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I am having a hard time proving this simple and natural identity of sets. what I do is go round and round in circles:

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

Now what? I apply the distributive property again and reach the first expression. How can I show this using set properties (distributive, idempotent, associative, de morgan etc)?

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    $\begingroup$ Hint: $A\cap B\subset A$. $\endgroup$
    – gniourf_gniourf
    Feb 10, 2016 at 17:07
  • $\begingroup$ Do you have the absorption laws in your toolkit? This is one of them. $\endgroup$
    – Brian M. Scott
    Feb 10, 2016 at 17:08
  • $\begingroup$ @gniourf_gniourf I have done it by element chasing method, but wanted to use set properties instead. You mean proving each other the subsets of each other to show equality right? $\endgroup$
    – jonsno
    Feb 10, 2016 at 17:08
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    $\begingroup$ @DrHAL But that is still not using the set properties. Although it is valid proof. $\endgroup$
    – jonsno
    Feb 10, 2016 at 17:17
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    $\begingroup$ @BrianM.Scott Yes they are in my toolkit! $\endgroup$
    – jonsno
    Feb 10, 2016 at 17:18

3 Answers 3

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Assume that $A$ and $B$ are subsets of some universal set $X$. Then

$$\begin{align*} A\cup(A\cap B)&=(\color{red}A\cap X)\cup(\color{red}A\cap B)\\ &=\color{red}A\cap(X\cup B)\\ &=A\cap X\\ &=A\;. \end{align*}$$

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  • $\begingroup$ Thanks but I cant get how you got to second step $(A\cap(X\cup B))$ from line 1 $\endgroup$
    – jonsno
    Feb 10, 2016 at 17:20
  • $\begingroup$ @samjoe: One of the distributive laws. $\endgroup$
    – Brian M. Scott
    Feb 10, 2016 at 17:21
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Answer 1. Clearly $A \subseteq A\cup (A\cap B).$
For the converse, Note that $\color{maroon}A \subseteq A$ and $\color{lime}{A\cap B} \subseteq A$. So $\color{maroon}A\cup \color{lime}{(A\cap B)} \subseteq A.$

Answer 2. If $x\in A$, clearly $x\in A\cup (A\cap B).$
Let $x\in A\cup (A\cap B).$ So either $\color{red}{x\in A}$ or $\color{blue}{x\in (A\cap B)}.$ The blue means $\color{blue}{x\in A},$ and $\color{blue}{x\in B}.$ So in both cases (blue&red) we have $x\in A$.

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$A∪(A∩B)=(A∪A)∩(A∪B)$=$A∩(A∪B)$=$A$, the first equality using the distribution law and the last equality since $A\subset A∪B$.

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