RelationsWatch

#1
0
4 years ago
#2
a and b can be the same number. For example think about the = relation on the reals. Clearly a~b if and only if a=b.

You are correct in saying it is reflexive as for any element in the set we have a~a.

Given the relation a~b then we know since the relation set is 1,1 2,2 3,3 4,4 that if a~b then a=b and it must be symmetric by the first property.

I believe you can apply the same argument for the proof of transitivity i.e if a~b and b~c then a=b=c and so a~c by the first property.
0
4 years ago
#3
(Original post by Monaa123)
In fact, this has defined the finest equivalence relation on any set (in the sense that it has the most equivalence classes). is an equivalence relation on any set, which is kind of the aim of this question. Intuitively it's clear, because we can quotient out by equivalence relations, and there's only one obvious thing we can quotient by to get the original set: namely, "identify every element with itself only".

ETA: "For all x,y in X" does mean "x may equal y". By convention, "for all distinct x,y" means "x may not equal y", but I prefer "for all x, y in X with x not equal to y", being the clearest.
1
#4
(Original post by Smaug123)
In fact, this has defined the finest equivalence relation on any set (in the sense that it has the most equivalence classes). is an equivalence relation on any set, which is kind of the aim of this question. Intuitively it's clear, because we can quotient out by equivalence relations, and there's only one obvious thing we can quotient by to get the original set: namely, "identify every element with itself only".

ETA: "For all x,y in X" does mean "x may equal y". By convention, "for all distinct x,y" means "x may not equal y", but I prefer "for all x, y in X with x not equal to y", being the clearest.
Aaah okay, I get you. Thanks!!
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