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$(\mathbb{R}^{n},d)$ 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!

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    $\begingroup$ What is $\operatorname{diam} G_m$? $\endgroup$ Feb 13, 2015 at 11:42
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    $\begingroup$ Why all this generalizing? It doesn't even hold one dimensionally I think. [a,a) is the empty set right? $\endgroup$
    – user198044
    Feb 13, 2015 at 11:46
  • $\begingroup$ @Daniel Fischer: diam$G_{m}$=sup$\lbrace d(x^{'},x^{''}):x^{'},x^{''} \in G_{m} \rbrace$ $\endgroup$
    – user202406
    Feb 13, 2015 at 11:50
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    $\begingroup$ I intended that you find the value. Knowing the definition of course helps with that. $\endgroup$ Feb 13, 2015 at 11:57
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    $\begingroup$ No--just answer Daniel's question and you will see. $\endgroup$
    – Did
    Feb 13, 2015 at 12:17

2 Answers 2

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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.

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  • $\begingroup$ I made a stupid mistake . I'm glad for the help! $\endgroup$
    – user202406
    Feb 13, 2015 at 13:16
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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.

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