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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 }


  • X ∈ A ∩ B => { x ∈ A and x ∈ B }


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 $A \cap B^c$. – Rudy the Reindeer 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 $S_2 \subset S_1$. – Rudy the Reindeer Jan 18 '11 at 18:09
$B^c$ (read "B complement") means all elements that are not in B. – Rudy the Reindeer 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
up vote 5 down vote accepted

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.

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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…, and then Chapter 3 of – Rahul Jan 18 '11 at 19:10

To show set equality you show $\supset$, $\subset$ respectively.


Let $x \in A$. Then $x$ either in $A \cap B$ or in $A \cap B^c = A - B$, so $x \in (A \cap B) \cup (A - B)$.


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 $x \in A$.

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$\rm\ A\backslash B\ =\ A\cap\overline B\ \ \:$ so $\rm\ \: (A\backslash B)\cup (A\cap B)\ =\ (A\cap\overline B)\cup(A\cap B)\ =\ A\cap(\overline B\cup B)\ =\ A$

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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 $x \in A$.

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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

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