Science:Math Exam Resources/Courses/MATH220/December 2010/Question 02 (b)

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MATH220 December 2010

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Question 02 (b)
Let $A$, $B$, and $C$ be sets. Prove the following distributive law from first principles: $A\cap (B\cup {C})=(A\cap {}B)\cup (A\cap {}C).$

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Proving this from first principles would mean to show that an element of the set on the left must be an element of the set on the right and vice versa.

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

We prove double inclusion of sets. First suppose that $\displaystyle x\in A\cap (B\cup C)$. To be a member of the set on the right, we must show that either $x\in \displaystyle A\cap B$ or that $\displaystyle x\in A\cap C$. If $\displaystyle x\in A\cap B$, then we are done. So suppose otherwise, that is, $\displaystyle x\notin A\cap B$. Since $\displaystyle x\in A\cap (B\cup C)$ we know that $\displaystyle x\in A$ and $\displaystyle x\in (B\cup C)$. However, as $\displaystyle x\notin A\cap B$, we must have that $\displaystyle x\in C$. Hence, as $\displaystyle x\in A$ and $\displaystyle x\in C$, we see that $\displaystyle x\in A\cap C$ completing the proof in this direction. For the reverse direction, suppose that $x\in \displaystyle (A\cap B)\cup (A\cap C)$. If $x\in \displaystyle A\cap B$, then we have that $x\in \displaystyle A$ and $x\in \displaystyle B\subseteq B\cup C$. Thus, $\displaystyle x\in A\cap (B\cup C)$. Now, if instead $x\notin \displaystyle A\cap B$, then necessarily we must have that $x\in \displaystyle A\cap C$. Hence $x\in \displaystyle A$ and $x\in \displaystyle C\subseteq B\cup C$. Thus, $\displaystyle x\in A\cap (B\cup C)$. This completes the proof.

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