Supremum of sets containing inequalities. Let $r \in \mathbb{R}$ be fixed. Prove that:
a) $\sup\{q \in \mathbb{Q}: q \leq r\} = r$.
b) $\inf\{q \in \mathbb{Q}: q \leq r\} = r$.
It looks so strange for me that the supremum equals the infimum, Could anyone help me or at least say that the problem contains a mistake? 

The statement is saying that if $r$ is a real number, you can find rational numbers as close to $r$ as you want, either a little bit greater than $r$ or a little bit less. This is easy to see if you consider the decimal expansion of $r$. For example,
$$\sqrt{2} = 1.41421\ldots$$
Consider the rational numbers
$$1.5,1.42,1.415,1.4143,1.41422,\ldots > \sqrt{2}$$
and
$$1.3,1.40,1.413,1.4141,1.41420,\ldots < \sqrt{2}.$$
