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My first year Analysis textbook at university includes examples to grasp the concepts of infimum, supremum, maximum, minimum, lower bound and upper bound in set theory for subsets of the real numbers $\mathbb{R}$. One of such examples is the set: $$S=\{x\in \mathbb{Q}\;|\;x^2<2\}$$

for which the textbook says that "there does not exist a maximum, the upper bounds are all real numbers greater than or equal to $\sqrt{2}$, and the supremum is $\sqrt2$".

Now, it also says that for this same subset, "there is no minimum, there are no lower bounds and thus there is no infimum": to me, it would seem like there is no minimum (since one could approach $-\sqrt2$ infinitely by smaller and smaller rational numbers part of the set), with the infimum being $-\sqrt2$ and lower bounds all real numbers smaller than or equal to $-\sqrt2$. Am I wrong, or is the textbook wrong?

Asaf Karagila
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Mew
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2 Answers2

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Looking at $S$ as a subset of $\mathbb Q$ it has lower bounds, but has no infimum (i.e. greatest lower bound).

Looking at $S$ as a subset of $\mathbb R$ it does have an infimum which is $-\sqrt2\in\mathbb R$.

drhab
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  • Would the first case not mean there is also no supremum, though? – Mew Dec 23 '18 at 09:39
  • Indeed if $S$ is looked at as a subset of $\mathbb Q$ (ordered by $\leq$) it has no supremum either. If in the textbook they say that there is a supremum but no infimum then they are not consistent. – drhab Dec 23 '18 at 09:41
  • Thus, as it is asserted that there exists a supremum, I may conclude that the textbook contradicts itself and is wrong. – Mew Dec 23 '18 at 09:44
  • I think so, yes. – drhab Dec 23 '18 at 09:44
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To me it looks like the textbook is wrong, and you have it right. The set does have lower bounds, and therefore a greatest lower bound (among the reals), and that's $-\sqrt2$.

Arthur
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