Section 1: Problem 1 Solution

Working problems is a crucial part of learning mathematics. No one can learn topology merely by poring over the definitions, theorems, and examples that are worked out in the text. One must work part of it out for oneself. To provide that opportunity is the purpose of the exercises.

James R. Munkres

Check the distributive laws for $\cup$ and $\cap$ and DeMorgan’s laws.

To show that two sets $M$ and $N$ are equal, we need to show that for any $x$ , $x\in M$ iff $x\in N$ .

Distributive laws:
1. $x\in A\cap(B\cup C)$ iff $x\in A$ and $x\in B\cup C$ iff $x\in A$ and ( $x\in B$ or $x\in C$ ) iff ( $x\in A$ and $x\in B$ ) or ( $x\in A$ and $x\in C$ ) iff $x\in A\cap B$ or $x\in A\cap C$ iff $x\in(A\cap B)\cup(A\cap C)$ .
2. Similarly, for the second distributive law, $x\in A\cup(B\cap C)$ iff $x\in A$ or $x\in B\cap C$ iff $x\in A$ or ( $x\in B$ and $x\in C$ ) iff ( $x\in A$ or $x\in B$ ) and ( $x\in A$ or $x\in C$ ) iff $x\in A\cup B$ and $x\in A\cup C$ iff $x\in(A\cup B)\cap(A\cup C)$ .
DeMorgan’s laws:
1. $x\in A-(B\cup C)$ iff $x\in A\mbox{ and }x\notin B\cup C$ iff $x\in A\mbox{ and }x\notin B\mbox{ and }x\notin C$ iff $x\in A-B\mbox{ and }x\in A-C$ iff $x\in(A-B)\cap(A-C)$ .
2. Similarly, for the second DeMorgan’s law, $x\in A-(B\cap C)$ iff $x\in A\mbox{ and }x\notin B\cap C$ iff $x\in A\mbox{ and }(x\notin B\mbox{ or }x\notin C)$ iff $x\in A-B\mbox{ or }x\in A-C$ iff $x\in(A-B)\cup(A-C)$ .

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Section 1: Problem 1 Solution

Section 1: Problem 1 Solution