For any events $A$ and $B$, prove that $A \cup B = (AB^c)\cup(AB)\cup(A^cB)$.

Question

For any events $A$ and $B$, prove that $A \cup B = (AB^c)\cup(AB)\cup(A^cB)$. This result is used to prove that $P(A \cup B) = P(A) + P(B) - P(AB)$ for any events $A$ and $B$.

The result is apparent when one draws a Venn diagram, but how can it be proven mathematically?

EDIT:

I suppose this is a more general form of: if $\omega \in \Omega$, then either $\omega \in A$ or $\omega \in A^c$. Going a step further, $A = AB^c \bigcup AB$ because if $\omega \in A$, it is also either in $B$ or $B^c$ . Are statements such as these obvious, or are they true due to "insert theorem name here?"

Answer 1

You can just use:

Adjacency

$AB \cup AB^C=A$

and you also have:

Idempotence

$A \cup A =A$

So:

$AB^C \cup AB \cup A^CB \overset{Idempotence}= AB^C \cup AB \cup AB \cup A^CB \overset{Adjacency \ x \ 2}= A \cup B$

Answer 2

If $x\in A\bigcup B$ than there are three case:

1. $x\in A $ and not in $B$ , so $x\in AB^c$;

2. $x\in A$ and $x\in B$ , so $x\in AB$;

3. $x\in B$ and not in $A$, so $x\in A^cB$

Then $x\in AB^c\cup AB \cup A^cB$ and so $A\bigcup B\subset AB^c\cup AB \cup A^cB$

If $x\in AB^c\cup AB \cup A^cB$ than $x\in A$ or $x\in B$ then $x\in A\bigcup B$ and so you have

$AB^c\cup AB \cup A^cB\subset A\bigcup B $

$P(A\bigcup B)=P (AB^c\cup AB \cup A^cB)= $

$=P(AB^c)+P(AB) +P(A^cB)= $

$=(P(AB^c)+P(AB) )+(P(A^cB)+ P(AB) )-P(AB)=$

$=P(A(B^c\cup B))+P((A^c\cup A)B)-$

$-P(AB)=P(A)+P(B)-P(AB)$

Answer 3

Let $\Omega$ be the global universe $$A \cup B = (A\cap \Omega)\cup (B\cap\Omega)$$ $$ = (A\cap(B\cup B^c))\cup (B\cap(A \cup A^c))$$ $$ = [(A\cap B)\cup (A\cap B^c)]\cup [(B\cap A) \cup (A^c\cap B)]$$ $$ = (A\cap B^c)\cup (A\cap B) \cup (A^c\cap B) $$ $$\space because \space A\cap B=B\cap A\space and \space (A\cap B)\cup (A\cap B)=A\cap B$$