Prove that P (A ) ⊆ P ( B ) if and only if A ⊆ B . [duplicate] up vote 8 down vote favorite This question already has an answer here: • Prove/Disprove that if two sets have the same power set then they the same set 3 answers Here is my proof, I would appreciate it if someone could critique it for To prove this statement true, we must proof that the two conditional statements !If P( A )⊆P( B ) , then A ⊆ B ,! and, If A ⊆ B , then P( A)⊆P( B ) " are true# $ontrapositive of the first statement: If A⊈ B , then P( A)⊈P( B ) If A ⊈ B , then there must %e some element in A , call it x , that is not in B : x ∈ A , and x ∉ B # &ince x ∈ A , then { x }∈P( A) ' moreover, since x ∉ B , then { x }∉P( B ) , which proves that, if A⊈ B , then P( A )⊈P( B ) # (y provin) the contrapositive true, the ori)inal proposition must %e true#
Prove that P(A)P(B) if and only if AB. [duplicate]up vote 8 down
vote favorite This question already has an answer here:
Prove/Disprove that if two sets have the same power set then they
are the same set 3 answers Here is my proof, I would appreciate it
if someone could critique it for me:To prove this statement true,
we must proof that the two conditional statements ("If P(A)P(B),
then AB," and, If AB, then P(A)P(B)) are true.
Contrapositive of the first statement: If AB, then P(A)P(B)If
AB, then there must be some element in A, call it x, that is not in
B: xA, and xB. Since xA, then {x}P(A); moreover, since xB, then
{x}P(B), which proves that, if AB, then P(A)P(B). By proving the
contrapositive true, the original proposition must be true.
