I'm having trouble understanding whether or not relations are reflexive, symmetric and transitive. I know that for a relation to be any of those it must satisfy the conditions:

  • reflexive: for every $s \in S$ $sRs$ (s is related to itself and therefore reflexive)
  • symmetric: for every $s,t \in S$, if $sRt$ then $tRs$
  • transitive: for every $s,t,u \in S$, if $sRt$ and $tRu$ then $sRu$

However I don't quite understand how to apply these conditions to problems. For example how would I solve something like this:

a) $x\sim y$ if $x$ and $y$ are people and there exists a country $C$ such that $x$ has been to country $C$ and $y$ has been to country $C$.

b) $x\sim y$ if $x$ and $y$ are strings which contain a common character.

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  • $\begingroup$ Well, can you figure out anything about either of these? $\endgroup$ – dfeuer Oct 29 '13 at 2:47

Perhaps it will help if you literally write out what each statement means in each case.

Let's try another example then you can do your examples. Let our overlying set be the set of all animals.

$x \sim y$ if $x$ is from the same species as $y$.

1) Reflexive: We are asking if $x$ is from the same species as $x$.

2) Symmetric: Assuming $x$ is from the same species as $y$. Is $y$ from the same species as $x$?

3) Transitive: Assuming $x$ and $y$ are from the same species, and $y$ and $z$ are from the same species. Are $x$ and $z$ from the same species?

Lets try to answer property three by thinking intuitively.

We assume $x$ and $y$ are of the same species. Without loss of generality, lets say that species is tigers. So we know that $x$ and $y$ are both tigers.

Then we also assume that $y$ and $z$ are of the same species. But we know $y$ is a tiger, so what species is $z$? is this the same species as $x$?

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  • $\begingroup$ I see so $z$ is also a tiger and therefore the same species as $x$. So for my country example: Reflexive: $x$ is a person and therefore $x$ is the same as $x$?? Symmetry: Assuming $x$ has been to country $C$ and $y$ has been to country $C$, then $x$ and $y$ have both been to country $C$. Transitive: Assuming $y$ has been to country $C$ and $z$ has been to country $C$, then $x$ and $z$ have been to country $z$. Yes? I could be completely wrong about this.. $\endgroup$ – JustAnotherUser Oct 29 '13 at 4:19
  • $\begingroup$ You are good up until the last one. The problem is what you know is: Both $x$ and $y$ have been to Argentina. Both $y$ and $z$ have been to Brazil. Then you can see that we can't say anything about $x$ and $z$, hence this need not be transitive. $\endgroup$ – Deven Ware Oct 29 '13 at 4:31
  • $\begingroup$ Ohhh, ok. So just trying to solve my second example in my head. Would that mean that transitivity could be false using the same reasoning that if $x$ and $y$ contain a common character and $y$ and $z$ contain a common character, that does not necessarily mean that $x$ and $z$ which are strings contain a common character. $\endgroup$ – JustAnotherUser Oct 29 '13 at 4:35
  • $\begingroup$ Yes that is correct. For example, ab and bc contain a common character b. Similarly bc and cd contain a common character c. But ab and cd contain no common characters. $\endgroup$ – Deven Ware Oct 29 '13 at 4:47
  • $\begingroup$ Ok, I'm starting to get it now thank you. $\endgroup$ – JustAnotherUser Oct 29 '13 at 4:54

Maybe you can consider the relation $\sim$ to be friendship and that would help? Let $S$ be the collection of all people living in your city.

Reflexive: Is everyone their own friend?

Symmetry: If Bob is a friend Alice, is Alice necessarily a friend of Bob?

Transitive: If Kyle is a friend of Judith, and Judith is a friend of Joe, is Kyle necessarily a friend of Joe?

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  • $\begingroup$ So would only the reflexive part be true? Or would they all be false. $\endgroup$ – JustAnotherUser Oct 29 '13 at 3:20
  • $\begingroup$ I'm sorry. I guess this was a bad example and it depends on your possibly depressing view of friendship. In general though, I think it is easy to see that friendship is not transitive. Since it fails one of them, the relation $\sim$ is not an equivalence relation. \\ Edit: Maybe try this again in the context of friendship on Facebook. In that case, you cannot friend yourself and Bob couldn't be Alices friend if Alice isn't Bobs friend. So reflexivity and transitivity would fail. $\endgroup$ – JessicaK Oct 29 '13 at 3:30

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