Prove that $\bigcap\limits^{\infty}_{n=1} \left(0, \frac{1}{n} \right)= \emptyset$

Question

$\def\emptyset{\varnothing}$Prove that $$\bigcap^{\infty}_{n=1} \left(0, \frac{1}{n} \right)= \emptyset.$$

I need help writing a formal proof. So far I reason that for all $n$, $\dfrac{1}{n}$ is an upper bound for $\left(0,\dfrac{1}{n} \right)$ but the supremum of all such intervals is $0$. Similarly, $0$ is the infimum for all such intervals as well. This means the interval common to all intervals of the form $\left(0,\dfrac{1}{n} \right)$ is $(0,0)$ (interval containing the supremum and infimum) which is empty.

Thanks for help in advance.

Answer 1

Your reasoning is correct; to write a formal proof (and use your reasoning):

Suppose $S=\bigcap_{n=1}^\infty\left(0,\frac{1}{n}\right)\neq\varnothing$. Then let $x\in S$, hence $x>0$ by definition of open intervals. Now, by the archimedean principle, for $N$ sufficiently large, $\frac1N<x$, thus $x\notin (0,\frac1N)$. This means $x\notin\bigcap_{n=1}^\infty(0,\frac1n)$. This is a contradiction, so $S$ is empty.

Answer 2

BWOC, suppose $\exists x \in \cap_{n=1}^\infty (0, \frac{1}{n})$. Then $x>0$, but $\forall n \in N, x<\frac{1}{n} $. By the archimedean property, $\exists n, x > \frac{1}{n}$, so this cannot be true. Thus the set must be empty.