Question

What is wrong with this proof of the following incorrect theorem?

P(A U B) ⊆ P(A) U P(B)

let W ∈ P(A U B), then W ⊆ A U B.
Let x ∈ W, then x ∈ A U B

Case 1: suppose x∈A.
Since x∈W and x∈A, W⊆A.
W ∈ P(A)
W ∈ P(A) U P(B)

Case 2: suppose x∈B.
Since x∈W and x∈B, W⊆B.
W ∈ P(B)
W ∈ P(A) U P(B)

Therefor W ∈ P(A) U P(B). Hence,
P(A U B) ⊆ P(A) U P(B)

Answer 1

btw, P(A) means the power set of A

Answer 2

What is your understanding of a power set? State the definition.

Answer 3

set of all subsets

Answer 4

when it says w is in A, you never talk about little w

Answer 5

what is little w

Answer 6

oh that's a typo. w and W are the same. I just forgot to capitalize it.

Answer 7

that line does not make sense to me
\(x\in W, W\in A, W\subset A\)

Answer 8

are you trying to prove this or find what is wrong with the proof?

Answer 9

>.< that's also a typo. It was supposed to be Since x∈W and x∈A, W⊆A.

Answer 10

let me edit the question real quick

Answer 11

but P(A U B) ⊆ P(A) U P(B) is an incorrect theorem. I'm supposed to find where the mistake is.

Answer 12

sec

Answer 13

P(A) U P(B) ⊆ P(A U B) is true
but, P(A U B) ⊆ P(A) U P(B) is false

Answer 14

yah yah I see

Answer 15

ok :D

Answer 16

x in A and x in W does not imply W subset of A, W could have more than A in it

Answer 17

ah I see

Answer 18

Like this perhaps?

Answer 19

well take \(\{1,2,3,4,5,\}\cup \{6,5,7,8,9\}\)
Let \(W = {\{4,5,6\}}\)
then let \(x\in W\) then \(x\in A\) but \(W\cancel\subset A\)

Answer 20

so yes like that:)

Answer 21

ok makes more sense now. Thank you (^.^)b

Answer 22

npz