# set problems

• Jun 24th 2007, 12:05 PM
Discrete
set problems
Let A and B be sets. Show that

$a)\; A { \subseteq } (A { \cup} B)$

$a)\; A { \cap } (B-A) = { \oslash}$
• Jun 24th 2007, 12:51 PM
Ilaggoodly
there is a simple methodology to showing things about sets, ill show you how to do the second one, $A \subseteq (A \cup B)$
let x \in A, we need to show $x \in A \rightarrow x \in (A \cup B)$ , by the definition of union, all elements in A must be in $A \cup B$.

well maybe that one was the wrong one to werk, since it was rather easy... :P
• Jun 24th 2007, 12:53 PM
Discrete
that's an easy way to proof it, is there any way that you can show me?
• Jun 24th 2007, 01:18 PM
Ilaggoodly
if you want to prove something weth sets, the general procedure is to examine each element in the sets, so what can you say about an element in the difference of (B-A) (recall the definition of difference being, all elements in B that are NOT elements in A)
• Jun 24th 2007, 01:37 PM
Plato
Do you know that if P is true then PvQ is also true for any Q?
If so, then $x \in A \Rightarrow \left[ {x \in A \vee x \in B} \right]$ must be a true statement. Thus that proves that $A \subseteq \left( {A \cup B} \right)$.
• Jun 24th 2007, 01:48 PM
Soroban
Hello, Discrete!

The second one can be proved with various set theorems.
I don't know the names that you've been taught,
. . so I'll let you supply some of the reasons.

Quote:

Let A and B be sets. Show that: $(b)\; A\cap(B-A) \:= \:\emptyset$

We have: . $A \cap(B \cap \overline{A})$ . . . . definition of set subtraction

. . . . . . $= \;(A \cap \overline{A}) \cap B$ . . . . associative and commutative properties of intersection

. . . . . . $= \;\emptyset \cap B$

. . . . . . $=\;\emptyset$

• Jun 24th 2007, 08:31 PM
TheRekz
is there any way to change the subset operator into a union or intersection logical operator, to prove part a??
• Jun 25th 2007, 02:53 AM
Plato
Quote:

Originally Posted by TheRekz
is there any way to change the subset operator into a union or intersection logical operator, to prove part a??

Yes, that is exactly what I did above. The statement that $P \subseteq Q$ means if P then Q. If something is in P then it is in Q.