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$$A = \{x \in \Bbb Q: x^2 < 11\}$$

I am looking for the upper and lower bounds of set $A$. I know they should be the greatest and the smallest rational number within $(- \sqrt{11}, \sqrt{11})$

So the difficulty is how to get those rational numbers in exact value. Thanks

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  • $\begingroup$ You just found supremum and infimum of $A$, because there are infinitely many rationals arbitrarily close to $\sqrt{11}$ $\endgroup$
    – A.P.
    Commented Apr 22, 2015 at 17:30

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As you correctly noted, $A = (-\sqrt{11}, \sqrt{11}) \cap \mathbb Q$. However $\pm\sqrt{11}\notin \mathbb Q$ and $\mathbb Q$ is dense in $\mathbb R$. Therefor $A$ is dense in $(-\sqrt{11}, \sqrt{11})$ and we can find a sequence in $A$ converging to any of $\pm\sqrt{11}$. This proves that no upper or lower bound exists in the set (i.e. $\min A$ and $\max A$ are undefined). However, $\sup A$ and $\inf A$ are defined and are least upper and greatest lower bounds for $A$. Since we already have a convergent sequence to $\pm\sqrt{11}$ and both are bounds on $A$, we can see that $$\sup A = \sqrt{11}\\ \inf A = -\sqrt{11}$$

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