Proving Power Set Notation: Let B \subseteq U

[sssssssssss]

my actual problem is to let B be a subset of the set U and prove

P(B$$^{C}_{U}$$) $$\neq$$ (P(B))$$^{C}_{P(U)}$$

but I am confused on the scripts and not quite sure what they are wanting me to do

i have Let B $$\subseteq$$ U where B = {b} and U = {B}
I know P(B) = {empty set, {b}} and P(U) = {empty set, {B}}

i know superscript c means compliment, but i don't know what the subscript u means. Is it similar to an index?
am i suposed to assume that U means universal. i just don't know the next thought that i need.

 

[AKG]

That's not a standard notation, but I assume that the subscript tells you the universal set with respect to which you are supposed to take complements. In other words, they want you to show:

$$\mathcal{P}(U \setminus B) \neq \mathcal{P}(U) \setminus \mathcal{P}(B)$$

 

[sssssssssss]

Alright I figured it out i think.

Let B $$\subseteq$$ U
Set B = {b} and U = {P(B), u}
P(B) = {empty set, {b}}
P(U) = {empty set, {empty set, {b}}, {u}, {{empty set, {b}}, u}}

Then i figured It was asking for the elements that are in set U that arent in set B

P(B$$^{c}_{U}$$) = {empty set, {u}}

And then i figured this was asking for the elements that are in P(U) that arent in P(B)

(P(B))$$^{c}_{P(U)}$$ = {{empty set, {b}}, {u}, {{empty set, {b}}, u}}

So therefore, {empty set, {u}} $$\neq$$ {{empty set, {b}}, {u}, {{empty set, {b}}, u}}

correct me if I am wrong por favor.

 

[sssssssssss]

That's not a standard notation, but I assume that the subscript tells you the universal set with respect to which you are supposed to take complements. In other words, they want you to show:

$$\mathcal{P}(U \setminus B) \neq \mathcal{P}(U) \setminus \mathcal{P}(B)$$

i know! its so frustrating because my math teacher uses his own notation which is extremely dificult to decipher cause the book he assigned is different, and the internt has been consistently different!

(i actually got it before I saw you posted but thanks for the reassurance!)

 

[AKG]

Alright I figured it out i think.

Let B $$\subseteq$$ U
Set B = {b} and U = {P(B), u}

You can't just set B = {b}. It's like being asked to prove "the square of an even integer is even" and starting your proof by saying, "let's set n = 4." You also can't set U = {P(B), u}. Firstly, for the same reason as before, since you need to prove things in general, and not in some particular case where you set things to be very simple. Secondly, because for U to be the universal set, you need B to be a subset of U, and B is (generally) not a subset of {P(B), u}.

 

[disregardthat]

Hint: Find a general common element of P(B) and P(U/B). The exercise is probably what AKG suggests.