So again I was studying topology and came across this problem:
Show that given two metric spaces $(X, d_1), (X, d_2)$ such that $d_2(x,y)=\frac{d_1(x,y)}{1+d_1(x,y)}$. A sequence will be a Cauchy sequence in the first space iff is a Cauchy sequence in the second.
So here's my attempt:
$\Rightarrow x_n$ is a Cauchy sequence in $(X, d_1)$ we want to prove that $x_n$ is also Cauchy in $(X, d_2)$
Since $x_n$ is Cauchy in $(X, d_1) \rightarrow d_1(x_n, x_m)<\epsilon$ for all $n,m \geq 0$ and $n, m \in Z_+$ and $\epsilon >0$.
$d_1(x_n, x_m)< \epsilon$
$\implies d_1(x_n, x_m) < \epsilon + \epsilon d_1(x_n, x_m)$
$\implies d_1(x_n, x_m)< \epsilon (1+ d_1(x_n,x_m))$
$\implies \frac{d_1(x_n,x_m)}{1+d_1(x_n,x_m)}<\epsilon$
And the inequalities don't change because everything added, multiplied or divided is positive.
$\Leftarrow x_n$ is a Cauchy sequence in $(X, d_2)$ we want to prove that $x_n$ is also Cauchy in $(X, d_1)$
Since $x_n$ is Cauchy in $(X, d_2) \rightarrow d_1(x_n, x_m)<\epsilon$ for all $n,m \geq 0$ and $n, m \in Z_+$ and $\epsilon >0$.
$\implies \frac{d_1(x_n, x_m)}{1+d_1(x_n,x_m)}<\epsilon$
$\implies d_1(x_n, x_m)< \epsilon + \epsilon d_1(x_m,x_m)$
$\implies d_1(x_n, x_m) -\epsilon d_1(x_nm x_m) < \epsilon$
$\implies d_1(x_n, x_m) (1-\epsilon) < \epsilon$
Since $\epsilon$ here cannot be greater than 1, we take $\epsilon \in (0,1)$
$\implies d_1(x_n,x_m) < \frac{\epsilon}{1-\epsilon}$
$\implies d_1(x_n, x_m) < \delta$ And $\delta >0$
That's my work, is it correct?
The first part is fine. The second part has two errors. You cannot say $\epsilon$ cannot be greater than $1$. Also you did not complete the proof.
Let $\epsilon' >0$ be arbitrary. Choose $\epsilon >0$ such that $\frac {\epsilon} {1-\epsilon} =\epsilon'$ which means $\epsilon= \frac {\epsilon'} {1+\epsilon'} $. Then we get $d_1(x_n,x_m) <\epsilon'$ finishing the proof. For the case $\epsilon \geq 1 $ find $n_0$ such that $d_1(x_n,x_m) <\frac 1 2$ for $n,m \geq n_0$ and note that $d_1(x_n,x_m) <\epsilon$ for $n,m \geq n_0$.