Solutions to Problem Set 17

Math 310-01/02

10-20-2017

1. Prove that $[0,2] \cap (1,3) =
   (1,2]$ .

Let $x \in [0,2] \cap (1,3)$ . Then $x \in [0,2]$ and $x \in
   (1,3)$ . Therefore, $0 \le x \le 2$ and $1 < x < 3$ , so

$$0 \le x \quad\hbox{and}\quad x \le 2 \quad\hbox{and}\quad 1 < x \quad\hbox{and}\quad x < 3.$$

In particular, $1 < x$ and $x
   \le 2$ , so $1 < x \le 2$ , and hence $x
   \in (1,2]$ .

Conversely, suppose $x \in
   (1,2]$ . Then $1 < x \le 2$ , so

$$1 < x \quad\hbox{and}\quad x \le 2.$$

Now $0 \le 1$ , so combining this with $1 < x$ gives $0 \le x$ . Likewise, $2 < 3$ , and combining this with $x \le 2$ gives $x < 3$ . All together,

$$0 \le x \quad\hbox{and}\quad x \le 2 \quad\hbox{and}\quad 1 < x \quad\hbox{and}\quad x < 3.$$

Thus, $0 \le x \le 2$ and $1
   < x < 3$ , so $x \in [0,2]$ and $x \in
   (1,3)$ . Consequently, $x \in [0,2] \cap (1,3)$ .

Since $[0,2] \cap (1,3) \subset
   (1,2]$ and $(1,2] \subset [0,2] \cap (1,3)$ , it follows that $[0,2] \cap (1,3) = (1,2]$ .


2. Let A, B, and C be sets. Prove the following statement using the definitions of union, intersection, and complement and the rules of logic. Justify each step.

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

$$\matrix{x \in A - (B \cap C) & \iff & x \in A \land \lnot x \in (B \cap C) & \hbox{Definition of complement} \cr & \iff & x \in A \land \lnot(x \in B \land x \in C) & \hbox{Definition of intersection} \cr & \iff & x \in A \land (\lnot x \in B \lor \lnot x \in C) & \hbox{DeMorgan} \cr & \iff & (x \in A \land \lnot x \in B) \lor (x \in A \land \lnot x \in C) & \hbox{Distributivity} \cr & \iff & (x \in A - B) \lor (x \in A - C) & \hbox{Definition of complement} \cr & \iff & x \in (A - B) \cup (A - C) & \hbox{Definition of union} \quad\halmos\cr}$$


We'd be wise to question why we hold a grudge as if it were going to make us happy and ease our pain. It's rather like eating rat poison and thinking the rat will die. - Pema Ch\"odr\"on


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