Prove that the infinite intersection of $R_n = [0,\frac{1}{n}] \times [0,\frac{1}{2n}]$ equals $(0,0)$

Question

I understand that $R_n$ is a subset of $R_{n-1}$ due to the Archimedean Property, however I don't understand why the intersection is $(0,0)$.

Answer 1

Start with $(a,b) \in \bigcap_{n=1}^\infty R_n$ Then you will get 2 inequalities from the problem you pose. Then use the Archimedean property to proceed with y=0 and x=0.

Answer 2

Guide:

Let $(a,b) \in \bigcap_{n=1}^\infty R_n$, suppose $a>0$, can you find $m$ such that $\frac1{m}<a$?

Answer 3

If $(x,y) \in \bigcap_{n=1}^\infty R_n$, then

$(1) \quad 0 \le x \le \frac{1}{n}$ for all $n \in \mathbb N$

and

$(2) \quad0 \le y \le \frac{1}{2n}$ for all $n \in \mathbb N$ .

Suppose that $x>0$. By the Archimedean Property there is $m \in \mathbb N$ such that $\frac{1}{m}<x$. But this contradicts (1). Hence $x=0$.

A similar argument, use $(2)$, gives $y=0$.