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ANSWERS: SOLUTIONS AND HINTS FOR SELECTED EXERCISES

Chapter 1

Section 1.1

5. Suppose xST. Then xS and xT, which means xS and x$\overline{T}$, that is xS$\overline{T}$. So STS$\overline{T}$. Conversely, if xS$\overline{T}$, then xS and xT, which means xST, that is S$\overline{T}$ST. Therefore ST = S$\overline{T}$.

6. For Equation (1.2), suppose x$\overline{{S}_{1}\cup {S}_{2}}$. Then xS1S2, which means that x cannot be in S1 or in S2, that is x${\overline{S}}_{1}\cap {\overline{S}}_{2}$. So $\overline{{S}_{1}\cup {S}_{2}}$${\overline{S}}_{1}\cap {\overline{S}}_{2}$. Conversely, if x${\overline{S}}_{1}\cap {\overline{S}}_{2}$, then x is not in S1 and x is not in S2, that is, x$\overline{{S}_{1}\cup {S}_{2}}$. So ${\overline{S}}_{1}\cap {\overline{S}}_{2}$$\overline{{S}_{1}\cup {S}_{2}}$. Therefore $\overline{{S}_{1}\cup {S}_{2}}$ = ${\overline{S}}_{1}\cap {\overline{S}}_{2}$.

12. (a) reflexivity: Since |S1| = |S1|, so reflexivity holds.

(b) symmetry: if |S1| = |S2| then|S2| = |S1| so symmetry is satisfied.

(c) transitivity: Since ...

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