# difficult question

#### pka

##### Elite Member
Here is your post: $$x,y$$ aren't empty, infinity series and $$x\cap y=\emptyset$$.
Show us $$f(x\cup y)\approx f(x)\times f(y)$$.
That makes no sense whatsoever. You need to post a well formed question in English.
You must explain the terms used and the object of the question.

#### KerpetenAli

##### New member
I'm sorry I couldn't explain the question correctly. Original question: (P= power set)

#### Jomo

##### Elite Member
So what have you tried, where are you stuck? What does P( X U Y) mean? How about P(X)xP(Y)?

#### pka

##### Elite Member
I'm sorry I couldn't explain the question correctly. Original question: (P= power set)
Well that is somewhat better but only by a bit.
In the LaTeX markup language $$\approx$$ means approximately.
Moreover, in almost all western mathematics if $$A~\&~B$$ are sets, the notation $$A\times B$$ is the cross product of the two sets.
As posted that makes no sense. Please define all your terms.
If necessary go to some standard website such as this or even this one.
Type in key words to find equivalent standard western notation.

#### Dr.Peterson

##### Elite Member
I'm sorry I couldn't explain the question correctly. Original question: (P= power set)
I presume you are to show that $$\displaystyle P(X\cup Y)\approx P(X)\times P(Y)$$ in some sense, saying that the power set of the union is ... what? ... to the Cartesian product of the individual power sets. What does "$$\displaystyle \approx$$" mean to you here? Please define it. I can see a couple kinds of relationship that would apply, but not "approximately equal".

#### pka

##### Elite Member
I'm sorry I couldn't explain the question correctly. Original question: (P= power set)
This is a dangerous thing to do: I am going to guess that you are asked to show that
$$\mathscr{P}(X\cup Y)$$ equipotent with $$\mathscr{P}(X)\times\mathscr{P}(Y)$$. [Those are power sets.]
Sets $$U~\&~V$$ are equipotent iff there are injections $$U\to V~\&~V\to U$$.
If $$A\in\mathscr{P}(X\cup Y)$$ then $$A\subseteq X\cup Y$$. Let us define $$A_X=A\cap X~\&~A_Y=A\cap Y$$.
Can you argue that $$A_X\cap A_Y=\emptyset~?$$
Can you show that $$\Phi(A)=(A_X,A_Y)$$ is an injection $$\mathscr{P}(X\cup Y)\to\mathscr{P}(X)\times\mathscr{P}(Y)$$?
Now can YOU find an injection $$\mathscr{P}(X)\times\mathscr{P}(Y) \to \mathscr{P}(X\cup Y)~?$$