I want to prove this two things:

1) $(\mathbb{R},d_B)$ is not totally bounded.

where $d_B=\frac{|x-y|}{1+|x-y|}$ and $d_E$ is the Euclidean metric.

2) $B_M(0)$ is totally bounded in $\mathbb{R}^n$.

For the first one, we know that every norm in $\mathbb{R}^d$ are equivalent so $(\mathbb{R},d_B)$ is homeomorphic to $(\mathbb{R},d_E)$ so If we assume that $ (\mathbb{R},d_B)$ is totally bounded, then so is $(\mathbb{R},d_E)$ but we know that this is not even bounded, so the result follows.

Well I think that the norm that induce that metric is $$\frac{||x||}{1+||x||}$$ where $||x||=\sqrt{|x^{2}|}$

Am I right?, and if it wrong can you help me to fix it please?

For two, I don't how to prove it, my definition of totally bounded is the following:

For every $\varepsilon>0$ there exists a finite set of points $x_1,\dots,x_n\in A$ such that $A=\cup_{i=1}^n B_{\varepsilon}(x_i)$, where $B_{\varepsilon}(x_i)$ denotes the open $\varepsilon$-ball with center $x_i$.

So I have to pick an arbitrary $\varepsilon$ and give general points to bound the ball of radius $M$.

Can someone help me with that verification and the part two please?

Thanks a lot in advance.


1 Answer 1


For #1, that doesn't work, because you don't know that $d_B$ is induced by a norm. In fact it is not, because all norms on positive dimensional spaces are unbounded, but that metric is bounded. (I might edit if you post an attempt and show where you get stuck.)

For #2, one open cover that you always have of a set $A$ in a metric space is

$$\{ B_\epsilon(x) : x \in A \}.$$

Now use compactness to get a finite subcover.

  • $\begingroup$ Thanks a lot for this :), so How can I fix 1?, and for 2 why do I use compactness? $\endgroup$
    – user162343
    Aug 27, 2015 at 1:08
  • $\begingroup$ @user162343 You don't have to use compactness, but this is an easy way to do it. You could also do it explicitly, if you want. $\endgroup$
    – Ian
    Aug 27, 2015 at 1:10
  • 1
    $\begingroup$ @user162343 So you have an uncountable open cover of the closed ball. You extract a finite subcover using the fact that the closed ball is compact. This subcover is also a cover of the open ball. $\endgroup$
    – Ian
    Aug 27, 2015 at 1:15
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    $\begingroup$ @user162343 Your proposed norm is not a norm, in particular it is not homogeneous. Homogeneity is the easiest way to prove that the metric induced by a norm is always unbounded, except in the zero dimensional case. And it is easy to see that your metric is bounded. $\endgroup$
    – Ian
    Aug 27, 2015 at 13:42
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    $\begingroup$ It might help to know whether your definition of total boundedness is relative to an ambient space, or only in the metric space in question. If it is relative to an ambient space, then there is no problem. If not, then you are right, some of the centers of the balls that I gave will be "on the boundary", which will require some correction to get an appropriate cover. Here's a suggestion for the modification: if you have a ball at center $Mx$ and radius $\epsilon$, replace it with a ball at center $(M-\epsilon/2)x$ and radius $2 \epsilon$. $\endgroup$
    – Ian
    Aug 28, 2015 at 14:20

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