- #1
evinda
Gold Member
MHB
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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?
$$x \in \mathcal{P}A \cup \mathcal{P} B \rightarrow x \in \mathcal{P}A \lor x \in \mathcal{P}B \rightarrow x \subset A \lor x \subset B \rightarrow x \subset A \cup B \rightarrow x \in \mathcal{P} (A \cup B)$$
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?