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I have the following question to answer, $$\text{Let } A_i = \left(1,1+\frac{1}{i}\right) \text{ for }i\in\mathbb{Z}^+\text{. Compute } \bigcup_{i=1}^{\infty}A_i\text{ and }\bigcap_{i=1}^{\infty}A_i$$ and came with $\bigcup_{i=1}^{\infty}A_i = (1,2)$ and $\bigcap_{i=1}^{\infty}A_i = (1,1)$ through iteration and noticing a pattern. I was curious if the assumption $(1,1) = \emptyset$ is a acceptable one?

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    Why not write directly as $$\bigcap_{i=1}^{\infty}A_i=\emptyset?$$ – 2017-02-15
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    I just wasn't sure if that was correct – 2017-02-15
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    Are you familiar with the Archimedean Property? – 2017-02-15
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    Some what but not within sets – 2017-02-15
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    Well, either the answer is the empty set or it isn't. If it is write that it is the empty set. If it isn't, find out what it is and write that. – 2017-02-15
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    How was $(a,b) $ defined? Some texts say for any $a$(1,1) $ is undefined because $1 \not < 1$. Other texts leave off the requirement that $a < b $ so that $(k,j)=\emptyset $ if $j\le k$. It's sometimes convenient to use that notation (as it is here) but it is not standard. Either way you *must* explain *why* the intersection is (1,1) and *what* you mean by it. – 2017-02-15
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    One thing to state is. If $x\in Interesction $ then i) x > 1 and ii) x < 1+ 1/i for all natural i, iii) there is no such number that does both i and ii. So x doesn't exist. So the intersection is empty. – 2017-02-15
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    @fleablood In the textbook $(a,b)$ is defined as $(a.b) = \{x|a$\emptyset$. Was just looking for confirmation. – 2017-02-15
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    In that definition than yes, the set of all 1 $\cap_{i=1}^n =(1,1+1/n)$ so $\cap_{i=1}^{\infty}=(1,1+1/\infty)=(1,1) $ isn't *quite* right. It's the right idea, but it's... abusive. – 2017-02-15
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    @fleablood I kinda did it the way you showed in the example, but used limits. It looked like $\bigcap_{i=1}^{n}A_i = (1,1+1/i) \rightarrow \lim_{n\to\infty}\bigcap_{i=1}^{n}A_i = \lim_{n\to\infty}(1,1+1/n)$ and then calculated $(1,1)$ – 2017-02-15
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    I'll need to review my text and it's too late for that... yeah. I don't know if it's okay to say $\cap_{\infty} = \lim \cap_n $. In fact, I'm pretty sure it is not. But I'll let someone else say for sure. – 2017-02-15

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Suppose that $$\bigcap_{i=1}^{\infty}A_i\neq\emptyset.$$ Let $x\in\bigcap_{i=1}^{\infty}A_i$. Then $x>1$ and $x<1+\frac{1}{i}$ for all $i\in\Bbb N$. Since $x-1>0$, we can find $n\in\Bbb N$ such that $\frac{1}{n}1+\frac{1}{n}$ for some $n\in\Bbb N$. Did you see the contradiction here? This proves that $$\bigcap_{i=1}^{\infty}A_i=\emptyset.$$