@anyone please edit the power set .as

{ø, {a},{b}, {a,b}}

{ø, {a},{b}, {a,b}}

14 votes

$S=\{(1,2), (2,1)\}$ 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 ($\square(S), \subseteq$)

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 ($\square(S), \subseteq$)

19 votes

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

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)}

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)}

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 ?

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 ?