1

I am looking for nice ways of proving that 1 is the least upper bound of the set $A=\{\frac{1}{y+x};x>1\}$ where y>0 is fixed.

One (not so nice and unachieved) way is to first prove that 1 is a upper bound of any element of A and then use the characterization of the least upper bound (i) ($\forall\varepsilon>0;\exists a\in A:1-\varepsilon<a$) :

Let $\varepsilon>0 ; n\in \mathbb{N}\setminus\lbrace{0}\rbrace$

We put y=1 and suppose there exists x such that (i) is verified, then we put: $1-\varepsilon<\frac{1}{1+(\frac{1}{n}+1)}$

Thus:$$\begin{align*}1-\varepsilon<\frac{n}{2n+1}\\ 1-\varepsilon<\frac{1}{2}(1-\frac{1}{2n+1})\\ (2n+1)(2\varepsilon-1)>1\end{align*}$$ If $\varepsilon>\frac{1}{2}$, then $2\varepsilon-1>0$. As $\mathbb{R}$ is archimedean: $\exists n_{0}\in\mathbb{N}\setminus\lbrace{0}\rbrace:(2n_{0}+1)\in\mathbb{N}\setminus\lbrace{0}\rbrace, (2n_{0}+1)(2\varepsilon-1)>1$. So (i) is verified in this case ($\varepsilon>\frac{1}{2}$).

If $0<\varepsilon<\frac{1}{2}$ then $$\begin{align*}2n+1<\frac{1}{2\varepsilon-1}<-1\\ 2n+1<-1\\ n<-1\end{align*}$$ which is impossible. So for $0<\varepsilon<\frac{1}{2}$ we must choose another x different from $1+\frac{1}{n}$ to prove the condition (i). I tried with x=n+1 but it's a little bit complicated. THANK FOR HELPING!

Yagami
  • 101

1 Answers1

0

For any $y \not\in \{-x,-1\}$, \begin{align*} x &> 1 \\ y + x &> y + 1 > 1 \\ \frac{1}{y + x} &< \frac{1}{y + 1} < \frac{1}{1} \\ \end{align*} So $1$ is an upper bound. However, it is not the least upper bound of $A$ once $y$ is fixed, because $y \neq 0$. $\frac{1}{y+1}$ is the least upper bound, which you shold be able to show by using the derivative to show $\frac{1}{y+x}$ is monotonically decreasing in $x$ then taking $x \in (1,1+\varepsilon)$.

Eric Towers
  • 67,037
  • How do we prove the condition (i) with $x\in(1;1+\varepsilon)$? Or is it a different way from using (i)? @EricTowers – Yagami Sep 20 '20 at 03:53
  • @MuruemDiop : If this were elementary calculus, we would take $$\lim_{x \rightarrow 1^+} \frac{1}{y+x} \text{.} $$ That is, we would study $x \in (1,1+\varepsilon)$... – Eric Towers Sep 20 '20 at 03:57
  • I understand now, so it seems just more "natural" using limits to prove the supremum. Thanks a lot @EricTowers – Yagami Sep 20 '20 at 04:06