If two metrics $d_i$ on the same set $X$ have the same Cauchy sequences (i.e. if a sequence is Cauchy for the first metric, it is also Cauchy for the other one and vice versa), does this imply that the two metric spaces are equivalent (i.e. generate the same topology)?
Please help!
Note that if $d_1$ and $d_2$ have the same Cauchy sequences, then they also have the same limits. If $\{a_n\}\rightarrow b$ in $d_1$, then consider the sequence $a_0, b, a_1, b, a_2, b, . . . $. Clearly this is Cauchy in $d_1$, so Cauchy in $d_2$, but that means it (and so the original sequence) must approach $b$ in $d_2$.
Now suppose $C$ is not closed with respect to $d_2$ but closed with respect to $d_1$. Taking $a$ in the $d_2$-closure of $C$ but not in $C$; then we can get a Cauchy sequence of elements of $C$ which $d_2$-approaches $a$ . . .
By symmetry, we get that every $d_2$-closed set is $d_1$-closed. Since $d_1$ and $d_2$ have the same notion of "closedness," they induce the same topologies on $X$.