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The following proof is given to show that $\mathcal P(A) \cup \mathcal P(B) \subseteq \mathcal P(A\cup B)$ ($\mathcal P$ is the power set):

Let $X \in (\mathcal{P}(A) \cup \mathcal{P}(B))$, by definition of union: $$X \in \mathcal{P}(A) \lor X \in \mathcal{P}(B)$$ $\implies$By definition of power set $$X \subseteq A \lor X \subseteq B$$ $\implies$By union transitivity$$X \subseteq A\cup B$$ $\implies$By definition of power set $$X \in \mathcal{P}(A \cup B)$$ Then $\mathcal{P}(A) \cup \mathcal{P}(B) \subseteq \mathcal{P}(A \cup B) \blacksquare$

Now, I thought I could apply the opposite logic to show that $\mathcal P(A) \cup \mathcal P(B) \supseteq \mathcal P(A\cup B)$, and thus prove that $\mathcal P(A) \cup \mathcal P(B)= \mathcal P(A\cup B)$:

Let $X \in \mathcal{P}(A \cup B)$, By definition of power set: $$X \subseteq A\cup B$$ $\implies$ By union transitivity: $$X \subseteq A \lor X \subseteq B$$ $\implies$ By definition of powerset: $$X \in \mathcal{P}(A) \lor X \in \mathcal{P}(B)$$ $\implies$ by definition of union: $$X \in (\mathcal{P}(A) \cup \mathcal{P}(B))$$ Then $\mathcal {P}(A \cup B) \subseteq \mathcal{P}(A) \cup \mathcal{P}(B) \blacksquare$


Multiple posts here prove that for any sets $A$ or $B$, if $\mathcal P(A) \cup \mathcal P(B)= \mathcal P(A\cup B)$ then either $A \subseteq B$ or $B \subseteq A$, so clearly my opposite direction proof can't be true.

What I'd like to understand is which step is based on a false assumtion/implication?

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$X \subseteq A \cup B$ doesn't imply $X \subseteq A$ or $X \subseteq B$.

For example let $A=\{1\}$ and $B=\{2\}$ and $X = A \cup B$.

$X$ is neither subset of $A$ nor subset of $B$.

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Your "union transitivity" step is wrong. That is, if you have $X \subset A \cup B$, then you don't necessarily have $X \subset A$ or $A \subset B$. For example, let $A =\{1\},B=\{2\},X=\{1,2\}$.

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I have lot of books in my shelf: Some of them I bought and the rest are borrowed from others and never returned.

A general subset of books from my shelf need not consist exclusively of borrowed books, or of own books.

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