What Is the Standing of a Relation in Set Theory?

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Discussion Overview

The discussion revolves around the properties of relations in set theory, specifically examining the inclusion of a relation \( R \) within the Cartesian product of its domain and range. Participants explore the implications of this relationship and seek examples to illustrate their points.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • Some participants propose that if \( \in R \), then \( x \in dom(R) \) and \( y \in rng(R) \), leading to the conclusion that \( R \subset dom(R) \times rng(R) \).
  • Others suggest that this inclusion holds for any relation \( R \), providing an example \( R=\{\langle0,2\rangle,\langle1,3\rangle,\langle0,3\rangle\} \) to illustrate the point.
  • One participant confirms the domain and range for the provided example, stating \( dom(R)=\{0,1\} \) and \( rng(R)=\{2,3\} \).
  • Another participant calculates \( dom(R) \times rng(R) \) and shows that \( R \) is indeed a subset of this product.

Areas of Agreement / Disagreement

Participants generally agree on the inclusion of \( R \) within \( dom(R) \times rng(R) \) and confirm the calculations related to the example provided. However, there is no explicit consensus on the necessity of examples beyond the initial proof.

Contextual Notes

The discussion does not resolve the broader implications of these relationships or the potential limitations of the examples provided. There is an implicit assumption that the properties discussed hold universally for all relations.

evinda
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Hello! (Wave)

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

If $R$ is a relation, then:
$$R \subset dom R \times rng R \subset fld R \times fld R$$
 
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Consider an example and try to generalize.
 
Evgeny.Makarov said:
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)
 
evinda said:
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.
evinda said:
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.
 
Evgeny.Makarov said:
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)
 
evinda said:
Is it in this case: $dom(R)=\{0,1\}$ and $rng(R)=\{2,3\}$ ?
You are right.
 
Evgeny.Makarov said:
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)
 
Yes.
 
Evgeny.Makarov said:
Yes.

Nice, thanks a lot! (Clapping)
 
  • #10
Thanks...
I'll try it and talk about my thoughts.
 

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