# Thread: Proof with Simple Problem

1. ## Proof with Simple Problem

Hey guys. I don't understand how to prove this:

(a)
$\displaystyle \text A \cup (null) = A$

(b)
$\displaystyle \text A \cap U = A$

2. Originally Posted by Foink
Hey guys. I don't understand how to prove this:

(a)
$\displaystyle \text A \cup (null) = A$
one method: Let $\displaystyle x \in A \cup \emptyset$ ......we assume A is non-empty, it's trivial otherwise

$\displaystyle \Rightarrow x \in A$ or $\displaystyle x \in \emptyset$

we cannot have $\displaystyle x \in \emptyset$, thus, $\displaystyle x \in A$

therefore, $\displaystyle (A \cup \emptyset ) \subset A$

now assume $\displaystyle x \in A$. this implies $\displaystyle x \in A \cup \emptyset$ (i'm sure there's some kind of axiom or something that allows us to say this). thus $\displaystyle A \subset (A \cup \emptyset )$

since $\displaystyle (A \cup \emptyset ) \subset A$ and $\displaystyle A \subset (A \cup \emptyset )$, we have $\displaystyle A \cup \emptyset = A$

3. Ohhh okay. I guess I was on the right track then. I didn't know you could just say that you could'nt have $\displaystyle x \in \emptyset$ lol Thanks

4. Originally Posted by Foink
Ohhh okay. I guess I was on the right track then. I didn't know you could just say that you could'nt have $\displaystyle x \in \emptyset$ lol Thanks
well, it's the empty set, by definition it has no elements...or it wouldn't be empty, and the name would just be stupid...

try a similar approach for the second

5. yea the second one was easier because I didn't have to deal with the emptysets lol thanks

$\displaystyle \text (a) A - \emptyset = A$

$\displaystyle \text (b) \emptyset - A = \emptyset$

The damn emptysets

7. Originally Posted by Foink
$\displaystyle \text (a) A - \emptyset = A$
$\displaystyle \text (b) \emptyset - A = \emptyset$
by definition: $\displaystyle A - B = A \cap B'$ where $\displaystyle A, B$ are sets, and $\displaystyle B'$ means $\displaystyle B$-compliment