Suppose $A = \{a, b, c, d\}$ and $\Pi_1$ is the following partition of A
$\Pi_1 = \left\{\left\{a, b, c\right\}\left\{d\right\}\right\}$
List the ordered pairs of the equivalence relations induced by $\Pi_1$.
Draw the graph of the above equivalence relation.
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, $\left\langle\left\{\Pi_1, \Pi_2, \Pi_3, \Pi_4\right\}, \text{ refines } \right\rangle$.
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Interchange
$\Pi_2\ <-> \Pi_4\\ \Pi_1\ <-> \Pi_3$
Tuhin Dutta $\prod$_{3 }refines both $\prod$4 and $\prod$1
am i right?
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?
Hasse diagram
You can interchange $\prod_1$ and $\prod_4$ as they are unrelated
for C)
https://gateoverflow.in/1224/gate2007-26 check this
Gatecse