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In our homework we are asked to prove $\mathcal P(A) \cup \mathcal P(B) \subseteq \mathcal P(A\cup B)$. Both using the element argument proof method and algebraic proof method (using set identities). Is there a way to use set identities with power sets?

Christoph
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Mustafa Çığ Gökpınar
  • 13

2 Answers2

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We assume the following identities for arbitrary sets $X,Y,Z$:

  1. It holds that $X\subseteq X\cup Y$.
  2. If $X\subseteq Z$ and $Y\subseteq Z$ then $X\cup Y\subseteq Z$.
  3. If $X\subseteq Y$ then $\mathcal P(X)\subseteq \mathcal P(Y)$.

From this we can prove that for arbitrary sets $A,B$ we have $\mathcal P(A)\cup\mathcal P(B)\subseteq \mathcal P(A\cup B)$ the following way:

By 1. we have both $A\subseteq A\cup B$ and $B\subseteq A\cup B$. Hence, by 3. it follows that both $\mathcal P(A)\subseteq \mathcal P(A\cup B)$ and $\mathcal P(B)\subseteq \mathcal P(A\cup B)$. By 2. this then implies that $$ \mathcal P(A)\cup\mathcal P(B) \subseteq \mathcal P(A\cup B). $$

Christoph
  • 24,912
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Hint:

You just have to prove that $\mathcal P(A)\subseteq\mathcal P(A\cup B)$ and similarly for $\mathcal P(B)$. Just observe that $A, B\subseteq A\cup B$ and remember inclusion is transitive.

Bernard
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  • Isn't that using the "element argument proof method", i.e. take an element of $\mathcal P(A)$ and show it is in $\mathcal P(A\cup B)$? – Christoph Mar 02 '20 at 09:46
  • Not sure what the O.P. exactly meant by ‘element argument proof’. I'd think he meant considering the elements of a subset of $A$, not subsets of $A$ as elements of $\mathcal P(A)$. – Bernard Mar 02 '20 at 09:51