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I want to know if different scenarios in relations must satisfy all the value in the relation.

In mathematical relations, a given set relation is reflexive if all the elements in the set exhibit (a,a).

What about the rest of the relations such as symmetric, transitive, anti-symmetric, and so on? Can the properties be satisfied once, or does it need to be valid for any given values in a relation?

N. F. Taussig
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user3227275
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1 Answers1

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Most of the conditions are universal, so they must be true for all values. Note that the set where things are coming from does make a difference in how the properties behave. Let $R$ be a relation on $A$ (so $R\subseteq A\times A$).

Reflexive: $$ \forall a\in A, (a,a)\in R $$

Irreflexive: $$ \forall a\in A, (a,a)\not\in R $$

Symmetric: $$ \forall (a,b)\in R, (b,a)\in R $$

Antisymmetric: $$ \forall (a,b)\in R\cap R^{-1}, a=b $$ (I've written this in an atypical way, usually, it's written ``If $(a,b)$ and $(b,a)$ are in $R$, then $a=b$.)

Transitive: $$ \forall (a,b),(b,c)\in R, (a,c)\in R $$ (Transitivity is often written as an if-then as well, ``if $(a,b)$ and $(b,c)$ are in $R$, then $(a,c)$ is in $R$.)

Michael Burr
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  • Thank you for the clarification. On a side note, take this example: R= {(1, 2), (2, 2)}

    (a,b) and (b,c) are (1,2) and (2,2), so (a,c) must be (1,2). Which already exists, so can we use that as a proof for transitive?, or do we need to find another (1,2) in the set?

     – user3227275 Mar 31 '15 at 12:58
  • Transitivity is a universal property, so one example is not enough. You must check all possible $(a,b),(b,c)\in R$. Therefore you must also check $(a,b)=(2,2)$ and $(b,c)=(2,2)$. In fact, you must check all pairs $(a,b)$ and $(b,c)$, it's just that for most pairs, that element isn't in $R$. – Michael Burr Mar 31 '15 at 13:05