Given $K = \mathbb N - \{0,1\}$ for all $n \in K$ and $B_n=\{x \in \mathbb R:\frac{1}{n}\le x<1+\frac{1}{n}\}$. Find $\bigcap_{n\in K}B_n$.
It is evident to me that the intersection is the interval $\big[0.5,1\big)$. In order to prove it I know I need to prove two directions.
First, $\bigcap_{n\in K}B_n \subseteq \big[0.5,1\big)$. If we plug in $n = 2$ then: $$ 0.5\le x <1.5 $$ and of course there're $x$ in $\big[0.5,1\big)$ when $n=2$. If $n$ is very large then: $$ 0<x<1 $$ so still there're $x$ in $\big[0.5,1\big)$.
I'm not sure how to do the second direction especially when the Archimedean principle is should be used (the principle is that for all $x > 0$, there exists $n \in \mathbb N$ such that $0<\frac{1}{n}<x$).
I think we need to substitute maybe $1+\frac{1}{n}$ as $\alpha$ and then somehow redefine $x$ but I'm not really seeing this.