0

I am sure that this question has already been asked, but I didn’t want to look at the full answer straight away. I just want a hint to help me get out of this stalemate. So far this is what I got:

Suppose $x<y$ and let $S=\{q\in\mathbb{Q}: q<y\}$. By definition, $y$ is an upper bound for S, therefore, $S$ is bounded above. Moreover, the Least Upper Bound property states that $S$ has a supremum in $\mathbb{R}$, call it sup$(S)$.

Here is where I am stuck. If I can prove that sup$(S)=y$. I can conclude the proof, since by definition, $\forall\epsilon>0,\exists r\in S:r>y-\epsilon$. Let $\epsilon = y-x$ and $\exists r\in S: r>y-(y-x)=x.$ Moreover, since $r\in S$, then $r\in\mathbb{Q}: r<y$. So $x<r<y$. QED

But I cannot prove that $y=sup(S)$ for the life of me. It seems to require that $x<r<y$, which is what i want to prove in the first place.

Igor Mello
  • 429
  • 1
    here is a duplicate question. Well, one of them. There is no shortage. – lulu Dec 31 '22 at 13:49
  • As you remark, knowing that the sup is $y$ is more or less equivalent to what you are being asked. – lulu Dec 31 '22 at 13:54
  • If you want a hint to get started, first prove that there are arbitrarily small positive rationals numbers. Deduce that there is a positive rational less than $\frac {y-x}{10}$. Why does that help? (Note: the $10$ isn't particularly important, I just picked a "sufficiently large" denominator). – lulu Dec 31 '22 at 13:56
  • 1
    WLOG $x>0$. The usual approach is to use Archimedean property twice. First to find $n\in\mathbb N$ such that $\frac{1}{n}<y-x$ and then to find the smallest $m\in\mathbb N$ such that $m\frac{1}{n}>x$. The proof that $x<\frac{m}{n}<y$ is then easy. –  Dec 31 '22 at 13:57
  • Take the average of $x$ and $y$. Then take a decimal approximation of that average that is greater than $x$ and less than $y$. – John Douma Jan 01 '23 at 02:03

0 Answers0