Proof or disprove that the following statement is true I have a problem I need help with:
Prove or disprove that P(M $\cup$ N) $\neq$ P(M) $\cup$ P(N).
What I have tried so  far
Consider: M = $\{ a,b,c\}$ and N = $\{\text{red},\text{blue},\text{yellow}\}$ 
$\Rightarrow$ P(M) ${}= 3^3 = 27$ and P(N) ${} = 3^3$
Combine P(M)$\cup$P(N)
$\Rightarrow \bigcup \{M,N\} = \{x\mid(x\in M) \text{ and } (x\in N)\}$
$\Rightarrow$ P(M ${}\cup{}$ N) ${}= \{a,b,c,\text{red}, \text{blue}, \text{yellow} \}$
$\Rightarrow$ P(M,N) ${}= 6^6 = 46656$
$\Rightarrow$ P(M $\cup$ N) ${}\neq{}$ P(M) ${}\cup{}$ P(N)
Is my answer correct? 
 A: Your answer has a good idea but there are some technical mistakes in it. 
For example, you say $P(M)=3^3$, which is not true. 


*

*First of all, $P(M)$ is a set, while $3^3$ is a number. The two are not equal.

*What you probably want to say $|P(M)|=3^3$, i.e. that the size of the set $P(M)$ is $3^3$. But even then, no, that is not true.



To try the proof again, I suggest you try to construct a smaller counterexample, and ditch the idea of set sizes. Simply look at the power sets directly.

For an added challenge, try proving this statement:

$P(A\cup B)=P(A)\cup P(B)$ if and only if $A=\emptyset$ or $B=\emptyset$.

A: Let $A=\{1,2,3\}$ and $B=\{a,b,c\}.$
Then $\{2,3,a,b\}$ is a member of $P(A\cup B)$ but is not a member of $P(A)\cup P(B).$
A: Take $M$ and $N$ to be any two disjoint nonempty finite sets of cardinality $m$ and $n$ respectively. Then $|P(M \cup N)|=2^{m+n}$. $P(M) \cap P(N)=\{ \{ \} \}$, so $$|P(M) \cup P(N)|=|P(M)|+|P(N)|-1=2^m + 2^n -1,$$
and since this number is odd, it can never be true that $P(M \cup N)=P(M) \cup P(N)$. 
