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Is $\bigcap\limits_{n=1}^\infty (0, 1 + \frac{1}{n})$ equal to $(0, 1)$ or $(0,1]$? Help is appreciated.

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$1\in (0,1+\frac 1 n)$ for all $n$ what does that tell you? –  azarel Jun 25 '12 at 19:43
[Low accept rate] discourage some from answering your posts. When a question has been answered to your satisfaction, it's considered good form to click the check mark below the voting arrows for that answer. The person whose answer you accept (you can only accept one) gets a small reputation boost, and you send a signal to others that you're no longer in need of an answer. BTW, it's not too late to go back to your old posts and accept your favorite answers. –  draks ... Jul 2 '12 at 8:14

3 Answers 3

up vote 2 down vote accepted

For all $n$, $1\in\left(0,1+\frac{1}{n}\right)$ and $\displaystyle\bigcap_{n=1}^{\infty}\left(0,1+\frac{1}{n}\right)=\left\{x,\,\forall n,\,x\in\left(0,1+\frac{1}{n}\right)\right\}$.
So $1\in\displaystyle\bigcap_{n=1}^{\infty}\left(0,1+\frac{1}{n}\right)$.

More generally :
The same way, we can show that $\displaystyle(0,1]\subset\bigcap_{n=1}^{\infty}\left(0,1+\frac{1}{n}\right)$.
If $x>1$, it exists $n$ such as $x\notin\left(0,1+\frac{1}{n}\right)$, so $\displaystyle x\notin\bigcap_{n=1}^{\infty}\left(0,1+\frac{1}{n}\right)$.
If $x\le 0$, $x\notin(0,\frac{1}{1})$, so $\displaystyle x\notin\bigcap_{n=1}^{\infty}\left(0,1+\frac{1}{n}\right)$.
Conclusion : $\displaystyle\bigcap_{n=1}^{\infty}\left(0,1+\frac{1}{n}\right)=(0,1]$.

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Well... is $1$ included in all sets being intersected, or is it missing from at least one? If it is missing from at least one, then it is not in the intersection. If it is included in all sets being intersected, then it is in the intersection.

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Hint: For all $n$ we have $1\in(0,1+\frac1n)$.

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