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Can someone help me . I need to find the supremum of $A= \bigcap_{n \ge 1} \left(0, 1 + \dfrac{1}{n}\right)$. I know it is an interval but then i dont know .

Umberto P.
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jan
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    Do you mean $\bigcap_{n \leq 1}$ or $\bigcap_{n \geq 1}$? – Arthur Oct 26 '15 at 14:12
  • i mean ⋂n≥1. oke – jan Oct 26 '15 at 14:14
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    Try writing out the set. What numbers belong to every interval $(0,1+1/n)$? – Umberto P. Oct 26 '15 at 14:25
  • Out of curiosity, the result of the intersection is the interval $ \left( 0, 1 \right]$ isn't? – user8469759 Oct 26 '15 at 14:29
  • @Lukkio Yes it is – R_D Oct 26 '15 at 14:29
  • Can you tell me how you find (0,1]? – jan Oct 26 '15 at 14:33
  • Well if you call $A_N = \bigcap_{n=1}^N \left( 0, 1 + \frac{1}{n} \right)$ then $A_N = \left( 0, 1 + \frac{1}{N} \right)$, so for each $N$ you have $1 \in A_N$, the statement specifically holds for $N \rightarrow \infty$. For each value $y$ greater than $1$ i guess you can easily find $N$ such that $y \notin A_N$, so basically this is the idea behind my reasoning. I don't know if it could be simpler or not. – user8469759 Oct 26 '15 at 14:45
  • At this point actually i think the answer is easy, since the least upper bound of $A$ is $1$. – user8469759 Oct 26 '15 at 14:48

1 Answers1

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Let $A = \bigcap_\limits{n\geq 1}\left(0,1+\dfrac{1}{n}\right)$ and $B=(0,1]$

$B\subseteq A$ is clear - If $x\in B$ then $0<x\leq1\leq1+\dfrac{1}{n}$ for all $n\geq 1$ . So $x\in A$

If $x\in A$ then $x\in \left(0,1+\dfrac{1}{n}\right)$ for all $n\geq 1$ , that is, $0<x<1+\dfrac{1}{n}$ for all $n\geq 1$.

Suppose $x>1$ then there exists an $\epsilon >0$ such that $x=1+\epsilon$ . So by Archimedean property there exists an $N \geq 1$ such that $\epsilon > \dfrac{1}{N}$ . So $x>1 + \dfrac{1}{N}$ which is a contradiction. Thus $x\leq 1$. So $x\in B$

Thus $A=B$ and we are done.

So your supremum is 1.

R_D
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  • Why is that exactly a contradiction? – jan Oct 26 '15 at 15:08
  • If $x\in A$ then $x< 1+\dfrac{1}{n}$ for all $n\geq 1$. But we have found an $N$ for which this is not true, that is for which $x>1+\dfrac{1}{N}$ . Hence the contradiction. – R_D Oct 26 '15 at 15:11
  • Why is the supremum not (0,2) because $ \left(0, 1 + \dfrac{1}{1}\right)=0,2$ – jan Oct 26 '15 at 15:15
  • You are only looking at the case $n=1$. That gives you $(0,2)$. But you need the intersection over all $n\geq 1$. Consider for example $n=2$ what is $(0,1+\dfrac{1}{1})\cap(0,1+\dfrac{1}{2})$? – R_D Oct 26 '15 at 15:19
  • Also, supremum cannot be an interval. It is a number. It is the least upper bound of a set. Perhaps you should re-look at the definition. – R_D Oct 26 '15 at 15:23