Seems obvious but I doubt how to show it step by step... I could prove the reverse statement (if $A\subseteq B$ or $B\subseteq A$ then $\mathcal{P}(A)\cup\mathcal{P}(B)=\mathcal{P}(A\cup B)$), but it doesn't help.
I saw a hint that it's easy to prove by negation, but I cant see how... Thanks for help at advance!

Edit: $\mathcal{P}(X)$ notes the powerset of $X$

  • $\begingroup$ Plus, the "reverse" would be a completely different result, maybe you were thinking about the contrapositive. $\endgroup$ – cronos2 Nov 6 '17 at 21:47
  • $\begingroup$ @learning union of two power sets $\endgroup$ – Marina Rappoport Nov 6 '17 at 21:50
  • $\begingroup$ @PeldePinda no, I wrote correct $\endgroup$ – Marina Rappoport Nov 6 '17 at 21:51
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    $\begingroup$ @cronos2 I don't know what is contrapositive statement... I mean that I could prove statement if change "if" and "then" parts. Reverse is not a negation $\endgroup$ – Marina Rappoport Nov 6 '17 at 21:53
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    $\begingroup$ math.stackexchange.com/questions/345978/… contains enough, no? $\endgroup$ – Asaf Karagila Nov 7 '17 at 0:04


If $A \not\subset B$ nor $B \not\subset A$ then $\exists a \in A \setminus B, b \in B \setminus A$. Think about some element of $\mathcal{P}(A \cup B)$ that couldn't possibly belong to $\mathcal{P}(A) \cup \mathcal{P}(B)$.


thinking about your comment it could actually constitute a proof (just for the finite case)

Clearly $|\mathcal{P}(A)| = 2^{|A|}, |\mathcal{P}(B)| = 2^{|B|}$.

$|\mathcal{P}(A) \cup \mathcal{P}(B)| = 2^{|A|} + 2^{|B|} - |\mathcal{P}(A) \cap \mathcal{P}(B)| = 2^{|A|} + 2^{|B|} - |\mathcal{P}(A \cap B)| = 2^{|A|} + 2^{|B|} - 2^{|A \cap B|}$ $|\mathcal{P}(A \cup B)| = 2^{|A| + |B| - |A \cap B|}$

Now if $A \not\subset B$ and $B \not\subset A$ then $|A \cap B| < \min \{|A|, |B|\}$. This way, $|\mathcal{P}(A) \cup \mathcal{P}(B)| = 2^{|A|} + 2^{|B|} - 2^{|A \cap B|}$ has, at least, three $1$s in its binary representation, while $|\mathcal{P}(A \cup B)| = 2^{|A| + |B| - |A \cap B|}$ has just one.

  • $\begingroup$ does it mean that I could prove it using size of P(A∪B) and P(A)∪P(B)? $\endgroup$ – Marina Rappoport Nov 6 '17 at 21:56
  • $\begingroup$ or it's simply means that in ∃{a,b} ∈ P(A∪B) that not belongs to P(A)∪P(B)? could I state it without any additional steps? $\endgroup$ – Marina Rappoport Nov 6 '17 at 22:15
  • $\begingroup$ The second comment is what I meant originally. See my edit for more about the first comment. $\endgroup$ – cronos2 Nov 6 '17 at 22:18

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