Example of Set for Relation Restriction to A

[evinda]

Hello! (Wave)

Let $R$ be a relation and $A$ a set.
The restriction of $R$ to $A$ is the set:

$$R\restriction A=\{ <x,y>: x \in A \wedge <x,y> \in R\}=\{ <x,y>: x \in A \wedge xRy\}$$

For a relation $R$ and a set $A$, it stands that:

$$dom(R \restriction A)=dom(R) \cap A$$

Could you give me an example of such a set, so that I can see that the above relation stands? (Thinking)

 

[Evgeny.Makarov]

If $R$ is a function from $\Bbb R$ to $\Bbb R$ defined by $xRy\iff y=x^2$, then $R\restriction\Bbb N$ is the restriction of function $R$ to $\Bbb N$ in this sense. Here $\operatorname{dom} R=\Bbb R$ and $\operatorname{dom}(R\restriction\Bbb N)=\Bbb R\cap\Bbb N=\Bbb N$.

If $mRn\iff m\text{ divides }n$ where $m,n\in\Bbb N$, then $R\restriction\{2\}=\{\langle2,n\rangle\mid n\text{ is even}\}$. Here $\operatorname{dom} R=\Bbb N$ and $\operatorname{dom}(R\restriction\{2\})=\Bbb N\cap\{2\}=\{2\}$.

If $mRn\iff n^2=m$ where $m,n\in\Bbb N$, then $\operatorname{dom} R$ is the set $\Bbb S$ of all square numbers. Let $\Bbb P=\{2^n\mid n\in\Bbb N\}$. Then $R\restriction\Bbb P=\{\langle 2^{2n},2^n\rangle\mid n\in\Bbb N\}$ and $\operatorname{dom}(R\restriction\Bbb P)=\Bbb S\cap\Bbb P=\{2^{2n}\mid n\in\Bbb N\}$.