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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, $\left\langle\left\{\Pi_1, \Pi_2, \Pi_3, \Pi_4\right\}, \text{ refines } \right\rangle$.

asked in Engineering Mathematics by Veteran (59.6k points)
recategorized by | 963 views
+6

.

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

Interchange

$\Pi_2\ <-> \Pi_4\\ \Pi_1\ <-> \Pi_3$

0

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

am i right?

2 Answers

+6 votes
(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 - $\phi$
answered by Active (2.5k points)
+13
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) }
0
what about (b) and (c) ?
0
c.) Make the graph according to the given connected vertices below.  
PI1----->PI3
PI4----->PI3
PI2----->PI4
PI2----->PI1
0

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?

+2

Hasse diagram

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

0
@Ayush I am not understanding the a) part of the question please explain it
+1
@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.
0

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

+3 votes
answered by Active (2.4k points)


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