Why $P(A) \cup P(B)$ is not equivalent to $P(A \cup B)$

Question

I know that they are equivalent if we have $\cap$ instead of $\cup$.

If we start from left, we have:

$X \in P(A) \cup P(B)$

$X \subset A \lor X \subset B$

If we start from right, we have:

$X \in P(A \cup B)$

$X \subset (A \cup B)$

And now, how do we proceed proving that they are not equivalent?

to prove that they are not equal, find a set which is in $P(A\cup B)$, but not in $P(A)\cup P(B)$. — sranthrop, Nov 08 '14 at 21:34
You are off to a good start. So $X \subseteq A \cup B$. Does this imply $, X \subseteq A$? If so, then $X \in \mathcal{P}(A)$. But we can't infer that $, X \subseteq A$ necessarily. Similarly, we can't infer $, X \subseteq B$ necessarily. Here is a specific example: Let $A = { 1, 2 }$, and $B = {2, 3 }$. Let $, X = {1, 3}$. $X$ is a subset of $A \cup B$, but it is neither a subset of $A$ nor a subset of $B$. — layman, Nov 08 '14 at 21:35
I assume we are talking power sets. I misread the title as probability. — Henry, Nov 08 '14 at 21:39

score 5 · Answer 1 · answered Nov 08 '14 at 21:34

5

Notice that $A\cup B$ is never in $P(A)\cup P(B)$ unless $A\cup B$ is a subset of $A$ or $B$ - which never happens unless $A\subseteq B$ or $B\subseteq A$.

answered Nov 08 '14 at 21:34

Milo Brandt

60,888

score 3 · Answer 2 · answered Nov 08 '14 at 21:36

3

HINT: Let $A=\{1,2\}$, $B=\{3,4\}$. What are $P(A)$, $P(B)$ and $P(A\cup B)$?

answered Nov 08 '14 at 21:36

Przemysław Scherwentke

13,668
5
35
56

score 2 · Answer 3 · answered Nov 08 '14 at 22:08

You can see it by the cardinality of the sets. Let $A=\{a_1,a_2,\dotsc,a_n\}$ and $B=\{b_1,b_2.\dotsc,b_m\}$ be disjoint sets.

We can see that $|A|=n$ and $|B|=m$. This implies $|\mathcal P(A)|=2^n$ and $|\mathcal P(B)|=2^m$. Now $|\mathcal P(A)\cup \mathcal P(B)|=2^n+2^m-1$ (because they both share $\emptyset$). However $|A\cup B|=m+n$ and $|\mathcal P(A\cup B)|=2^{m+n}$. Therefore $\mathcal P(A\cup B) \neq \mathcal P(A)\cup \mathcal P(B)$.

Why $P(A) \cup P(B)$ is not equivalent to $P(A \cup B)$

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