Functions (Their Properties and Importance)

Hasn’t everyone has heard of what a function is? In this article, I define what a function is and discuss the domain and codomain in detail. I also cover the image and preimage of a function. I do not assume anything other than basic elementary set theory.

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*}

About David A. Smith aka Dave
About Dave

David Smith (Dave) has a B.S. and M.S. in Mathematics and has enjoyed teaching precalculus, calculus, linear algebra, and number theory at both the junior college and university levels for over 20 years. David is the founder and CEO of Dave4Math.

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