# Intersection of an Infinite Indexed Family of Sets

In a mathematics course, I came across the following problem:

Identify (with a short proof) the following set: $\bigcap_{n\in\mathbb{N}}\left(0,1+\frac{1}{n}\right)$, where $\left(a,b\right)=\left\{x\in\mathbb{R}:a<x<b\right\}$.

Clearly, we are working with $\mathbb{N}_1$.

Analytically, I worked out the problem as follows:

$\bigcap_{n\in\mathbb{N}}\left(0,1+\frac{1}{n}\right)=\left(0,2\right) \cap \left(0,\frac{3}{2}\right) \cap \left(0,\frac{4}{3}\right) \cap \left(0,\frac{5}{4}\right) \cap \dots$

Essentially, it is clear that the upper bound on the interval will tend to $1$, thus producing $\left(0,1\right]$ for the intersection; this can be shown quite easily with a limit. However, I am fairly sure that I am not permitted to make use of a limit in my proof.

Is there another way to prove this?

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I think it is clear that $(0,1]$ is contained in the intersection. I would show the reverse inclusion by contrapositive: Suppose $x\notin (0,1]$. Then either $x\leq 0$, in which case $x$ is clearly not in the intersection of sets, or $x>1$. If $x>1$ then you should be able to show (this is where the "limit" part of the proof comes in) there exists $n\in \mathbb N$ such that $1+1/n <x$. But then $x\notin (0,1+1/n)$, so it is not in the intersection of sets.

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1. Make an educated guess what $S:=\bigcap_{n\in N}\left(0,1+\frac1n\right)$ might be.
2. For every $x\in S$, show that $x$ is in the intersection, that is $\forall n\in \mathbb N\colon x\in\left(0,1+\frac1n\right)$.
3. For every $x\notin S$, show that $s$ is not in the intersection, that is $\exists n\in\mathbb N\colon x\notin \left(0,1+\frac1n\right)$.
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Your proof is quite acceptable. Clearly $$(0,1]\subset \bigcap_{n\in\mathbb N} \left(0,1+\frac{1}{n}\right).$$ One the other hand, $1+\frac{1}{n}\rightarrow 1$, so any number larger than $1$ will not be in the intersection (smaller than $0$ is trivial), $$(0,1]\supset \bigcap_{n\in\mathbb N} \left(0,1+\frac{1}{n}\right).$$

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The OP's proof would not be acceptable in most "intro to proofs and sets" courses, certainly not in the one I'm teaching now. The claim that $1 + \frac1n \to 1$ (which itself probably requires justification) implies no number larger than $1$ will be in the intersection requires a clear justification. – Santiago Canez Jul 10 '14 at 17:42
It doesn't look like the OP provided his full answer, rather simply a sketch of his proof. Under such circumstances, I would deem it quite reasonable to conclude directly that if $x_n\rightarrow 1$, then $x_n<a$ for some $n$ if $a>1$. – Jonas Dahlbæk Jul 10 '14 at 17:45