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How many equivalence relations on the set $\{1,2,3\}$ containing $(1,2)$ and $(2,1)$ are there in all? Justify your answer.

My try: Okay, so how can I possibly write all the possible equivalence relations and not just of the condition specified above? And then, how would my answer be reduced if I use the condition given in the question?

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  • $\begingroup$ You can actually write out all relations $(1,1),(1,2),(1,3),(2,1),\cdots$ and you can deduce from being an equivalence relation which of these are explicitly needed automatically. Knowing you have $(1,1),(2,2),(3,3),(1,2),(2,1)$ can you determine what relations contain all of these? $\endgroup$ – user233746 Jan 14 '17 at 3:44
  • $\begingroup$ Re-read the def'ns of equivalence relation and equivalence class. Always use the def'ns. $\endgroup$ – DanielWainfleet Jan 14 '17 at 6:44
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Since this is an equivalence relation it must be reflexive symmetric and transitive. So we now observe that in order that the relation is equivalent it will either contain both the given elements or neither of the elements. Also the relation must be reflexive, so it contains $(1,1)$,$(2,2)$ and $(3,3)$ compulsorily. Due to its equivalent condition, realise that there are only $3$ more choices as each choice involves the inclusion of a pair or a triplet, namely $(1,2)(2,1)$, $(2,3)(3,2)$ and $(1,3)(3,1)$.

And if one of the last $2$ choices happen along with the first choice, then all $3$ choices happen due to the equivalence condition.

So , analysing case by case, your answer is $2$.

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  • $\begingroup$ Thanks for help, but how to write an equivalence relation exactly? Okay I get $(1,1),(2,2),(3,3)$ in all cases as all equivalence relations are reflective. Similarly you should also get $(1,2),(2,1)$ and $(2,3),(3,2)$ and $(1,3),(3,1)$ in all cases as all symmetric relations are equivalent. So how is the answer two? $\endgroup$ – Zlatan Jan 14 '17 at 4:06
  • $\begingroup$ Would $(1,1),(2,2),(3,3)$ be an equivalence relation? Because I do not think it is transitive. So applying the condition of my question, Shouldn't the answer be : case 1 - $(1,1),(2,2),(3,3)$ along with $(1,2),(2,1)$. case 2: $(1,1),(2,2),(3,3)$ along with $(1,2),(2,1)$ and $(2,3),(3,2)$ and case 3:$(1,1),(2,2),(3,3)$ along with $(1,2),(2,1)$ and (2,3),(3,2)$ and $(1,3),(3,1)$. So shouldn't be my answer be three? $\endgroup$ – Zlatan Jan 14 '17 at 4:10
  • $\begingroup$ Let me answer you part by part: 1) Yes it will be an equivalent relation. Check why. 2) Case 2 and case 3 are not equivalent as they are not transitive. So the answer is 2. $\endgroup$ – SchrodingersCat Jan 14 '17 at 11:07
  • $\begingroup$ Okay if a set has only $(1,1),(2,2),(3,3)$ it will be reflexive,symmetric and transitive? Also, case 2 is missing $(1,3)$ as one I the essential terms so it isn't transitive, but case3 looks good to me? What's missing in that to call it non-transitive? $\endgroup$ – Zlatan Jan 14 '17 at 12:39
  • $\begingroup$ Again, lets go for a part by part answer. 1) Yes it is. If you are not sure why, I'll say check the definition of reflexive, symmetric and transitive. STRICTLY check the definitions. The answer lies there. 2) Okay perhaps I saw a different version of the case 3. It looks okay now. So you get the answer is 2. $\endgroup$ – SchrodingersCat Jan 14 '17 at 14:26
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1 and 2 are already in the same cell. So either there is only one cell (123) or else two cells (12)(3). Each of these partitions gives a unique equivalence relation by the standard definition. So there are 2 equivalence relations.

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  • $\begingroup$ Could you explain the reasons? And what's the standard definition exactly? Also have a look at my comments below schrodingerscat's answer. $\endgroup$ – Zlatan Jan 14 '17 at 4:12
  • $\begingroup$ A partition of a set $A$ is an expression of it as a pairwise disjoint union of nonempty subsets often called the "cells" of the partition. There may be only one cell, $A$ itself, or may be more than one cell. Then the standard definition of the equivalence relation for the given partition is that, given $x,y$ in $A,$ we define $(x,y) \in R$ to mean that $x,y$ are in the same cell of the partition. From this definition, reflexive, symmetric and transitive follow immediately. Using this partition approach for your problem shows only two possible partitions. $\endgroup$ – coffeemath Jan 14 '17 at 14:50
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I can imagine two possible avenues, one specific to equivalence relations and another introducing a generally useful mathematical concept.

For the first, recall that an equivalence relation specifies a partition of the ground set into the equivalence classes (the quotient set). Conversely, every partition of the set specifies an equivalence relation (prove this if necessary). Now, the problem is reduced to enumerating the partitions compatible with the equivalences you already have. [This is what coffeemath is suggesting.]

For the second, take the equivalences you already have, and then for each axiom, try to form more equivalences that are required to hold. Rinse and repeat until no more equivalences have to be added. What you have just formed is called an equivalence relation generated by a given set, and is (more or less by construction) the smallest equivalence relation containing it. (Excercise: The usual definition is “the intersection of all equivalence relations containing the given set”. Prove that such relations exist, that the result is an equivalence relation, that it is indeed minimal, and convince yourself that the definitions are equivalent.) This is one possible equivalence relation satisfying the given conditions. Now you can try adding some more pairs to the result and repeating the procedure, etc., until you can generate no more different equivalence relations.

A gizmo generated by a set of initial things is a very common idea. “Take some things and add more until axioms hold” is the intuitive approach, “intersect all the gizmos containing the given things” is the usual definition. The Tao is actually the universal property, “the minimal gizmo containing the given things”, but unlike these definitions, it is not immediately apparent that such a gizmo exists at all. (The first definition is just messy to state formally, and the second needs a proof some such gizmos exist and that their intersection is, indeed, a gizmo, for each kind of gizmo.)

The first examples encountered during one’s education are usually the linear span of a set of vectors and the direct sum of subspaces (see esp. the external direct sum). Later on, there are free groups, free vector spaces and so on; “free” and “generated by” are the usual terms. For inspiration and/or a scare, see the general definition of a “free object” on Wikipedia and nLab. Quotients of free objects are also very commonly used to construct gizmos not only containing something, but also satisfying additional properties, e. g. tensor products, exterior products, Clifford algebras and countless others.

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