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Mathematics LibreTexts

Preliminaries

  • Page ID
    10715
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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 \(A\).

    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 \(\begin{bmatrix}1 &1\\
    0 &0\end{bmatrix} \ne \begin{bmatrix}1 &0\\
    0 &0\end{bmatrix}\), \(\begin{bmatrix}1 & 0\\
    0 & 0\end{bmatrix} \notin A\).◻

     

    1. \(\begin{bmatrix}0 & 1\\
      0 & 0\end{bmatrix}\) 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 \(\begin{bmatrix}0 &1\\
    0 &0\end{bmatrix} = \begin{bmatrix}0 &1\\
    0 &0\end{bmatrix}\), \(\begin{bmatrix}1 & 0\\
    0 & 0\end{bmatrix} \in A\).◻

    1. \(\begin{bmatrix}0 & 0\\
      1 & 0\end{bmatrix}\) 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 \(\begin{bmatrix}0 &0\\
    1 &1\end{bmatrix} \ne \begin{bmatrix}1 &0\\
    1 &0\end{bmatrix}\), \(\begin{bmatrix}0 & 0\\
    1 & 0\end{bmatrix} \notin A\).◻

    1. \(\begin{bmatrix}0 & 0\\
      0 & 1\end{bmatrix}\) 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 \(\begin{bmatrix}0 &0\\
    0 &1\end{bmatrix} \ne \begin{bmatrix}0 &1\\
    0 &1\end{bmatrix}\), \(\begin{bmatrix}0 & 0\\
    0 & 1\end{bmatrix} \notin A\).◻

    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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