Is P(E) U P(F) Equal to P(E U F)?

[alexs1000]

Homework Statement

Prove that(power set) P(E) U P(F) is a subset of P(E U F)

Homework Equations

P(E) U P(F) is a subset of P(E U F)

The Attempt at a Solution

P(E)U P(F)={x:xεP(E) or xεP(F)}
but P(E)={X:X is a subset of E} or P(E)={x:xεX→xεE}
so we get P(E)U P(F)={x:xεX→xεE or xεX→xεF}
but logical implication p→q⇔non(p) or q
so we get P(E) U P(F)={x:non(xεX) or xεE or non(xεX) or xεF}
therefore by the idempotence property and comutativity of the logical operators
P(E) U P(F)={x:non(xεX) or xεE or xεF}
and we get P(E) U P(F)={x:non(xεX) or xεE U F}
which is exactly P(E U F).My question is why is it that only the inclusion stands, and why not equality also.Thank you.

 

[PeroK]

Homework Statement

Prove that(power set) P(E) U P(F) is a subset of P(E U F)

Homework Equations

P(E) U P(F) is a subset of P(E U F)

The Attempt at a Solution

P(E)U P(F)={x:xεP(E) or xεP(F)}
but P(E)={X:X is a subset of E} or P(E)={x:xεX→xεE} <== This is not correct.
so we get P(E)U P(F)={x:xεX→xεE or xεX→xεF} <== Nor this

You've gone wrong here. Power sets are collections of sets, so you have to be able to work with sets as elements to do this correctly.

To show equality does not hold, you could find an example where it doesn't hold.