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Let $R \subseteq A \times A$ and $S \subseteq A \times A$ be two arbitary equivalence relations. Prove or disprove that $R \cup S$ is an equivalence relation.

Reflexivity: Let $(x,x) \in R$ or $(x,x) \cup S \rightarrow (x,x) \in R \cup S$

Now I still have to prove or disprove that $R \cup S$ is symmetric and transitive. How can I do that?

My guess for symmetry is: R and S are equivalence relations, which means that $(x,y)\ and\ (y,x) \in R,S$ For each (x,y) in R and S there is an (y,x) in R and S so that (x,y) ~ (y,x). Is that correct?

Transitivity: ?

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up vote 8 down vote accepted

Transitivity fails. Let $A = \{1,2,3\}$, $R = \{(1,1), (2,2), (3,3), (1,2), (2,1)\}$ and $S = \{(1,1), (2,2), (3,3), (2,3), (3,2)\}$, then $R \cup S$ contains both $(1,2)$ and $(2,3)$, but not $(1,3)$.

The symmetry could be worded better, but is alright. The important thing is that if $(x,y) \in R \cup S$ then $(x,y) \in R$ or $(x,y) \in S$, but both $R$ and $S$ are symmetrical so $(y,x)$ must be contained in at least one of them (here I use "or" instead of "and" you have used).

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Thank you very much :) – user3001 Mar 7 '12 at 12:23
And just to add, though trivial, its reflexice too, I guess. – PardonMeForMySuperPoorMaths Jan 23 '15 at 8:57

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