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Suppose $A = \{a, b, c, d\}$ and $\Pi_1$ is the following partition of A

$\Pi_1 = \left\{\left\{a, b, c\right\}\left\{d\right\}\right\}$

1. List the ordered pairs of the equivalence relations induced by $\Pi_1$.

2. Draw the graph of the above equivalence relation.

3. Let $\Pi_2 = \left\{\left\{a\right\}, \left\{b\right\}, \left\{c\right\}, \left\{d\right\}\right\}$

$\Pi_3 = \left\{\left\{a, b, c, d\right\}\right\}$

and $\Pi_4 = \left\{\left\{a, b\right\}, \left\{c,d\right\}\right\}$

Draw a Poset diagram of the poset, $\left\langle\left\{\Pi_1, \Pi_2, \Pi_3, \Pi_4\right\}, \text{ refines } \right\rangle$.

recategorized | 1.8k views
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.

................................

Interchange

$\Pi_2\ <-> \Pi_4\\ \Pi_1\ <-> \Pi_3$

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Tuhin Dutta $\prod$3 refines both $\prod$4 and $\prod$1

am i right?

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a. For Calculating Ordered Pairs Just multiply the partition with itself.

$\{a,b,c\} * \{a,b,c\} = \{ (a,a),(a,b),(a,c),(b.a),(b,b),(b,c),(c,a),(c,b),(c,c)\}$

$\{d\}*\{d\} = \{(d,d)\}$

$\therefore$ The ordered pairs of the equivalence relations are $= \{ (a,a),(a,b), (a,c), (b,a), (b,b), (b,c), (c,a), (c,b), (c,c),(d,d) \}$

b. For each ordered pair $(a,b)$ in the equivalence relation just make a directed edge from a towards $b$.

c. Suppose we have two partitions of a set $S:P_1={A_1,A_2,A_3...}\ and\ P_2={B_1,B_2,B_3...}$

We say that $P_1$ is a refinement of $P_2$ if every $A_i$ is a subset of some $B_j$ . Or We can say that $P_1$ refines $P_2$

by Boss (19k points)
edited by
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@Satbir Not the literal meaning. What is the concept behind listing the ordered pairs as you did?

My main doubt here is why are we not multiplying {a,b,c} with {d}.

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Got it. In a partition, all the elements are related to each other and in two different partitions, no two elements are related to each other.
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@Satbir In the last statement where you have written If P1 is a refinement of P2 then it means P1 refines P2.

Is this correct or do you mean P2 refines P1?

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first statement
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partition of the given interval, Q, is defined as a refinement of the partition, P, when it contains all the points of P and possibly some other points as well; the partition Q is said to be “finer” than P. Given two partitions, P and Q, one can always form their common refinement, denoted P ∨ Q, which consists of all the points of P and Q, re-numbered in order.

So in last statement it should be p2 refines p1

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We say P1 refines P2 or P1 is finer than P2 if every partition of P1 is subset of some partition of P2.

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suppose P = {A,B} and Q = {{A},{B}}

then Q refines P or we can write Q is a refinement of P.
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@Satbir Yes, You can write so...because every partition of Q is subset of partitions of P. And now you have corrected the statement in your answer.

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YES ...i got little confused with harshpeswani's comment.
(a) The ordered pairs of the equivalence relations induced $= \{ (a,a),(a,b), (a,c), (b,a), (b,b), (b,c), (c,a), (c,b), (c,c),(d,d) \}$

PS: equivalence relations = each partition in power set - $\phi$
by Active (2.7k points)
+20
we can find the same by { {a,b,c} ${\times }$ {a,b,c} , {d} ${\times }$  {d} }

= { (a,a), (a,b), (a,c), (b,a), (b,b), (b,c), (c,a), (c,b) ,(c,c), (d,d) }
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what about (b) and (c) ?
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c.) Make the graph according to the given connected vertices below.
PI1----->PI3
PI4----->PI3
PI2----->PI4
PI2----->PI1
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Equivalence relation is reflexive, symmetric and transitive.

But that does not means that all symmetric and transitive pairs should be there, right? We can just have
reflexive pairs also, right? Like this: $\{(a,a),(b,b),(c,c),(d,d)\}$

Will it make difference if we notice that the question does not at all make use of word "equivalence class", but use word "equivalence relation", what you say? I guess it makes difference if we think of the following three points:

1. "equivalence relation" has no such restriction that each elements in the set should be related to each other, which is the restriction put by "equivalence class".

2. Set of all "equivalence classes" form partition, but the converse is not true, that is

3. Partition is made of sets which may or may not be "equivalence classes". The only criteria is that they should be disjoint and their union should be original set.

I guess after considering these facts, we are allowed to have only reflexive pairs, or even some symmetric and transitive pairs, if not all, right?

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Hasse diagram

You can interchange $\prod_1$ and $\prod_4$ as they are unrelated

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@Ayush I am not understanding the a) part of the question please explain it
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@Prince -You must be knowing the two way theorem

Every Equivalence relation induces a partition on a set if and only if Every partition of the set induces an equivalence relation on the set.

Now to find the ordered pairs from the partition, simply take cross product of the partitions among themselves.
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A partition α of a set X is a refinement of a partition ρ of X—and we say that α is finerthan ρ and that ρ is coarser than α—if every element of α is a subset of some element of ρ. Informally, this means that α is a further fragmentation of ρ. In that case, it is written that α ≤ ρ.

https://en.wikipedia.org/wiki/Partition_of_a_set#Refinement_of_partitions

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@Ayush Upadhyaya you mean to say cross product of equivalence classes? coz equivalence relation divide set into disjoint classes.

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Yes, and those classes form the biggest equivalence relation possible on the set.
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,how we will draw part (b)

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@codingo1234-Do you know what refines relation means?

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@,earlier I was not knowing what "refines relation" means ,but I saw 's comment and came to know about when a partition is refinement of other

"A partition α of a set X is a refinement of a partition ρ of X—and we say that α is finer than ρ and that ρ is coarser than α—if every element of α is a subset of some element of ρ. Informally, this means that α is a further fragmentation of ρ. In that case, it is written that α ≤ ρ." ,

but I was confused regarding drawing graph of equivalence relation of part(a) and I was thinking about hasse digaram(god knows  why everything mixes up in mind),I think it will be directed graph,am i correct?

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@Ayush Upadhyaya Can you please provide a resource to read two-way theorem

by Active (2.6k points)

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