supremum of $A= \bigcap_{n \ge 1} \left(0, 1 + \dfrac{1}{n}\right)$

Question

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 .

Answer 1

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.