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I have to solve a problem which involves sets, and I am unsure of how to interpret it:

$$\mathcal{P}(M) = \{A : A\subset M\}$$

$$f: \mathcal{P}(M\cup N) \to \mathcal{P}(M) \times \mathcal{P}(N)$$

$$A \to (A\cap M,A\cap N)$$

I know and understand the symbols of union, intersection and "is a subset of", but I don't grasp how to read them in this context. Can someone give me some help?

Thank you very much in advance.

J

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$\mathcal{P}(M) = \{A : A\subset M\}$ is the set equal to the subsets of $M$. I imagine that this is OK for you. Right?

Now, $f$ is a map that associates to a subset $A$ of the union of $M$ and $N$ a couple of subsets. The first element of the couple is a subset of $M$, the second one a subset of $N$. The most important is to convince yourself that $f$ is well defined. I mean by well defined that $f$ maps correctly a subset of $M \cup N$ to a couple of sets belonging to $\mathcal{P}(M) \times \mathcal{P}(N)$.

This is indeed the case because if $A$ is a subset of $M \cup N$, then $A \cap M$ is indeed a subset of $M$ and $A \cap N$ a subset of $M$.

Hope that it helps!

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    $\begingroup$ If the 3rd line is referring to the function f of the 2nd line then it means that $\forall A\in (M\cup N) , [ f(A)= (A\cap M , A\cap N) ].$ $\endgroup$
    – DanielWainfleet
    Oct 21, 2015 at 20:45
  • $\begingroup$ Thank you very much both for your answers, I appreciate it a lot. Am I right in this intepretation then? P(M∪N) = {A : A ⊂ (M∪N)} $\endgroup$
    – Pxx
    Oct 21, 2015 at 21:44
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    $\begingroup$ @Julien Yes, that's right. $\endgroup$
    – mathcounterexamples.net
    Oct 22, 2015 at 1:57
  • $\begingroup$ And so P(M) x P(N) = (A intersection with M) x ( A intersection with N) = (AxM) intersection with (AxN)? (Sorry for not using the symbols, I am writing from a tablet) $\endgroup$
    – Pxx
    Oct 22, 2015 at 7:45
  • $\begingroup$ I have to show that f is injective. Now I know I must show that for y = y' and f (x) = f (x'), x = x'. It is a bit more obscure for me when working with sets: can I just say that f (x) = {y € P (M) x P (N) : there exists at least one x € P (MuN) so that y = f (x)} = {f (x), x € P (MuN)}. I do the same thing with x',y': f (x') = {f (x'), x' € P (MuN)}. In my notes, it seems to be enough to prove that if f (x) = f(x'), then x = x', then the function is injective. What do you guys think? $\endgroup$
    – Pxx
    Oct 22, 2015 at 8:14

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