Let $X$ be an arbitrary set and $\mathcal{P}(X)$ be the power set of $X$. Prove that for every $A,B \in \mathcal{P}(X)$ we have $A\cup B \in \mathcal{P}(X)$.

By the definition of the power set we know that $A,B \subseteq X$. Let's use the definition of subset:

  • $A, B \subseteq X \iff (\forall \space x \in A, y \in B \implies x, y \in X)$

We also have per definition that $\forall A,B \subseteq X$:

  • $A \cup B = \{ x \in X | \space x \in A \lor x \in B\}$

Let $z \in A \cup B$. Per definition, we know that $\space z \in A \lor z \in B$. Since we have $(\forall \space x \in A, y \in B \implies x, y \in X)$, we also know that $z \in X$.

Here I'm stuck. I don't know how to continue with this proof. Intuitively the result is logical but I don't know how to make something useful out of $z$.


3 Answers 3


Don't forget what you're trying to prove. You're trying to prove that $A\cup B\in\mathcal{P}(X)$. By definition of $\mathcal{P}(X)$, this just means that $A\cup B\subseteq X$. And by definition of $\subseteq$, this means that for any $z\in A\cup B$, $z\in X$.

So what you need to prove is that for any $z\in A\cup B$, $z\in X$. But that's exactly what you did prove! So you're done.


You've shown that for every $z \in A\cup B$ we have $z\in X$. According to your definition of a subset, this means that $A\cup B\subseteq X$. Then according to your definition of the power set, $A\cup B \in \mathcal{P}(X)$.

Nice job!


Too much notation, which distracts. If $A$ and $B$ are subsets of $X$, then each element of their union lies in at least one of the two subsets, hence lies in $X$. Done.


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