Equivalence closure

When I'm trying to find the Equivalence closure of something, I usually find the reflexive first, then the symmetric then the transitive. But i'm unsure if i should fill out the symmetric closures that may form once I find the transitive closure. So If I do the Transitive closure and get a link (b,d) do I need to also add (d,b) even if there's not a transitive link for it.

Also when finding the Equivalence closure, where R is a relation of X, do you automatically make every element of set X reflexive even if there's no pair featuring that element in R?.

Solution 1:

If you already did the symmetric closure, let's assume that you have now $R$ a symmetric relation and the pair $(a,b)$ is going to be in its transitive closure, say because there are pairs $(a,c)$ and $(c,b)$ in $R$, then the transitive closure will also be forced to have the pair $(b,a)$. This is because in $R$ we have also the pairs $(b,c)$ and $(c,a)$, since $R$ is symmetric.

Solution 2:

If $R'$ is the symmetric closure of a relation $R$ and you then take the transitive closure of $R'$, you will get a symmetric relation, because for each pair $(a, b)$ you added in the second step you will also add $(b, a)$ (by symmetry). In more detail, when forming the transitive closure of $R'$, if you added in $(a, b)$ because there is a chain $a = a_1 \mathrel{R'} a_2 \mathrel{R'} \ldots \mathrel{R'} a_k = b$, then you will also have the chain $b = a_k \mathrel{R'} a_{k-1} \mathrel{R'} \ldots \mathrel{R'} a_1 = a$ (because $R'$ is symmetric), causing you to add in $(b, a)$.

The second part of your question depends on how you are defining the equivalence closure of $R$. If it means the smallest equivalence relation containing $R$ and having some specified set $X$ that contains the field of $R$ as its field, then, yes, you have to add in pairs $(x, x)$ for $x$ that are not in the domain or range of $R$. If it means the smallest relation containing $R$, then you don't have to add anything extra and you will get an equivalence relation whose domain and range are equal to the field of $R$.

[In case you haven't come across the term field of $R$, it means the union of the domain and range of $R$.]