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

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.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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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