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I have to demonstrate this formulae:

Prove $A = (A\setminus B) ∪ (A ∩ B)$

But it seems to me that it is false.

$(A\setminus B) ∪ (A ∩ B)$

  • $X \in A\setminus B \implies { x ∈ A \text{ and } x ∉ B }$

or

  • $X ∈ A ∩ B \implies { x ∈ A \text{ and } x ∈ B }$

so:

$x ∈ A ∩ B$

so: $A ≠ (A\setminus B) ∪ (A ∩ B)$

Did I solve the problem or I am just blind?

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    There are 2 theorems available here, one from set theory, one from logic. For sets X and Y, X=Y means X is Y, but the theorem from set theory says two sets are equal if each is a subset of the other. The theorem from logic says that if you have two statements S and T and you wish to show S or T is true, it's enough to show that if S is false then T is true. These 2 theorems make your problem easy. They are basic tools that should be in your toolbox. I remember a class stuck on showing two fancy topologies were equal, until someone said let's just show each topology is a subset of the other.2017-02-01

7 Answers 7

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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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    @Nerian: Have a look at http://meta.math.stackexchange.com/questions/107/faq-for-math-stackexchange/117#117, and then Chapter 3 of http://www.ctan.org/tex-archive/info/lshort/english/lshort.pdf2011-01-18
3

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 $x \in A$.

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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.2019-03-13
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Using your initial arguments:

$(A\setminus B) \cup (A \cap B)$

  • $X \in A\setminus B \implies { x \in A \text{ and } x\not\in B }$

or

  • $X \in A \cap B \implies { x \in A \text{ and } x \in B }$

So what we have is $x\in A\setminus B$ non-exclusive OR $x\in A \cap B$.

So $ { x \in A \text{ and } x\not\in B } \lor { x \in A \text{ and } x \in B }$.

In Boolean logic this becomes $A \land (B\lor ¬B)=A$.

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For all, each, any, and every $x$ we have $(x\in A)\iff ((x\in A\land x\in B)\lor (x\in A\land x\not \in B))\iff$ $\iff (x\in A\cap B)\lor (x\in A \backslash B).$