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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?

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    $\begingroup$ This is not an equivalence relation. $\endgroup$ – Mr. X Dec 3 '17 at 17:31
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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).

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  • $\begingroup$ 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 $$\text{reflexive $\land$ symmetric} \implies \lnot \text{transitive}$$ $\endgroup$ – Andrew Tawfeek Dec 3 '17 at 20:05
  • $\begingroup$ @AndrewTawfeek I can deduce it is not transitive, otherwise it would be an equivalence relation (added a parenthetical comment). Downvoters that don't tell why they downvoted a correct answer should be ashamed of themselves. But they won't, I'm afraid. $\endgroup$ – egreg Dec 3 '17 at 20:30
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    $\begingroup$ @AndrewTawfeek My aim was to tell the OP that his/her analysis is correct, contrary to the other answer that doesn't even attempt at it. $\endgroup$ – egreg Dec 3 '17 at 20:35
  • $\begingroup$ Ahh, my mistake! +1 :) $\endgroup$ – Andrew Tawfeek Dec 3 '17 at 20:39
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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.

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