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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

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    $\begingroup$ 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. $\endgroup$ – player3236 Sep 22 '20 at 12:11
  • $\begingroup$ $y\in\{-2,-1,0,1,2\}$, so $\sup A=1$ $\endgroup$ – J. W. Tanner Sep 22 '20 at 12:16
  • $\begingroup$ 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 $B=\{-2, - 1,0,1,2\},A=(0,1/2)$ $\endgroup$ – kingW3 Sep 22 '20 at 12:32

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