Sets and relation

Question

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?

Answer 1

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.

The relation is clearly reflexive and symmetric, so it is not transitive (otherwise it would be an equivalence relation).

Answer 2

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.