# Let A be $A=${$xy:x \in (0; 1/2),y \in \mathbb{Z}, |y|<3$}. Which is the value of $\inf A+\sup A$?

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 $$d)?$$

• For $x=\frac12$, $y=-3$, we have $xy = -1.5$. Also $x$ and $y$ cannot actually take boundary values, $\inf$ and $\sup$ are just the greater lower bound and least upper bound. – player3236 Sep 22 '20 at 12:11
• $y\in\{-2,-1,0,1,2\}$, so $\sup A=1$ – J. W. Tanner Sep 22 '20 at 12:16
• The supremum and infimum of a product are one of $\sup(A)\sup(B), \sup(A)\inf(B), \inf(A)\sup(B), \inf(A)\inf(B)$ Where $B=\{-2, - 1,0,1,2\},A=(0,1/2)$ – kingW3 Sep 22 '20 at 12:32