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What I've done so far is stated that

  1. $(A \cup A) \cap (A \cup B)$ by distribution
  2. $A\cap (A\cup B)$ by 1, definition of $\cup$
  3. $A\cap A$ by 2, definition of $\cup$
  4. $A$ by 3, definition of $\cap$

I'm not sure if I need to state that $A\cup B = \{x\mid x\in A\lor x\in B\}$ somewhere in there.

I'd really appreciate some help on this one. Thank you!

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up vote 8 down vote accepted

Without context, I would do this without applying rules. First I would show: $$ A \cup (A \cap B) \subset A$$

then I would show $$A \subset A \cup (A \cap B) .$$ These two together imply the relation you're trying to prove.

For the first statement, assume you have $a \in A \cup (A \cap B)$. Because $a$ is in this union, it is either in $A$ or in the intersection of $A$ and $B$... It must be in $A$ in either of these cases.

For the second statement, assume you have $a \in A$. Every element of $A$ is also in $A \cup C$ for any set C so $a \in A\cup (A \cap B)$.

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I'm still not exactly sure how to express this using the notation I gave. – Addison Feb 20 '13 at 2:44
What notation do you want to use? Do you want to use specific rules to prove this? – orlandpm Feb 20 '13 at 4:57
I believe so. I understand how to reach the conclusion, I'm just not sure how to prove the steps I take using rules. – Addison Feb 20 '13 at 5:15
I found an example of this proved in a book and it was very similar to your answer, together I believe I completed the proof in the notation I was looking for. Thank you! – Addison Feb 22 '13 at 13:16

Your third step is incorrect: it’s not true in general that $A=A\cup B$.

I would prove the result by element-chasing, i.e., showing that if $x\in A$, then $x\in A\cup(A\cap B)$, which is immediate from the definition of union, and that if $x\in A\cup(A\cap B)$, then $x\in A$, which is also pretty straightforward.

If you’re going to do it algebraically, by manipulating unions and intersections directly, the argument will depend on what rules of manipulation you have available. For instance, there is an absorption rule that says that if $X\subseteq Y$, then $X\cup Y=Y$. If you have that available, you can apply it with $X=A\cap B$ and $Y=A$ to get the result immediately. Or you might have the other absorption rule, that if $X\subseteq Y$, then $X\cap Y=X$; in that case you can apply one of the De Morgan laws to write $A\cup(A\cap B)$ as $(A\cup A)\cap(A\cup B)=A\cap(A\cup B)$ and apply the absorption rule with $X=A$ and $Y=A\cup B$.

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You can either do this algebraically, by manipulating $\cup$ and $\cap$ over the symbols. Or you can chase the elements.

  • $A\subseteq A\cup(A\cap B)$ because if $x\in A$ then, $x\in A$ or $x\in A\cap B$. Therefore if $x\in A$ then $x\in A\cup(A\cap B)$.

  • $A\cup(A\cap B)\subseteq A$ because whenever $x\in A\cup(A\cap B)$ either $x\in A$ and we are done, or $x\in A\cap B$ and then $x\in A$ and $x\in B$ so in particular $x\in A$. Either way if $x\in A\cup(A\cap B)$ we have that $x\in A$ and so the inclusion holds.

$\implies$ We have shown a two-sided inclusion and therefore the sets are equal.

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Note that if $C\subset A$, $A\cup C = A$. Since $A\cap B \subset A$, $A\cup(A\cap B) = A$.

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