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$S=\{(1,2), (2,1)\}$ is binary relation on set $A = \{1,2,3\}$. Is it irreflexive? Add the minimum number of ordered pairs to S to make it an equivalence relation. Give the modified $S$.

Let $S=\{a,b\}$ and let $\square(S)$ be the powerset of $S$. Consider the binary relation '$\subseteq$ (set inclusion)' on $\square(S)$. Draw the Hasse diagram corresponding to the lattice ($\square(S), \subseteq$)
in Set Theory & Algebra
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Best answer

$S=\{(1,2), (2,1)\}:$ This relation is Irreflexive, Symmetric, Not Transitive, Not Reflexive, Not Asymmetric, Not antisymmetric.

Equivalence Relation $\implies$ Symmetric, Transitive, Reflexive.

It is not transitive & Reflexive.

Reflexive closure of $S = \{ (1,1),(2,2),(3,3),(1,2),(2,1) \}$

After taking transitive closure relation does not change.

Answer: $S = \{ (1,1),(2,2),(3,3),(1,2),(2,1)\}$


$S = \{a,b\}$

$P(S) = \{ \emptyset,{a},{b},{a,b}\}$

Related Hasse Diagram

 


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3
@anyone please edit  the power set .as

{ø, {a},{b}, {a,b}}
0
Why are taking the closure like transitive, reflexive symetric.? is this set{(1,2),(2,1),(1,1)} not an equivalnce ?
1
I don't think Hasse diagram Has directions !
0
No its not because look at other case where in set {(2,1),(1,2),(1,1)} here 1 is 2 is related to 1 and 1 is related to 2 so for transitivity there should be a 2 related to 2..
10 votes
S={(1,2), (2,1)} is irreflexive as well as symmetric but not transitive..
to make equivalence relation we need to add Reflexive, symmetric, transitive closure to it.
reflexive closure = {(1,1,) (2,2) (3,3)}
Symmetric Closure = { }
Transitive Closure = {1,1 }
equivalence relation = {(1,1,) (2,2) (3,3) (1,2) (2,1)}

edited by
1
Hi @laser , is the original relation transitive ? ( i am not talking about modified relation ).

In the original relation , { (1,2), (2,1) } : it is (a) irreflexive , (b) symmetric , but is it transitive ?

I know null set is also transitive . But , in the given original set had there been an element (1,1) , it would have been transitive. Am I wrong ?
1
corrected !!!

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