I am asked to disprove that:
$$\mathcal P(A\cup B) \subseteq \mathcal P(A) \cup \mathcal P(B) …..(1)$$
I am also asked to prove that:
$$\mathcal P(A) \cup \mathcal P(B) \subseteq \mathcal P(A\cup B)…..(2)$$ $$\mathcal P(A\cap B) = \mathcal P(A) \cap \mathcal P(B)…..(3)$$
Where $\mathcal P(A)$ represents the Power Set of $A$
For (1), I have somewhat managed to prove it, so there must be a mistake in my proof:
Let $X \in \mathcal P(A\cup B)$ then by definition $X \subseteq A \cup B$
This means that $X$ is either a subset of $A$, $B$ or $A \cup B$. Therefore $X\subseteq A$ or $X\subseteq B$ and hence $X \in \mathcal P(A) \cup \mathcal P(B)$ (I find something really wrong with this line, but I am not sure where am I wrong at.)
For (2), I did this:
Let $X \in \mathcal P(A) \cup \mathcal P(B)$, then $X\in \mathcal P(A)$ or $X\in \mathcal P(B)$
Case 1: $X\in \mathcal P(A)$, this means $X\subseteq A \implies X\subseteq A\cup B \implies X \in \mathcal P(A\cup B)$
Case 2: $X\in \mathcal P(B)$, this means $X\subseteq B \implies X\subseteq A\cup B \implies X \in \mathcal P(A\cup B)$
So in both cases, the statements are true. The questtion is, do I need to consider $X\in \mathcal P(A) \cap \mathcal P(B)$?
For (3), I did this:
(Backward Direction) Let $X\in \mathcal P(A) \cap \mathcal P(B)$, then $X\in \mathcal P(A)$ and $X\in \mathcal P(B)$. Meaning that $X\subseteq A$ and $X\subseteq B$ so it must be true that $X\subseteq A\cap B$ and hence $\mathcal P(A \cap B)$
(Forward Direction) Let $X \in\mathcal P(A \cap B)$, then $X\subseteq A\cap B$. This means that $X\subseteq A$ and $X\subseteq B$ which implies that $X\in \mathcal P(A) \cap \mathcal P(B)$.
Any help or advice on my way of proving is really appreciated!
Best Answer
When you say
you cannot deduce that $X \subseteq A$ or $X \subseteq B$. You could have some of $X$ be a subset of $A$ and the rest be a subset of $B$. Take $A=\{0,1,2\}, B=\{2,3,4\}, X=\{1,4\}$ for example
For $2$ you don't need to worry about $X \subseteq A \cap B$. It is covered in both your cases.