I think option G is false because it has two same elements in the set, which disobeys the distinct property of sets.

3 votes

Best answer

a) true b) true c) false d)true e)true f)true g)false

few points

1) contains means set has element .

2) set A is said to be proper subset of B , if there exist element of B that is not in set A.(A$\subset$ B)

NOW in option a, b,c ,d asking about lhs element contain in rhs side set ...

in option c , lhs itself is element $\left \{ \Phi \right \}$ but rhs has set so its false

in option e ) proper subset of rhs will be $\left \{ \Phi \right \}, \left \{ \left \{ \Phi \right \} \right \}$ so its true

similar f ) is true

for g) we have to find proper subset of $\left \{ \left \{ \Phi \right \},\left \{ \Phi \right \} \right \}= \left \{ \left \{ \Phi \right \} \right \}$

so lhs will never be propersubset its itself set ...

few points

1) contains means set has element .

2) set A is said to be proper subset of B , if there exist element of B that is not in set A.(A$\subset$ B)

NOW in option a, b,c ,d asking about lhs element contain in rhs side set ...

in option c , lhs itself is element $\left \{ \Phi \right \}$ but rhs has set so its false

in option e ) proper subset of rhs will be $\left \{ \Phi \right \}, \left \{ \left \{ \Phi \right \} \right \}$ so its true

similar f ) is true

for g) we have to find proper subset of $\left \{ \left \{ \Phi \right \},\left \{ \Phi \right \} \right \}= \left \{ \left \{ \Phi \right \} \right \}$

so lhs will never be propersubset its itself set ...

0

I think option G is false because it has two same elements in the set, which disobeys the distinct property of sets.

0

(G) is not set according to Wikipedia definition of set is:

In mathematics, a

setis a collection of distinct objects.

Since it is not a set, so operations on sets are also meaningless.