MHB Why does the inequality stand if there are no common elements?

[evinda]

Hi! (Smirk)

$$x \in \mathcal{P}A \cup \mathcal{P} B \rightarrow x \in \mathcal{P}A \lor x \in \mathcal{P}B \rightarrow x \subset A \lor x \subset B \rightarrow x \subset A \cup B \rightarrow x \in \mathcal{P} (A \cup B)$$

So, $\mathcal{P}A \cup \mathcal{P}B \subset P(A \cup B) $.

The equality stands, if $A \cap B=\varnothing$.

Could you explain me why the equality stands, if $A,B$ have no common elements?

 

[Evgeny.Makarov]

So, $\mathcal{P}A \cup \mathcal{P}B \subset P(A \cup B) $.

The equality stands, if $A \cap B=\varnothing$.

You can check that this is not so by picking $A=\{1\}$ and $B=\{2\}$. In fact, $\mathcal{P}A\cup\mathcal{P}B=\mathcal{P}(A\cup B)$ iff $A\subseteq B$ or $B\subseteq A$.