prove that for the equivalence class : X/(X/E) = E











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Let E be an equivalence relation on the set X .
prove that: X / (X / E) = E



The definition of euivalence class: x/E = { for every y in the X | xEy }



The set of all equivalence classes: X/E = { x/E | x is a member of X }










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  • 1




    You defined x/E but did not define X/E.
    – William Elliot
    Nov 23 at 2:07










  • @WilliamElliot Yes you are right. I have edited the information and appended the definition of X/E.
    – Yahya
    Nov 23 at 8:03






  • 1




    Ok. So how is X/E, a collection of subsets of X, an equivalence relation. X/E subset P(X). Equivalence relations for X are subsets of X×X.
    – William Elliot
    Nov 23 at 11:39















up vote
0
down vote

favorite












Let E be an equivalence relation on the set X .
prove that: X / (X / E) = E



The definition of euivalence class: x/E = { for every y in the X | xEy }



The set of all equivalence classes: X/E = { x/E | x is a member of X }










share|cite|improve this question




















  • 1




    You defined x/E but did not define X/E.
    – William Elliot
    Nov 23 at 2:07










  • @WilliamElliot Yes you are right. I have edited the information and appended the definition of X/E.
    – Yahya
    Nov 23 at 8:03






  • 1




    Ok. So how is X/E, a collection of subsets of X, an equivalence relation. X/E subset P(X). Equivalence relations for X are subsets of X×X.
    – William Elliot
    Nov 23 at 11:39













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Let E be an equivalence relation on the set X .
prove that: X / (X / E) = E



The definition of euivalence class: x/E = { for every y in the X | xEy }



The set of all equivalence classes: X/E = { x/E | x is a member of X }










share|cite|improve this question















Let E be an equivalence relation on the set X .
prove that: X / (X / E) = E



The definition of euivalence class: x/E = { for every y in the X | xEy }



The set of all equivalence classes: X/E = { x/E | x is a member of X }







relations equivalence-relations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 23 at 8:01

























asked Nov 22 at 22:13









Yahya

175




175








  • 1




    You defined x/E but did not define X/E.
    – William Elliot
    Nov 23 at 2:07










  • @WilliamElliot Yes you are right. I have edited the information and appended the definition of X/E.
    – Yahya
    Nov 23 at 8:03






  • 1




    Ok. So how is X/E, a collection of subsets of X, an equivalence relation. X/E subset P(X). Equivalence relations for X are subsets of X×X.
    – William Elliot
    Nov 23 at 11:39














  • 1




    You defined x/E but did not define X/E.
    – William Elliot
    Nov 23 at 2:07










  • @WilliamElliot Yes you are right. I have edited the information and appended the definition of X/E.
    – Yahya
    Nov 23 at 8:03






  • 1




    Ok. So how is X/E, a collection of subsets of X, an equivalence relation. X/E subset P(X). Equivalence relations for X are subsets of X×X.
    – William Elliot
    Nov 23 at 11:39








1




1




You defined x/E but did not define X/E.
– William Elliot
Nov 23 at 2:07




You defined x/E but did not define X/E.
– William Elliot
Nov 23 at 2:07












@WilliamElliot Yes you are right. I have edited the information and appended the definition of X/E.
– Yahya
Nov 23 at 8:03




@WilliamElliot Yes you are right. I have edited the information and appended the definition of X/E.
– Yahya
Nov 23 at 8:03




1




1




Ok. So how is X/E, a collection of subsets of X, an equivalence relation. X/E subset P(X). Equivalence relations for X are subsets of X×X.
– William Elliot
Nov 23 at 11:39




Ok. So how is X/E, a collection of subsets of X, an equivalence relation. X/E subset P(X). Equivalence relations for X are subsets of X×X.
– William Elliot
Nov 23 at 11:39















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