Total boundedness for a non standard metric on $\mathbb{R}^n$

Question

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.

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.