Supremum and greatest element 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} | 0 < x < 3\}  $
and
$\{ x \in \mathbb{Q} | x^2 \leq  5\}$ 
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 don't understand the second one. If we solve  $x^2 \leq 5$ I believe we get $-\sqrt{5} \leq x \leq \sqrt{5}$. Wouldn't this mean that $\sqrt{5}$ is the greatest element in this subset?   
 A: It would, if $\sqrt{5}$ were in the subset. But it isn't, since $\sqrt{5}\notin \mathbb Q$.
A: 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 "$\le$")
