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How it can be shown that:

The equality equivalence relation is the finest equivalence relation relation.

Since the equality relation denoted $R$ is an equivalence relation hence it should be a subset from the Cartesian product of two same sets namely $X$,e.g.:

$$R⊆X×X$$

I know that we should pick an ordered pair $\left(x,y\right)∈R$ where $x,y∈X$ and show if it's an element of the equality equivalence relation $R$ then it's also contained in any other equivalence relations and using the definition of subset the result follows, but I don't know how exactly it can be done.

The other theorem states:

The Trivial relation that makes all pairs of elements related is the coarsest relation.

The trivial relation denoted $R$ is a subset of Cartesian product of two subsets $S,T$, e.g. $$R⊆S×T$$ and since every element in $R$ is related to the other elements so $R=S×T$

I don't know how to proof this one and any help is appreciated. (Trivial relation is not necessarily a subset of Cartesian product of two same sets, hence it's not necessarily a homogeneous binary relation)

  • Can you please define 'finest' and 'coarsest' in your question? – CardioidAss22 Jan 31 '20 at 14:57
  • For the first, recognize that every equivalence relation is reflexive. For the second, recognize that every equivalence relation is a relation and thus a subset of $S\times T$. – JMoravitz Jan 31 '20 at 14:58
  • @Cardioid_Ass_22, I will add a link for that –  Jan 31 '20 at 14:58
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    @Cardioid_Ass_22 these are very standard terms... if you have two relations, $A$ and $B$ such that $A\subseteq B$, then $A$ is "finer" than $B$ and $B$ is "coarser" than $A$. In layman's terms, $A$ is more specific than $B$. For example "starts with the same first three letters" is finer than "starts with the same letter." – JMoravitz Jan 31 '20 at 15:00
  • @JMoravitz,I don't know how to use them –  Jan 31 '20 at 15:26
  • @user715522 Every reflexive relation on $X$ must for every $x\in X$ include $(x,x)$ in the relation (among possibly many more pairs), otherwise it wouldn't be reflexive. Phrase it in a correct way as to imply that $Id_X \subseteq A$ for every possible equivalence relation $A$ over $X$. – JMoravitz Jan 31 '20 at 15:28
  • There can be no finer equivalence relation since equality is the finest possible reflexive relation on a set. There can be no coarser relation since the trivial relation is the coarsest possible relation. – Somos Jan 31 '20 at 15:30
  • @JMoravitz, I got the first one, thank you –  Jan 31 '20 at 15:40
  • @Somos, well the statement that claim " trivial relation is the coarsest possible relationt" need a proof –  Jan 31 '20 at 15:41
  • For the second, recall as I said in my first comment that every relation over $S\times T$ will be by definition a subset of $S\times T$ based on what it means to be a relation in the first place. You say you showed and understand that the trivial relation where everything is related to everything else, $R$, satisfies that $R=S\times T$. So... since every other relation is a subset of $S\times T$... don't you see why that means that every other relation is a subset of $R$? After all... $R$ and $S\times T$ are equal, aren't they? And then why this implies that $R$ is coarser than any $A$? – JMoravitz Jan 31 '20 at 15:43
  • @JMoravitz, I understand, thanks again –  Jan 31 '20 at 15:46
  • @JMoravitz, sorry but in the first one I guess you showed that identity relation is the finest equivalence relation but you did not show that equality is the finest (In my opinion identity relation is the finest), well I think since every ordered pair are in the form (x,x) this is actually the same as identity relation) –  Feb 02 '20 at 15:16
  • @user715522 the equality equivalence relation is the identity equivalence relation. They are two names for the same thing. – JMoravitz Feb 02 '20 at 18:23
  • thank you so much for your responding,intuitively I agree with that but in there proof of that? –  Feb 02 '20 at 18:51

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