at 12.52.27 PM.png$

Sets are collections of objects,
Sets are normally 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 $A$.

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

Answer

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

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

Answer

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

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

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

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

Answer

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

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

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

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

Answer

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

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

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

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

Solution

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

Notations:

$\cup$ 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.
$\mathbb{C}=$ set of all complex numbers.
$M_{mn}(\mathbb{R}):=$ set of all $m \times n$ matrices over real numbers.
$M_{mn}(\mathbb{C}):=$ set of all $m \times n$ matrices over complex numbers.
The cardinality of a set $A$ is written as $|A|$.

Example $\PageIndex{2}$

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

Example $\PageIndex{3}$

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

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Notations:

Example \(\PageIndex{2}\)