Prove there exist sets $A$, $B$ with $\mathcal P \left({A \cup B}\right) \not= \mathcal P \left({A}\right) \cup \mathcal P \left({B}\right)$

Question

Prove: There exist sets $A$, $B$ with $\mathcal P \left({A \cup B}\right) \not= \mathcal P \left({A}\right) \cup \mathcal P \left({B}\right)$

$\mathcal P$ here is the Power Set

Update: With @Simon S and @Stefanos help, I was able to write the following:

Proof:

Let $A = \{a\}$, $B = \{b\}$, where the element $a\not = b$

$\mathcal P \left({A}\right) = \{ \varnothing, \{a\}\}$

$\mathcal P \left({B}\right) = \{ \varnothing, \{b\}\}$

$\mathcal P \left({A}\right) \cup \mathcal P \left({B}\right) = \{\varnothing,\{a\},\{b\}\}$

$A \cup B = \{a,b\}$

$\mathcal P \left({A \cup B}\right) = \mathcal P \left({\{a,b\}}\right) = \{\varnothing, \{a\}, \{b\}, \{a, b\}\}$

We can see that $\mathcal P \left({A \cup B}\right) \ne \mathcal P \left({A}\right) \cup \mathcal P \left({B}\right)$

Does that look right?

What is $P$ here, the power set, the measure/probability? +/union? — Simon S, Nov 02 '14 at 22:11

Simon S · Accepted Answer · 2014-11-02T22:18:39.017

2

If $A = \{ 0 \}$ and $B = \{ 1 \}$, then $P(A) = \left\{ \emptyset, \{ 0 \} \right\}$ and $P(B) = \left\{ \emptyset, \{ 1 \} \right\}$ while

$$P( A \cup B ) = \left\{ \emptyset, \{ 0 \}, \{ 1 \}, \{ 0, 1 \} \right\} \ \neq \ P(A) \cup P(B)$$

edited Nov 02 '14 at 22:18

answered Nov 02 '14 at 22:14

Simon S

26,898

2

This works if $0\neq 1$. – Jonas Meyer Nov 02 '14 at 22:20
This is the same as what the OP did, no? – GFauxPas Nov 02 '14 at 22:33
@GFauxPas: Yes, but after Simon S & Stefanos helped me. – lucidgold Nov 02 '14 at 22:34
1

For what it's worth, I did post this before Stefanos or the OP lucidgold edited the post with a solution. – Simon S Nov 02 '14 at 22:35
Wouldn't this be the case for any two sets s.t. $A \cap B \neq \emptyset$? – Daniel Goldman Nov 02 '14 at 22:43

score 1 · Answer 2 · answered Nov 02 '14 at 22:15

Take two elements $a,c$ such that $a \in A$ but $a \notin B$ and $c \in B$, but $c \notin A$. Then $$\{a,c\}\in \mathcal P(A\cup B)$$ but $$\{a,c\}\notin \mathcal P(A) \quad \text{ and } \quad \{a,c\}\notin \mathcal P(B)$$ which implies that $$\{a,c\}\notin \mathcal P(A)\cup\mathcal P(B)$$

Prove there exist sets $A$, $B$ with $\mathcal P \left({A \cup B}\right) \not= \mathcal P \left({A}\right) \cup \mathcal P \left({B}\right)$

2 Answers2

Linked