Nov 18, 2017 · But that simplicity makes graph theory a good place to start understanding the idea of an equivalence relation, so that when you happen across it in a different context, where it might be more useful, you can recognize it and say “oh, we can model this with an equivalence relation, which means that it partitions the structure into components ...

Yes, you're on the right track. "Well-defined" just means that it should be an actual operation that takes 2 equivalence classes and sends them to a (uniquely determined) equivalence class.

If you can determine the equivalence classes geometrically (in this case, the straight lines having slope $1$), and that these sets partition the plane, then you have proved that you have an equivalence relation, without explicitly proving the reflexive, symmetric and transitive properties. More precisely, here is what you need to prove (or argue):

After all, equivalence relations need all three conditions. Your second attempt at symmetry doesn't make it clear where you obtained the constants from, or why they indeed work. Your second (?) attempt at transitivity suffers from a similar problem, but how it follows is a bit less out of the way in this case.

The desired function just needs to give the same output for every element of the same equivalence class. $\endgroup$ – user61527 Commented Aug 27, 2013 at 3:21

$\begingroup$ Any equivalence relation containing R is stable by reflexivity, symmetry and transitivity hence it must contain every such (x,y). On the other hand, the relation S thus defined is an equivalence relation hence S is the smallest equivalence relation containing R. No other reference (sorry). $\endgroup$ –

Dec 29, 2014 · Then the full subcategory of sets equipped with an equivalence relation (called setoids) is reflective in ${\mathscr C}$, with the left adjoint of the inclusion sending an object $(X,\sim)$ to $(X,\simeq)$ where $\simeq$ is the equivalence relation generated by $\sim$.

Jul 7, 2016 · An equivalence class IS the same as a partition, defined by using some equivalence relation. But the quotient is ALL of those equivalence classes (partitions) under that particular equivalence relation. You DO need an equivalence relation to build a quotient set, which is why the notation is S/~, which is read as "the quotient set of the set S ...

How do I approach proving that the relation holds? I understand that I need to prove that it is reflexive, symmetric, and transitive, but I don't entirely understand how to prove each case! equivalence-relations

Dec 9, 2016 · We define cardinality as an equivalence relation on sets. But the class of all sets is not a set, so how do we do that? In particular, I'm interested in the proposition that equivalence classes form a partition of the initial set. It seems like it can be translated to cardinality, but I do not know how, at least in ZFC (and I don't even know ...