Exploring the Standing of a Relation: A Journey of Discovery

[evinda]

Hello! (Wave)

Could you explain me why the following stands? (Thinking)

If $R$ is a relation, then:
$$R \subset dom R \times rng R \subset fld R \times fld R$$

 

[Evgeny.Makarov]

Consider an example and try to generalize.

 

[evinda]

Consider an example and try to generalize.

We could prove it like that:

If $<x,y> \in R$, then $x \in dom(R)$ and $y \in rng(R)$. So, $<x,y> \in dom(R) \times rng(R)$, and therefore, $R \subset dom(R) \times rng(R)$, right? (Thinking)
But, I can't think of an example... (Worried)

 

[Evgeny.Makarov]

If $<x,y> \in R$, then $x \in dom(R)$ and $y \in rng(R)$. So, $<x,y> \in dom(R) \times rng(R)$, and therefore, $R \subset dom(R) \times rng(R)$, right?

Yes, that's correct.

But, I can't think of an example...

It's not like you need to find something that holds only rarely. This fact holds for any $R$. Take $R=\{\langle0,2\rangle,\langle1,3\rangle,\langle0,3\rangle\}$, for example. And once you have a proof, you need examples only for elucidation.

 

[evinda]

It's not like you need to find something that holds only rarely. This fact holds for any $R$. Take $R=\{\langle0,2\rangle,\langle1,3\rangle,\langle0,3\rangle\}$, for example. And once you have a proof, you need examples only for elucidation.

Is it in this case: $dom(R)=\{0,1\}$ and $rng(R)=\{2,3\}$ ? Or am I wrong? (Thinking)

 

[Evgeny.Makarov]

Is it in this case: $dom(R)=\{0,1\}$ and $rng(R)=\{2,3\}$ ?

You are right.

 

[evinda]

You are right.

And $dom(R) \times rng(R)=\{ <0,2>,<0,3>,<1,2>,<1,3> \}$.
So, we see that $R=\{ <0,2>,<1,3>,<0,3> \} \subset \{ <0,2>,<0,3>,<1,2>,<1,3> \}$, right? (Smile)

 

[Evgeny.Makarov]

Yes.

 

[evinda]

Yes.

Nice, thanks a lot! (Clapping)

 

[arham1]

Thanks...
I'll try it and talk about my thoughts.