Let A and B be sets. Show that

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

$\displaystyle a)\; A { \cap } (B-A) = { \oslash} $

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- Jun 24th 2007, 12:05 PMDiscreteset problems
Let A and B be sets. Show that

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

$\displaystyle a)\; A { \cap } (B-A) = { \oslash} $ - Jun 24th 2007, 12:51 PMIlaggoodly
there is a simple methodology to showing things about sets, ill show you how to do the second one, $\displaystyle A \subseteq (A \cup B)$

let x \in A, we need to show $\displaystyle x \in A \rightarrow x \in (A \cup B)$ , by the definition of union, all elements in A must be in $\displaystyle A \cup B$.

well maybe that one was the wrong one to werk, since it was rather easy... :P - Jun 24th 2007, 12:53 PMDiscrete
that's an easy way to proof it, is there any way that you can show me?

- Jun 24th 2007, 01:18 PMIlaggoodly
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 PMPlato
Do you know that if

*P*is true then*PvQ*is also true for**any**?*Q*

If so, then $\displaystyle x \in A \Rightarrow \left[ {x \in A \vee x \in B} \right]$ must be a true statement. Thus that proves that $\displaystyle A \subseteq \left( {A \cup B} \right)$. - Jun 24th 2007, 01:48 PMSoroban
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: $\displaystyle (b)\; A\cap(B-A) \:= \:\emptyset$

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

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

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

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

- Jun 24th 2007, 08:31 PMTheRekz
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 AMPlato