Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 5 down vote accepted

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.

share|cite|improve this answer
  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 $\exists n\in\mathbb N\colon x\notin \left(0,1+\frac1n\right)$.
share|cite|improve this answer

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), $$ (0,1]\supset \bigcap_{n\in\mathbb N} \left(0,1+\frac{1}{n}\right). $$

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.