# Sets and relation

An equivalence relation is defined as{(1,1) (2,2) (3,3) (4,4) (5,5) (1,2) (2,1) (2,3) (3,2)}

[1]={1,2} [2]={2,1,3} Clearly [1] is not equal to[2] but they have intersection as 2 is common in both.How it is possible?

• This is not an equivalence relation. – Mr. X Dec 3 '17 at 17:31

Let's call this relation $R$ over the set $A=\{1,2,3,4,5\}$; if it is an equivalence relation, then two subsets of the form $[x]=\{a\in A:x\mathrel{R}a\}$ are either equal or disjoint. Now $$[1]=\{1,2\} \qquad [2]=\{2,3\} \qquad [3]=\{2,3\} \qquad [4]=\{4\} \qquad [5]=\{5\}$$ Since $[1]$ and $[2]$ are neither equal nor disjoint, we conclude that $R$ is not an equivalence relation.
• I don't see why this answer is downvoted. Anyhow, I recommend changing the phrase "so it is not transitive" to "but it is not transitive" in your final sentence because it seems like you're stating $$\text{reflexive \land symmetric} \implies \lnot \text{transitive}$$ – Andrew Tawfeek Dec 3 '17 at 20:05
It is not an equivalence relation. It is not transitive: We have $(1,2)$ and $(2,3)$ but not $(1,3)$. So, no contradiction. You just don't have an equivalence relation.