Does $\inf(E)$ exist in a case such as the following?
$$E = \{r \in \mathbb{Q} | r < \sqrt{12}\}$$
I know that we can construct sets which upper bound the integers, but which have no least upper bound on the integers. I assume this is also the case for lower bounds of all rational numbers?
Because the set $E=\{r\in\mathbb{Q}| r<\sqrt{12}\}$ has no lower bound, it also cannot have an infimum, (greatest lower bound). However if you are looking for a supremum (least upper bound), then $\sup (E)=\sqrt{12}.$ This is easy to prove, since $\sqrt{12}$ is clearly both an upper bound and a limit point of the set $E.$