# Thread: Proof with Simple Problem

1. ## Proof with Simple Problem

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

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

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

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

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

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

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

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

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

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

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 $x \in \emptyset$ 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 $x \in \emptyset$ 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

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

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

The damn emptysets

7. Originally Posted by Foink

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

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

The damn emptysets
see where this gets you:

by definition: $A - B = A \cap B'$ where $A, B$ are sets, and $B'$ means $B$-compliment

8. Ahh I see. I got it Thanks!!