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 ?

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    $\begingroup$ 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 $\inf$) $\endgroup$ – Gono Oct 21 '17 at 12:49
  • $\begingroup$ @G.Sassatelli Because I can say that $-4 < I < 4$ $\endgroup$ – DeltaWeb Oct 21 '17 at 12:50
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    $\begingroup$ 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! $\endgroup$ – 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, $\inf I=-\pi$.

  • $\begingroup$ 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 ? $\endgroup$ – DeltaWeb Oct 21 '17 at 12:56
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    $\begingroup$ @DeltaWeb No, you cannot, because between any two real numbers there's a rational number. $\endgroup$ – José Carlos Santos Oct 21 '17 at 12:58

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