The question is stated as:
Consider a binary operation on a set $A$ $$F:A \times A \rightarrow A$$ Assume $C \subseteq A$. By the restriction of the operaction $F$ to $C$, we mean the function $G$ such that $Dom(G)=C \times C$ and $G(x,y) = F(x,y)$ for all $x$ and $y$ in $C$.
a. Prove that the restriction of $G$ of $F$ to $C$ is an operation on $C$ if and only if $C$ is closed under $F$.
Here is my atempt:
- First assume $C$ is closed under $F$, thus we have $(\forall x)(\forall y)([x \in C \land y \in C] \Rightarrow F(x,y) \in C)$
As $F(x,y) = G(x,y)$ for any $x$ and $y$ and $Dom(G) = C \times C \subseteq Dom(F)$, we have that $(\forall x)(\forall y)([x \in C \land y \in C] \Rightarrow G(x,y) \in C)$, and therefore we have $G:C \times C \rightarrow C$, then $G$ is a binary operation on $C$.
- On other hand assume that the restriction $G$ is a binary opeation on $C$, thus we have $G:C \times C \rightarrow C$
We have $C \subseteq A$ and $F(x,y) = G(x,y)$ for any $x$ and $y$, thus $(\forall x)(\forall y)([x \in C] \land y \in C] \Rightarrow G(x,y)=F(x,y) \in C)$, then $C$ is closed under $F$.
As both sides implies in eachother I conclude that he restriction of $G$ of $F$ to $C$ is an operation on $C$ if and only if $C$ is closed under $C$.
This atempt is correct? If not, how it should be proved?
To make the argument more clear I will use the notation $F_{|C}$ to mean the function $F$ restricted to the set $C\times C$. It is easy to verify that the restriction of a function is a function. An operation on a set $A$ is simply a function $A\times A \to A$. Using this definition we have that $F_{|C}$ is an operation on $C$ if and only if it is a function from $C\times C \to C$. We know that it is a function with domain equal to $C\times C$ so $F_{|C}$ is an operation on $C$ if and only if it has range that is a subset of $C$. Finally we have that $C$ is closed under $F$ if and only if $F(C\times C)\subseteq C$ which is the same as $range(F_{|C})=C$ which is the required property for $F_{|C}$ to be an operation on $C$