So I am learning about proving intersection and union statements of sets, but the problem is I am never confident about my proofs, I never know when I am right. So if you could check my attempt, and maybe offer some help that would be great:

Prove: $A\cup \!\, (B\cap \!\ C)=(A\cup \!\, B)\cap \!\ (A\cup \!\, C)$

Here's my attempt:

$A\cup \!\, (B\cap \!\ C)=A\cup \!\,${$x:x\in \!\, B \ \text{and}\ x\in \!\, C $}

$=${$x:x\in \!\, A \ \text{or}\ x\in \!\, C\ \text{and} \ x\in \!\, B $}

(I think I need a step in between here, right?)

$=(A\cup \!\, B)\cap \!\ (A\cup \!\, C)$

  • $\begingroup$ Draw a Venn diagram. $\endgroup$ – copper.hat Jan 29 '15 at 18:37
  • $\begingroup$ @copper.hat so a Venn diagram would be a sufficient proof? $\endgroup$ – Math Major Jan 29 '15 at 18:37
  • $\begingroup$ No, but it will help show you what is going on. $\endgroup$ – copper.hat Jan 29 '15 at 18:38
  • $\begingroup$ @copper.hat I think I understand unions and intersections, but the problem I am having is constructing an actual proof $\endgroup$ – Math Major Jan 29 '15 at 18:45
  • $\begingroup$ I added an answer below outlining steps. This is one way of doing it. $\endgroup$ – copper.hat Jan 29 '15 at 18:48

There is no magic, just work.

One way is as follows, basically case analysis:

First show that $A \cup ( B \cap C) \subset (A\cup B) \cap (A\cup C)$.

Suppose $x \in A \cup ( B \cap C)$. Then either (i) $x \in A$ or (ii) $x \in B \cap C$.

In the first case, we see that $x \in A \subset A\cup B$ and $x \in A \subset A\cup C$, and so $x \in (A\cup B) \cap (A\cup C)$.

In the second case, we have $x \in B \cap C$, so $x \in B$ and $x \in C$. Then $x \in B \subset A \cup B$ and $x \in C \subset A \cup C$, so $x \in (A\cup B) \cap (A\cup C)$.

Now show $(A\cup B) \cap (A\cup C) \subset A \cup ( B \cap C)$.

Suppose $x \in (A\cup B) \cap (A\cup C)$, then we must have $x \in A\cup B$ and $x \in A\cup C$. Now we need to make judicious choices. I will split into (i) $x \in A$ or (ii) $x \notin A$.

In the first case, we see that $x \in A \subset A \cup ( B \cap C)$.

In the second case, since $x \in A\cup B$ and $x \notin A$ we must have $x \in B$. Similarly, since $x \in A\cup C$ and $x \notin A$ we must have $x \in C$. Hence $x \in B \cap C$ and then $x \in B \cap C \subset A \cup ( B \cap C)$.

  • 1
    $\begingroup$ this was very helpful! $\endgroup$ – Math Major Jan 29 '15 at 18:53
  • 1
    $\begingroup$ Glad to be able to help. You will become familiar with the techniques over time. $\endgroup$ – copper.hat Jan 29 '15 at 18:56

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