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Chapter 2 Unions and Intersections

Exercises Practice Problems

1. Union and Intersection.

The concepts of union and intersection naturally correspond to the concepts of "or" and "and." Explain why using words and/or pictures. Solution.

\begin{equation*} \begin{array}{rcl} A \cup B &=& \{ x \mid x \in A \mbox{ OR } x \in B \} \\ A \cap B &=& \{ x \mid x \in A \mbox{ AND } x \in B \} \end{array} \end{equation*}

2. Counting Subsets.

Let \(A\) be any set. In this problem, we will see why mathematicians decided that is it "right" to say that \(\emptyset \subset A\) and \(A \subset A\text{.}\)

  1. List all of the subsets of \(\{ 1 \}\text{.}\) Solution.

    \(\emptyset, \{ 1 \}\)

  2. List all of the subsets of \(\{ 1,2 \}\text{.}\) Solution.

    \(\emptyset, \{ 1 \}, \{ 2 \} , \{ 1,2 \}\)

  3. List all of the subsets of \(\{ 1,2,3 \}\text{.}\) Solution.

    \(\emptyset, \{ 1 \}, \{ 2 \}, \{ 3 \}, \{ 1,2 \}, \{ 1,3 \} , \{ 2,3 \}, \{ 1,2,3 \}\)

  4. Look at the pattern you see so far. Guess how many subsets are there of the set \(\{ 1,2,3,4 \}\text{.}\) How about the set \(\{ 1,2,3, \ldots , 10 \}\text{?}\) Solution.

    \(\{ 1,2,3,4 \}\) has \(2^4\) subsets and \(\{ 1,2,3, \ldots , 10 \}\) has \(2^{10}\) subsets.

  5. Your answers in part (d) are very simple expressions. This is one reason why mathematicians defined "subset" to include the cases \(\emptyset \subset A\) and \(A \subset A\text{.}\) The power set of a set \(A\) is the set of all subsets of \(A\text{.}\) This power set is denoted \(\mathcal{P}(A)\text{.}\) If \(|A|=n\text{,}\) what is \(| \mathcal{P}(A)|\text{?}\) (You do not have to prove your answer.) Solution.

    If \(|A|=n\) then \(| \mathcal{P}(A)|=2^n\)

3. Distributive Laws.

Prove the following distributive laws in two ways. First, using a Venn diagram, and then using a double inclusion proof.

  1. \(A \cap (B \cup C) = (A \cap B) \cup (A \cap C)\) Solution.

    This argument works backwards and forward. So this proves the double inclusion of these sets.

    \begin{equation*} \begin{array}{rcl} x \in A \cap (B \cup C) &\Longleftrightarrow& x \in A \mbox{ AND } (x \in B \mbox{ OR } x \in C) \\ &\Longleftrightarrow& x \in A \cap B \mbox{ OR } x \in A \cap C \\ &\Longleftrightarrow& x \in (A \cap B) \cup ( A \cap C) \\ \end{array} \end{equation*}

  2. \(A \cup (B \cap C) = (A \cup B) \cap (A \cup C)\) Solution.

    Once again, this argument works backwards and forward. So this proves the double inclusion of these sets.

    \begin{equation*} \begin{array}{rcl} x \in A \cup (B \cap C) &\Longleftrightarrow& x \in A \mbox{ OR } ( x \in B \mbox{ AND } x \in C) \\ &\Longleftrightarrow& (x \in A \mbox{ OR } x \in B) \mbox{ AND } (x \in A \mbox{ OR } x \in C) \\ &\Longleftrightarrow& x \in (A \cup B) \cap (A \cup C) \end{array} \end{equation*}

    This argument does work, but it does have one subtlety. If \(x \notin A\) then each condition forces \(x\) to be an element of both \(B\) and \(C\text{.}\)