# Prove A = (A\B) ∪ (A ∩ B)

I have to demonstrate this formulae:

Prove A = (A\B) ∪ (A ∩ B)

But it seems to me that it is false.

(A\B) ∪ (A ∩ B)

• X ∈ A/B => { x ∈ A and x ∉ B }

or

• X ∈ A ∩ B => { x ∈ A and x ∈ B }

so:

x ∈ A ∩ B

so: A ≠ (A\B) ∪ (A ∩ B)

Did I solve the problem or I am just blind?

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Think of $(A\backslash B)$ as $A \cap B^c$. –  Matt N. Jan 18 '11 at 18:07
Matt: Forgive my elementary knowledge, but what does it mean B^c$? – Nerian Jan 18 '11 at 18:08 You did not solve the problem. If you want to show two sets$S_1, S_2$are equal, i.e.$S_1 = S_2$then you show$S_1 \subset S_2$and$S_2 \subset S_1$. – Matt N. Jan 18 '11 at 18:09$B^c$(read "B complement") means all elements that are not in B. – Matt N. Jan 18 '11 at 18:09 I suppose the question is how you got to$x \in A \cap B$from what you wrote above that. How did you conclude that? – Aryabhata Jan 18 '11 at 18:11 add comment ## 5 Answers To show that two sets are equal, you show they have the same elements. Suppose first$x\in A$. There are two cases: Either$x\in B$, or$x\notin B$. In the first case,$x\in A$and$x\in B$, so$x\in A\cap B$(by definition of intersection). In the second case,$x\in A$and$x\notin B$, so$x\in A\setminus B$(again, by definition). This shows that if$x\in A$, then$x\in A\cap B$or$x\in A\setminus B$, i.e.,$x\in (A\setminus B)\cup(A\cap B)$. Now we have to show, conversely, that if$x\in (A\setminus B)\cup(A\cap B)$, then$x\in A$. Note that$x\in(A\setminus B)\cup(A\cap B)$means that either$x\in A\setminus B$or$x\in A\cap B$. In the first case,$x\in A$(and also,$x\notin B$). In the second case,$x\in A$(and also,$x\in B$). In either case,$x\in A$, but this is what we needed. In summary: We have shown both$A\subseteq (A\setminus B)\cup(A\cap B)$and$(A\setminus B)\cup(A\cap B)\subseteq A$. But this means the two sets are equal. - Excellent explanation, thanks! I used unicode characters for the mathematical symbols, but I see that you all use something else. Where can I read the syntax? – Nerian Jan 18 '11 at 18:38 @Nerian: Have a look at meta.math.stackexchange.com/questions/107/…, and then Chapter 3 of ctan.org/tex-archive/info/lshort/english/lshort.pdf – Rahul Jan 18 '11 at 19:10 add comment To show set equality you show$\supset$,$\subset$respectively.$\subset$: Let$x \in A$. Then$x$either in$A \cap B$or in$A \cap B^c = A - B$, so$x \in (A \cap B) \cup (A - B)$.$\supset$: Let$x \in (A \cap B) \cup (A - B)$. Then either$x$in$ A \cap B$or x in$A \cap B^c$. But in both cases$x \in A$, therefore$x \in A$. - add comment$\rm\ A\backslash B\ =\ A\cap\overline B\ \ \:$so$\rm\ \: (A\backslash B)\cup (A\cap B)\ =\ (A\cap\overline B)\cup(A\cap B)\ =\ A\cap(\overline B\cup B)\ =\ A$- add comment Let$x \in A$. Then$x \in A \backslash B$or$x \in A \cap B$. Likewise, if$x \in A \backslash B$or$x \in A \cap B$then$x \in A$. - add comment Working inside a universe$X\$: $$A = A \cap X = A \cap (B \cup (X \setminus B)) = ( A \cap B ) \cup ( A \cap (X \setminus B)) = (A \cap B) \cup (A \setminus B)$$

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That's precisely the reverse of my answer. –  Bill Dubuque Jan 18 '11 at 19:40