What's the supremum of $]-\pi; \pi[ \cap Q$

Question

I'm trying to find the supremum and infimum of this set .

$$I = ]-\pi; \pi[ \cap Q$$

The problem is that it accept a least upper bound (resp : greatest lower bound) since it's a bounded set and non empty . However there's an endless rational numbers that are bigger than all elements of this set due to the density of $Q$ in $R$ , also from the archimdean property we know that for every $\epsilon > 0$ there exist some rational number smaller than $\epsilon$ . Which mean that I can always find a smaller number then any rational I suppose to be the supremum

Can anyone explain why this has or hasn't any $sup$ or $inf$ while it's a non empty set and bounded ?

Answer 1

I am assuming that you meant $(-1,1)\cap\mathbb Q$.

Since $\pi$ is an upper bound of $I$, $\sup I\leqslant\pi$. Actually, $\sup I=\pi$, because if $x\in\mathbb R$ is such that $x<\pi$, there's a rational $q\in(x,\pi)$ and, in fact a rational greater than $0$. Therefore $x$ is not $\sup I$. So, $\pi$ is the least upper bound of $I$.

By a similar argument, $\inf I=-\pi$.

Answer 2

Set $A:=(-π,π)\cap \Bbb Q$. First of all, $\forall x\in A: x< π$, so $π$ is an upper bound of $A$. Secondly, suppose an $\epsilon>0$; from the density of $\Bbb Q$ in $\Bbb R$ there exists a $q\in \Bbb Q$ (an interpretation of density is that between any two real numbers there exists a rational), such that $π-\epsilon<q<π$, so $q\in A$. Consequently, by the characterization of supremum we get $supA=π$. q.e.d.