$(+\infty,+\infty,\cdots,+\infty)$ exists in $\mathbb{R^{n}}$? $(\mathbb{R}^{n},d)$ 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! 
 A: For all $m$ you have $diam (G_m) = + \infty$, so the condition
$$\lim_m diam (G_m) = 0$$
is not satisfied. In particular you are not allowed to use the proposition in this case.
A: Let's walk through the proof of the proposition you're appealing to and see where the argument breaks down.  So suppose we have a sequence of closed, nonempty sets $F_n \subset X$ so that $F_n \supset F_{n+1}$ or all $n$ and $diam(F_n) \to 0$ as $n \to \infty$.  For each $n$, choose arbitrarily a point $x_n \in F_n$.  Then I claim that the sequence $x_n$ is Cauchy; indeed, let $\epsilon > 0$.  Since $diam(F_n) \to 0$, choose $n$ large enough that $diam(F_n) < \epsilon$.  Let $k, m > n$.  By the superset condition, $F_n \supset F_m, F_k$; so, we have that $d(x_m, x_k) \le d(x_m, x_n) + d(x_n, x_k) \le diam(F_n) + diam(F_n) = 2diam(F_n) < 2 \epsilon$. 
note the nontrivial usage of the assumption that the diameters decrease to 0
So the sequence is Cauchy.  Since $X$ is a complete metric space, there's a point $x \in X$ so that $x_n \to x$.  At this point, the proof proceeds by showing that $x \in F_n$ for all $n$ (this should be clear) and that no other point $y \in X$ can lie in $F_n$ for all $n$ (y would have to be closer to x than the diameter of any $F_n$ so....)
But this is all I need, so I'll stop here.  What happens when we apply this construction to your example?  We start picking points $x_m \in G_m$.  Great.  But $diam(G_m) = \infty$.  So I get this great sequence of points, but I have no upper bound on their distances from one another.  They can lie basically anywhere! Suppose I picked the point $x_m = (m, ..., m)$.  The distance between successive points is always 1.  The sequence isn't Cauchy.
