4. 함수 (i): 정의와 기본 성질Functions (i): Definitions and Basic Properties

(⟹) Assume that (F2) holds. Let $x_1, x_2 \in \mathit{dom}\ f$ with $x_1 = x_2$. Since $x_1 \in \mathit{dom}\ f$, by the definition of the domain (Definition 3.63), there exists $y_1$ such that $(x_1, y_1) \in f$, and by (4.2), $y_1 = f(x_1)$. Similarly, since $x_2 \in \mathit{dom}\ f$, there exists $y_2$ such that $(x_2, y_2) \in f$, and $y_2 = f(x_2)$. Since $x_1 = x_2$, we have $\{x_1\} = \{x_2\}$ and $\{x_1, y_1\} = \{x_2, y_1\}$, so by the axiom of extent,

i.e. $(x_2, y_1) \in f$. Now $(x_2, y_1) \in f$ and $(x_2, y_2) \in f$, so by (F2),

Therefore

and (F2’) is proved.

(⟸) Assume that (F2’) holds. Let $(x, y_1) \in f$ and $(x, y_2) \in f$. Since $(x, y_1) \in f$, by the definition of the domain, $x \in \mathit{dom}\ f$, and by (4.2), $y_1 = f(x)$. Similarly, since $(x, y_2) \in f$, $x \in \mathit{dom}\ f$ and $y_2 = f(x)$. Setting $x_1 := x$ and $x_2 := x$, we have $x_1 = x_2 \in \mathit{dom}\ f$, so by (F2’), $f(x_1) = f(x_2)$. Therefore

and (F2) is proved.

Combining both directions, (F2) and (F2’) are equivalent.

1. This statement, i.e. “(F2) only if (F2’)” (in other words (F2) $\implies$ (F2’)), is true. Assume that $G$ satisfies (F2) (Remark 4.4). Let $x_1, x_2 \in \mathit{dom}\ G$ with $x_1 = x_2$. Since $x_1 \in \mathit{dom}\ G$, by the definition of the domain (Definition 3.63), there exists $y_1$ such that $(x_1, y_1) \in G$, and by (4.3), $G(x_1) = y_1$. Similarly, since $x_2 \in \mathit{dom}\ G$, there exists $y_2$ such that $(x_2, y_2) \in G$, and $G(x_2) = y_2$. Since $x_1 = x_2$, by the axiom of extent $(x_1, y_1) = (x_2, y_1)$, so $(x_2, y_1) \in G$. Now $(x_2, y_1) \in G$ and $(x_2, y_2) \in G$, so by (F2), $y_1 = y_2$. Therefore
   $$G(x_1) = y_1 = y_2 = G(x_2)$$
   and (F2’) is proved.
2. This statement, i.e. “(F2) if (F2’)” (in other words (F2’) $\implies$ (F2)), is false. As a counterexample, consider
   $$G = \{(a, b), (a, c)\}, \qquad b \neq c.$$
   First we check that $G$ does not satisfy (F2). Since $(a, b) \in G$ and $(a, c) \in G$ but $b \neq c$, $G$ does not satisfy (F2).

   On the other hand, by (4.3), since $(a, b) \in G$, the reading “$G(a) = b$” holds, and since $(a, c) \in G$, the reading “$G(a) = c$” also holds (again by (4.3)). That is, for a graph on which (F2) fails, the notation $G(x)$ does not designate a unique value, and at $x = a$ the expression “$G(a)$” itself is ambiguous, admissible as either $b$ or $c$. Nevertheless $G$ does satisfy (F2’). Indeed $\mathit{dom}\ G = \{a\}$, so the only instance of (F2’) is $x_1 = x_2 = a$, and there the conclusion of (F2’) reads “$G(a) = G(a)$”. Whichever of the two admissible values $b$, $c$ one fixes for $G(a)$, the two sides are the same expression, so this holds. Hence $G$ satisfies (F2’) but not (F2), and (F2’) $\implies$ (F2) is false.

   In short, the notation $G(x)$ of (4.3) designates a unique value only when $G$ satisfies (F2). Without (F2) the expression “$G(x)$” can be read as several different values at once, and (F2’) — which speaks only of $G(x_1)$ and $G(x_2)$ for $x_1 = x_2$ — degenerates into the triviality that one and the same expression equals itself. This is exactly why (F2’) carries no information about $G$ and cannot recover (F2); that is, the converse of (1) fails.

Let $y \in \mathit{ran}\ G$. By the definition of the range (Definition 3.63), there exists $x$ such that $(x,y) \in G$. Then by the definition of the domain, $x \in \mathit{dom}\ G$, and by (4.3), $y = G(x)$. Since $y \in \mathit{ran}\ G$ was arbitrary, we have shown that for every $y \in \mathit{ran}\ G$, there exists $x \in \mathit{dom}\ G$ such that $y = G(x)$. Therefore, by the definition in Remark 4.15(5) (that a graph is surjective to a class $B$, applied with $B := \mathit{ran}\ G$), $G$ is surjective to $\mathit{ran}\ G$.

Note. For the stated claims to hold, (1) requires $B \neq \emptyset$ and (2) requires $B$ to have two distinct elements. If $B = \emptyset$, then (F1) forces $A = \emptyset$, and since $C \supsetneq A = \emptyset$, $C$ is nonempty; but no function from a nonempty $C$ to $B = \emptyset$ can exist, by (F1), so there is no extension at all. On the other hand, if $B = \{b\}$ has only one element, then there is exactly one function from $C$ to $B$ (the constant function), so the extension is unique. Below we prove (1) under the assumption $B \neq \emptyset$, and (2) under the assumption that $B$ has two distinct elements $b \neq b'$.

1. Since $B \neq \emptyset$, we may choose $b \in B$. As in the constant function of Example 4.18, define the graph
   $$K := \{(x, b) : x \in C - A\}.$$
   Let $g := f \cup K$. We show that $g : C \to B$ is an extension of $f$ to $C$.

   First we show $g \subset C \times B$. Let $(x, y) \in g$. If $(x, y) \in f$, then since $f \subset A \times B$ and $A \subset C$, $(x, y) \in C \times B$. If $(x, y) \in K$, then $x \in C - A \subset C$ and $y = b \in B$, so $(x, y) \in C \times B$. Either way, $(x, y) \in C \times B$.

   Next we show (F2) holds. Let $(x, y_1) \in g$ and $(x, y_2) \in g$. Since $\mathit{dom}\ f = A$, $\mathit{dom}\ K = C - A$, and $A \cap (C - A) = \emptyset$, we split into the cases $x \in A$ and $x \in C - A$.

   • If $x \in A$, then $x \notin C - A = \mathit{dom}\ K$, so $(x,y_1)$ and $(x,y_2)$ are elements of $f$, not $K$. That is, $(x,y_1), (x,y_2) \in f$, so by (F2) of $f$, $y_1 = y_2$.
   • If $x \in C - A$, then for the same reason $(x,y_1), (x,y_2) \in K$. By the definition of $K$, $y_1 = y_2 = b$.

   Either way $y_1 = y_2$, so $g$ satisfies (F2).

   Next we show $\mathit{dom}\ g = C$. By Theorem 3.77(1),

   $$\mathit{dom}\ g = \mathit{dom}\ (f \cup K) = \mathit{dom}\ f \cup \mathit{dom}\ K = A \cup (C - A) = C$$

   (the last equality holds since $A \subset C$).

   We also show $\mathit{ran}\ g \subset B$. By Theorem 3.77(2),

   $$\mathit{ran}\ g = \mathit{ran}\ f \cup \mathit{ran}\ K.$$

   Since $f : A \to B$ is a function, $\mathit{ran}\ f \subset B$ by Lemma 4.5(2). Also, since $A \subsetneq C$, $C - A \neq \emptyset$, and by the definition of $K$, $\mathit{ran}\ K = \{b\} \subset B$. Therefore $\mathit{ran}\ g \subset B$.

   We have now shown that $g$ satisfies (F2), $\mathit{dom}\ g = C$, and $\mathit{ran}\ g \subset B$, so by Theorem 4.8, $g : C \to B$ is a function.

   Finally we show $g$ agrees with $f$ on $A$. Let $x \in A$. Then $(x, f(x)) \in f \subset g$, so by (4.2),

   $$g(x) = f(x).$$

   Therefore $g : C \to B$ is an extension of $f : A \to B$ to $C$, and (1) is proved.

2. Assume that $B$ has two distinct elements $b \neq b'$. Applying the construction of (1) to $b$ and $b'$ separately, let
   $$K := \{(x, b) : x \in C - A\}, \qquad K' := \{(x, b') : x \in C - A\}$$
   and $g := f \cup K$, $g' := f \cup K'$. By (1), both $g : C \to B$ and $g' : C \to B$ are extensions of $f$ to $C$. It suffices to show $g \neq g'$.

   Since $A \subsetneq C$, $C - A \neq \emptyset$, so there exists $x_0 \in C - A$. By the definitions of $K$ and $K'$, $(x_0, b) \in K \subset g$ and $(x_0, b') \in K' \subset g'$, so by (4.2),

   $$g(x_0) = b, \qquad g'(x_0) = b'.$$

   Since $b \neq b'$, $g(x_0) \neq g'(x_0)$. But $g, g' : C \to B$ are functions with the same domain and codomain, so by Theorem 4.13, $g = g'$ would require $g(x) = g'(x)$ for all $x \in C$. Since $g(x_0) \neq g'(x_0)$, this fails, and therefore $g \neq g'$.

   Hence the extensions $g$ and $g'$ of $f : A \to B$ to $C$ are distinct, and (2) is proved.