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I'm to prove or disprove $ {\cal P}(A\cup B) = {\cal P}(A) \cup {\cal P}(B)$

Let $A = \{1,2\}$ and $B = \{2,3\}$

Then ${\cal P}(A) = \{\{\emptyset\}, \{1\}, \{2\}, \{1,2\}\}$ and ${\cal P}(B) = \{\{\emptyset\}, \{2\}, \{3\}, \{2,3\}\}$

And ${\cal P}(A) \cup {\cal P}(B) =\{\{\emptyset\}, \{1\}, \{2\}, \{3\}, \{1,2\},\{2,3\}\}$

$A\cup B = \{1,2,3\}$ and

${\cal P}(A\cup B) = \{\{\emptyset\}, \{1\}, \{2\}, \{3\} \{1,2\},\{1,3\},\{2,3\},\{1,2,3\}\}$

We see that ${\cal P}(A) \cup {\cal P}(B) \subseteq {\cal P}(A\cup B)$, but ${\cal P}(A\cup B)$ isn't a subset of

${\cal P}(A) \cup {\cal P}(B)$.

${\cal P}(A\cup B)$ contains $\{1,2,3\}$ and $\{1,3\}$,

whereas ${\cal P}(A) \cup {\cal P}(B)$ doesn't.

$\to {\cal P}(A\cup B) \neq {\cal P}(A) \cup {\cal P}(B)$

Is this correct/sufficient?

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  • $\begingroup$ $\varnothing\subseteq A$ so that $\varnothing\in\wp(A)$, but not $\{\varnothing\}\in\wp(A)$. Same mistake for $B$ and $A\cup B$. $\endgroup$ – drhab Oct 3 '17 at 8:55
  • $\begingroup$ @Raffaele also if the sets are disjoint then the claim is not valid. If e.g. $A$ and $B$ are disjoint and have $3$ elements then cardinality of $\wp(A)\cup\wp(B)$ is $2^3+2^3=16$ and cardinality of $\wp(A\cup B)$ is $2^6>16$. $\endgroup$ – drhab Oct 3 '17 at 9:01
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It's almost correct. The only mistake you made was when you wrote $${\cal P}(A) = \{\{\emptyset\}, \{1\}, \{2\}, \{1,2\}\}$$ when you should have written $${\cal P}(A) = \{\emptyset, \{1\}, \{2\}, \{1,2\}\}$$

And you repeated the same mistake everywhere you wrote the power set.


It can, however, be shortened significantly:

Using the same two sets for a counterexample:

  • Take $A=\{1,2\}, B=\{2,3\}$
  • Then, $A\cup B =\{1,2,3\}\in\mathcal P(A\cup B)$
  • However, (1) $A\cup B\not\subseteq A$, so $A\cup B\notin \mathcal P(A)$, and $A\cup B\not\subseteq B$, so $A\cup B\notin \mathcal P(B)$
  • Therefore, $A\cup B\notin \mathcal P(A)\cup \mathcal P(B)$

So, $A\cup B\in \mathcal P(A\cup B)$ and $A\cup B\notin\mathcal P(A)\cup\mathcal P(B)$, which means $\mathcal P(A\cup B)\neq \mathcal P(A)\cup\mathcal P(B)$


An even shorter argument:

  • Take $A=\{1,2\}, B=\{3,4\}$, so $A\cup B$ has $4$ elements.
  • Then $\mathcal P(A)$ has $2^2$ elements, $\mathcal P(B)$ has $2^2$ elements, so $\mathcal P(A)\cup\mathcal P(B)$ can have at most $2^2+2^2=8$ elements
  • However, $\mathcal P(A\cup B)$ has $2^4=16>8$ elements.
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