Functions (Their Properties and Importance)

Direct Knowlege Contributor David A. Smith
  • By David A. Smith, Founder & CEO, Direct Knowledge
  • David Smith has a B.S. and M.S. in Mathematics and has enjoyed teaching calculus, linear algebra, and number theory at both Tarrant County College and the University of Texas at Arlington. David is the Founder and current CEO of Direct Knowledge.

Let $X$ and $Y$ be sets, we say $f$ is a function from $X$ to $Y$ if $f$ is a subset of $X\times Y$ such that the domain of $f$ is $X$ and $f$ has the property:  if $(x,y)\in f$ and $(x,z)\in f$ then $x=z$. For each $x\in X$, the unique $y\in Y$ such that $(x,y)\in f$ is denoted by $f(x)$.  The element $y$ is called the value of $f$ at the argument $x$. 

The importance of functions in mathematics is discussed here.

Domain and Codomain

If we do not specify a function with the notation $f:X\to Y$ we will use $D(f)$ and $R(f)$ to denote the domain and range of $f$, respectively.

Image

Definition. Let $X$ and $Y$ be sets. The image of $A\subseteq X$ is the set $$f(A)=\{y\in Y : \exists x\in A, y=f(x)\}. $$

Theorem. Let $f:X\to Y$ be a function. If $A_1, A_2\subseteq X$, then  $$ f(A_1)\cup f(A_2)=f(A_1\cup A_2). $$

Proof. \begin{align*} & y\in f(A_1)\cup f(A_2)  \Longleftrightarrow y\in f(A_1) \lor y\in f(A_2)  \\& \qquad  \Longleftrightarrow \exists x_1\in A_1, y=f(x_1) \lor \exists x_2\in A_2, y=f(x_2)  \\& \qquad  \Longleftrightarrow \exists x\in X, (x\in A_1 \lor x\in A_2) \land y=f(x)  \\& \qquad  \Longleftrightarrow \exists x \in A_1 \cup A_2, y=f(x)  \Longleftrightarrow y\in f(A_1\cup A_2)  \end{align*}

Theorem. Let $f:X\to Y$ be a function. If $A_1, A_2\subseteq X$, then  $$ f(A_1\cap A_2) \subseteq f(A_1)\cap f(A_2). $$

Proof. \begin{align*} & y\in f(A_1 \cap  A_2)  \Longrightarrow \exists x\in A_1 \cap A_2, y=f(x) \\& \qquad  \Longrightarrow  \exists x\in A_1, y=f(x) \land \exists x\in A_2, y=f(x) \\& \qquad  \Longrightarrow  y\in f(A_1)\land y\in f(A_2)  \Longrightarrow  y\in f(A_1)\cap f(A_2) \end{align*}

Theorem. Let $f:X\to Y$ be a function. If $A_1, A_2\subseteq X$, then  $$ f(A_2)\setminus f(A_1) \subseteq f(A_2\setminus A_1). $$

Proof. \begin{align*} & y\in f(A_2)\setminus f(A_1)  \Longleftrightarrow y\in f(A_2) \land y\notin f(A_1)  \\& \qquad  \Longleftrightarrow (\exists x\in A_2, y=f(x) )\land \neg (y\in f(A)) \\& \qquad  \Longleftrightarrow (\exists x\in A_2, y=f(x) )\land \neg (\exists z\in A_1, y=f(z)) \\& \qquad  \Longleftrightarrow (\exists x\in A_2, y=f(x) )\land (\forall z\in A_1, y\neq f(z)) \\& \qquad  \Longrightarrow \exists x\in A_2\setminus A_1, y=f(x) \Longleftrightarrow y\in f(A_2\setminus A_1)  \end{align*}

Preimage

Definition. Let $X$ and $Y$ be sets. The preimage of $B\subseteq Y$ is the set $$f^{-1}(B)=\{x\in X : f(x)\in B\}. $$

Theorem. Let $f:X\to Y$ be a function. If $B_1, B_2\subseteq Y$, then  $$ f^{-1}(B_1\cup B_2)=f^{-1}(B_1)\cup f^{-1}(B_2). $$ 

Proof. \begin{align*} & x\in f^{-1}(B_1\cup B_2)  \Longleftrightarrow   f(x)\in B_1 \cup B_2  \\ & \qquad  \Longleftrightarrow   f(x) \in B_1 \lor f(x)\in B_2  \\ & \qquad  \Longleftrightarrow  x\in f^{-1}(B_1) \lor x\in f^{-1}(B_2)  \\ & \qquad  \Longleftrightarrow   x\in f^{-1}(B_1)\cup f^{-1}(B_2) \end{align*}

Theorem. Let $f:X\to Y$ be a function. If $B_1, B_2\subseteq Y$, then  $$ f^{-1}(B_1\cap B_2)=f^{-1}(B_1)\cap f^{-1}(B_2). $$

Proof. \begin{align*} & x\in f^{-1}(B_1\cup B_2)  \Longleftrightarrow   f(x)\in B_1\cup B_2  \\ & \qquad  \Longleftrightarrow  f(x)\in B_1 \land f(x)\in B_2 \\ & \qquad  \Longleftrightarrow  f(x)\in B_1 \land f(x)\in B_2   \\ & \qquad  \Longleftrightarrow   x\in f^{-1}(B_1) \land x\in f^{-1}(B_2) \\ & \qquad  \Longleftrightarrow   x\in f^{-1}(B_1)\cap f^{-1}(B_2) \end{align*}

Theorem. Let $f:X\to Y$ be a function. If $B_1, B_2\subseteq Y$, then  $$ f^{-1}(B_2\setminus B_1)=f^{-1}(B_2)\setminus f^{-1}(B_1). $$

Proof. \begin{align*} & x\in f^{-1}(B_2\setminus B_1) \Longleftrightarrow   f(x)\in B_2\setminus B_1  \\ & \qquad  \Longleftrightarrow  f(x)\in B_2 \land \neg(f(x)\in B_1) \\ & \qquad  \Longleftrightarrow   x\in f^{-1}(B_2) \land \neg (x\in f^{-1}(B_1)) \\ & \qquad  \Longleftrightarrow   x\in f^{-1}(B_2) \land x\not\in f^{-1}(B_1) \\ & \qquad  \Longleftrightarrow   x\in f^{-1}(B_2) \setminus f^{-1}(B_1) \end{align*}

Theorem. Let $f:X\to Y$ be a function. Then  $$ A_1\subseteq A_2\subseteq X\implies f(A_1)\subseteq f(A_2). $$

Proof. \begin{align*} & y\in f(A_1) \Longleftrightarrow \exists x\in A_1, y=f(x)  \\ & \qquad  \implies \exists x\in A_2, y=f(x) \implies y\in f(A_2)  \end{align*}

Theorem. Let $f:X\to Y$ be a function. Then  $$ A_1\subseteq A_2\subseteq X\implies f(A_1)\subseteq f(A_2). $$

Proof. \begin{align*} x\in f^{-1}(B_1)  \Longleftrightarrow   f(x)\in B_1 \Longrightarrow   f(x)\in B_2 \Longleftrightarrow   x\in f^{-1}(B_2) \end{align*}

Theorem. Let $f:X\to Y$ be a function. Then  $$ B\subseteq Y  \implies  f(f^{-1}(B))\subseteq B $$

Proof. \begin{align*} & y\in f(f^{-1}(B)) \Longleftrightarrow   \exists x\in f^{-1}(B), y=f(x)  \\ & \qquad  \Longrightarrow   \exists x\in X, f(x)\in B \land y=f(x) \Longrightarrow  y\in B  \end{align*}

Theorem. Let $f:X\to Y$ be a function. Then  $$ A\subseteq X \implies A\subseteq f^{-1}(f(A)). $$

Proof. \begin{align*} x\in A \implies \exists y\in Y, y=f(x) \Longleftrightarrow y\in f(A) \implies x\in f^{-1}(f(A)) \end{align*}

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