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I have just learned what the definition is of a supremum, and I am confused to something my textbooks says:

Subsets with a supremum don't have to have a greatest element, for example:$$(0,3): = \{x \in \mathbb{R} \mid 0 < x < 3\}$$and$$\{ x \in \mathbb{Q} \mid x^2 \le 5\}$$

I understand the first example since we know that the supremum is $3$ but the subset doesn't have a greatest element since it must be less than $3$. I however do not understand the second one. If we solve $x^2 \le 5$ I believe we get $-\sqrt{5} \le x \le \sqrt{5}$. Wouldn't this mean that $\sqrt{5}$ is the greatest element in this subset?

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    $\begingroup$ Note that $\sqrt{5}$ is not a rational number, though, so it doesn't lie in your set. $\endgroup$
    – froggie
    Dec 22, 2012 at 22:13
  • $\begingroup$ notice that the set consist only of rational numbers, but the square root of 5 is not rational. $\endgroup$ Dec 22, 2012 at 22:13

2 Answers 2

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It would, if $\sqrt{5}$ were in the subset. But it isn't, since $\sqrt{5}\notin \mathbb Q$.

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  • $\begingroup$ Oh, I see. careless mistake, sorry. $\endgroup$ Dec 22, 2012 at 22:19
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But in your second example, $x \in \mathbb{Q}$, BUT $\sqrt{5} \notin \mathbb{Q}$, so the greatest (maximal) element in the second set cannot be $\sqrt{5}$.

$\sqrt{5}$ is the supremum of the set, but it is not the greatest element (i.e., maximal element) in the set (as a maximal element of a set must be IN the set). The supremum of a set need not be in the set, and a set may have a supremum, without a greatest element, as you note in the first set.) In this case, like the first example you provide, there is no maximal element in the set.

$$(2)\quad\{x \in \mathbb{Q} \mid x^2 \le 5\} \iff \{x \in \mathbb{Q}: -\sqrt{5} \le x \le \sqrt{5}\} \iff \{x \in \mathbb{Q}\;\; \land \;\;x \in [-\sqrt{5}, \sqrt{5}]\}.$$ So $(2)$ excludes irrationals like $-\sqrt{5}$ and $\sqrt{5}$.

Note that $(2)$ is precisely the same set as $\{x \in \mathbb{Q}: \sqrt{5} < x < \sqrt{5}\}$ (using the "strict" inequality, "$<$" to replace "$\le$")

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  • $\begingroup$ Yes it is, thank you. $\endgroup$ Dec 24, 2012 at 10:53

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