Does $M:=\{x \in \mathbb{Q}: x^2<7\}$ $M\subseteq\mathbb{R}$ have an infimum, supremum, min, max? $M\subseteq\mathbb{R}$
$M:=\{x \in \mathbb{Q}: x^2<7\}$ 
Does $M$ have an infimum, supremum, min, max?
My answer would be that it doesn't because $\sqrt{7}$ and $-\sqrt{7}$ are $\not\in \mathbb{Q}$
Is that correct?
 A: Definition: An upper bound $u$ of $M \subset \mathbb{R}$ is an element $u \in \mathbb{R}$ such that $u \geq m$ for all $m \in M$.
Definition: A supremum of $M \subset \mathbb{R}$ is an element $x \in \mathbb{R}$ such that


*

*$x$ is an upper bound of $M$

*for each upper bound $u$ of $M$, $x \leq u$


According to the above definition, $x$ does not have to be in $M$, which is perhaps where you have gotten confused.
Example: $M = [0, 1)$ does not have a maximum, since for any $x \in [0, 1)$, we can define $y = (x + 1)/2 \in M$, noting that $y > x$.
Example: $M = [0, 1)$ has a supremum, namely $1$. Can you prove this?
Once you have tackled the above example, you should be able to answer your original question (hint: use also the fact that any real number can be defined as the limit of a sequence of rational numbers).
A: The set does indeed have a supremum and infimum (use the density of $\Bbb Q$ in $\Bbb R$). However, it does not have a max/min as you say, because $M$ has a maximum if and only if $\sup M\in M$ and $M$ has a minimum if and only if $\inf M\in M$.
