I'm confused on a question... I was thinking about it because I'm wondering if $R$ is transitive if and only if $ R \circ R \subseteq R$.
I don't think it is because for R to be transitive, we need
$(\forall x,y,z \in S)[(x,y) \in R \land (y,z) \in R) \rightarrow (x,z) \in R$
that has three elements x, y, and z

The composite definition for $R_2 \circ R_1$ is
$(x,y) \in S \times S: (\exists v \in S)( (x,v) \in R_1 \land (v,y) \in R_2$

but since I just have $R \circ R$, then
$(x,y) \in S \times S: (\exists v \in S)( (x,v) \in R \land (v,y) \in R$

$R$ is transitive if and only if $ R \circ R \subseteq R$



I was thinking about it because I'm wondering if R is transitive if and only if R∘R⊆R.
I don't think it is because for R to be transitive, we need
(∀x,y,z∈S)[(x,y)∈R∧(y,z)∈R)→(x,z)∈R
that has three elements x, y, and z

The composite definition for R2∘R1 is
(x,y)∈S×S:(∃v∈S)((x,v)∈R1∧(v,y)∈R2

but since I just have R∘R, then
(x,y)∈S×S:(∃v∈S)((x,v)∈R∧(v,y)∈R

Add to the fact that R∘R⊆R so R∘R belongs in R thanks to (∀x)[x∈R∘R→x∈R]

The R is the equivalence relation on a set S. For each element x∈S, the set [x]=[y∈S:(x,y)∈R] is the e…