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

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 ?

• You are sure you mean $$]-\pi;\pi[\; \cup\; \Bbb Q$$ (what's not bounded) and not $$]-\pi;\pi[ \;\cap \;\Bbb Q$$ (what is bounded but has a $\sup$ and $\inf$) – Gono Oct 21 '17 at 12:49
• @G.Sassatelli Because I can say that $-4 < I < 4$ – DeltaWeb Oct 21 '17 at 12:50
• No, not for $$]-\pi;\pi[\; \cup\; \Bbb Q$$. Obviously $$5 \in ]-\pi;\pi[\; \cup\; \Bbb Q$$ because $5 \in \Bbb Q$... So i assume you meant $$]-\pi;\pi[\; \cap\; \Bbb Q$$, right? AND: Do you mean a $\sup$ in $\Bbb R$ or a $\sup$ in $\Bbb Q$? This is important! – Gono Oct 21 '17 at 12:52

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, $\inf I=-\pi$.
• Can't I find an irrational number smaller than Pi and bigger than all elements of this set ? Let's say the difference between $\pi$ and this irrational is extremly small ? – DeltaWeb Oct 21 '17 at 12:56