Completeness of the $Q$ given the set $\{ x \in Q | x < \pi \}$ and its usual order

Question

Is the set of rational numbers $\mathbb{Q}$ complete given the set $\{ x \in \mathbb{Q} | x < \pi \} $ and the usual order?

I have stumbled upon multiple resources saying that it is not, because there is no least upper bound for this set in $\mathbb{Q}$. However, even though I am familiar with the definition for "the least upper bound" I don't get why 4 is not simply the least upper bound for this particular example.

I am a mathematics student in my second year, and all help would be much appreciated as I am trying to understand this concept more thoroughly.

Answer 1

There are smaller rational upper bounds than 4, like 3.2, 3.15, 3.142, 3.1416, and so forth. None of them is the smallest, though.

This is a bit harder to see for $\pi$ than it is for $\sqrt{2}$, for which one can easily give a more explicit construction (given any rational upper bound $x$, $(x+2/x)/2$ is a smaller rational upper bound).

Answer 2

Consider the following sequence of rational numbers,

$$3,3.1,3.14,3.141,3.1415,.......$$

That is an increasing sequence of rational numbers which converges to $\pi$

Its's supremum is $\pi$ which is not a rational number, therefore that set of rationals less than $\pi$ is not complete.