1
$\begingroup$

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 ?

$\endgroup$
  • 3
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
2
$\begingroup$

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, $\inf I=-\pi$.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.