$(\mathbb{R}^{n},d)$ is a metric space and $d$ is the standard metric on $\mathbb{R^{n}}.$ Let $(\mathbb{R^{n},\tau_{d}})$ is the topology space induced by metric space $(\mathbb{R}^{n},d)$ .We can prove that $(\mathbb{R^{n},\tau_{d}})$ is complete.
Propostion
A metric space $(X,d)$ is complete $\Rightarrow$ whenever $\lbrace F_{n} \rbrace$ is a sequence of nonempty subsets in $(X,d)$ satisfying:
$(a)$ each $F_{n}$ is closed; $(b)$ $F_{1}\supseteq F_{2}\supseteq \cdots ;$ $(c)$ diam$F_{n}\rightarrow 0,$ then $\bigcap_{n=1}^{\infty}F_{n}$ is a single point.
I structure $ G_{m}=[m,+\infty)^{n}=\underset{n\quad times}{\underbrace{[m,+\infty)\times[m,+\infty)\times\cdots\times[m,+\infty)}}.$ Obviously,$\lbrace G_{m}\rbrace$ satisfied the above three conditions, but $\bigcap_{m=1}^{\infty}G_{m}=\varnothing$. This is contradicted against $\bigcap_{m=1}^{\infty}G_{m}$ is a single point. If we define $(+\infty,+\infty,\cdots,+\infty)$ is also a point in $(\mathbb{R^{n},\tau_{d}})$ ,we can eliminate the contradiction,because $\bigcap_{m=1}^{\infty}G_{m}=(+\infty,+\infty,\cdots,+\infty)$. From the above statement, whether this definition is correct? Who can give me some details? Any of your help will be appreciated!