How i can show that this set is compact?

Question

Let $(M,d)$ a metric space.

The following metric space $(M,d_A)$ is defined as:

$d_A(x,y)=\frac{d(x,y)}{1+d(x,y)}$

Show that if a set $F$ is compact on $(M,d_A)$ then is compact on $(M,d)$.

How i can show that?. I can't find a correct cover for $F$.

Answer 1

Hint: As Arnaud D pointed out, compactness is a topological property and it therefore suffices to prove that the topologies induced by these two norms coincide. It suffices to prove (why?) that for every $x\in M$ and every $R>0$ there exists $r_1, r_2>0$ such that $$ B_{d_A}(x, r_1) \subseteq B_d(x, R) \subseteq B_{d_A}(x, r_2). $$ Now write this down in terms of the two metrics and use for the first inclusion the fact that $x\mapsto \frac{x}{1+x}$ is continuous and for the second $d_A(x,y)\leq d(x,y)$.