Function

Function

Definition 1. Let $X, Y$ be sets. A function from $X$ to $Y$ is a relation $f$ from $X$ to $Y$ satisfying 
(a) Dom($f$) = $X$,
(b) If $(x, y) \in f$ and $(x, z) \in f$, then $y = z$.

$(x, y) \in f$는 관습상 $xfy$가 아닌 $y = f(x)$라고 쓴다. 또한 $X$에서 $Y$로의 관계인 함수 $f$는 $f : X \longrightarrow Y$와 같이 표기한다.

The Condition For Functions To Be Equal

Theorem 1. Let $f, g : X \longrightarrow Y$ be functions. Then $f = g \iff f(x) = g(x), \forall x \in X$.

The Union of Functions

Theorem 2. Let $f : A \longrightarrow C$ and $g : B \longrightarrow D$ be functions such that $f(x) = g(x), \forall x \in A \cap B$. Then the union $h$ of $f$ and $g$ defines the function $$h = f \cup g : A \cup B \longrightarrow C \cup D$$ where $$h(x) = \begin{cases} f(x), \, \text{ if } x \in A \\ g(x), \, \text{ if } x \in B. \end{cases}$$

Image and Inverse Imange

Definition 1. Let $f : X \longrightarrow Y$ be a function, and let $A \subseteq X$ and $B \subseteq Y$, respectively. Then
(a) The image of $A$ under $f$, which we denote $f(A)$, is the set $f(A) = \{ f(x)\, | \, x \in A \}$.
(b) The inverse image of $B$ under $f$, which we denote $f^{-1}(B)$, is the set $\{ x\, | \, f(x) \in B \}$.

Theorem 3

Theorem 3. Let $f : X \longrightarrow Y$ be a function. Then
(a) $f(\emptyset) = \emptyset$.
(b) $f(\{x\}) = \{ f(x) \}, \forall x \in X$.
(c) $A \subseteq B \subseteq X \Longrightarrow f(A) \subseteq f(B)$.
(d) $C \subseteq D \subseteq Y \Longrightarrow f^{-1}(C) \subseteq f^{-1}(D)$.

Theorem 4

Theorem 4. Let $f : X \longrightarrow Y$ be a function. Then $$\begin{align*} & \text{(a)} \,\, f(\bigcup_{\gamma \in \Gamma} A_{\gamma}) = \bigcup_{\gamma \in \Gamma} f(A_{\gamma}). \hspace{15cm} \\ & \text{(b)} \,\, f(\bigcap_{\gamma \in \Gamma} A_{\gamma}) \subseteq \bigcap_{\gamma \in \Gamma} f(A_{\gamma}). \end{align*}$$

(b)에서 등호가 성립하지 않는 이유는 상수 함수의 존재 때문이다. 좌변은 상수 하나의 singleton인데, 우변은 그보다 항상 크거나 같기 마련이다.

Theorem 5

Theorem 5. Let $f : X \longrightarrow Y$ be a function. Then $$\begin{align*} & \text{(a)} \,\, f^{-1}(\bigcup_{\gamma \in \Gamma} B_{\gamma}) = \bigcup_{\gamma \in \Gamma} f^{-1}(B_{\gamma}). \hspace{15cm} \\ & \text{(b)} \,\, f^{-1}(\bigcap_{\gamma \in \Gamma} B_{\gamma}) = \bigcap_{\gamma \in \Gamma} f^{-1}(A_{\gamma}). \end{align*}$$

Theorem 6

Theorem 6. Let $f : X \longrightarrow Y$ be a function and let $B, C \subseteq Y$. Then $$f^{-1}(B - C) = f^{-1}(B) - f^{-1}(C).$$