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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

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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