Let $A=\{xy:x \in (0, \frac{1}{2}):y \in \mathbb{Z}, |y|<3\}$. Which is the value of $\inf A+\sup A$?
- a) $1$
- b) $-1$
- c) $\frac{1}{2}$
- d) $0$
I got an answer which is none of the alternatives. I thought about it this way: For $\sup A$ the biggest value that $x$ can take would be $\frac{1}{2}$ and $y$ can take $3.$ So it would be $\frac{1}{2}*3=\frac{3}{2}$ . On the other hand the smallest value for $x$ would be $0$ and for y would be $-3$ so the Infimum would be $0$. Therefore the sum of the two would be $\frac{3}{2}?$
Apparently the correct answer is $d)$ $0$ . Clearly there is a flaw with my argument since it's none of the alternatives and it would be really helpful if somebody could say me which is that flaw. And also how come $d)?$
Thank you in advance for your help, Annalisa