- Thread starter
- #1

- Thread starter solakis
- Start date

- Thread starter
- #1

- Mar 31, 2013

- 1,334

To prove the same we have $A \cup B = A $ iff $B \subseteq A$

Let us take 2 sets $A_1,A_2$ which are disjoint and because it is true for every set $A_1 \cup B = A_1 $ so $B \subseteq A_1$

and $A_2 \cup B = A_2 $ so $B \subseteq A_2$

so from above 2 we have

$B \subseteq A_1 \cap A_2$

because $A_1,A_2$ are disjoint sets so we have $A_1 \cap A_2= \emptyset$

so $B = \emptyset$

- Thread starter
- #3

put \(\displaystyle A=0\) and 1 becomes \(\displaystyle 0\cup B=0\)

And \(\displaystyle B=0\) since \(\displaystyle 0\cup B=B\)

Note 0 is the empty set

However somebody sujested the following counter example:

A={1,2,3}...........B={1,2} so we have :\(\displaystyle A\cup B=A\) and \(\displaystyle \neg(B=0)\)