Continuous linear operator preserving Cauchy sequence in metric vector spaces Let $T: X \rightarrow Y$ be a continuous linear operator. $(X,\rho), (Y,\xi)$ are linear metric spaces and $\{x_n\} \subset X$ is a Cauchy sequence.
I need to show that $\{Tx_n\}$ is a Cauchy sequence.
For $T$ to be continuous means:
$\forall \varepsilon>0\,\, \exists \delta>0$ such that $\rho(x,y)<\delta \implies \xi(Tx,Ty)<\varepsilon.$
And for sequence to be Cauchy means:
$\rho(x_n,x_m)\rightarrow 0$ for $m,n \rightarrow \infty.$
How do I show that $\xi(Tx_n,Tx_m)\rightarrow 0$ for $m,n \rightarrow \infty?$ So far I suspect that the operator needs to be linear, because there are counterexamples for nonlinear continuous mappings (Namely $\{\frac{1}{n}\}$ and $T(x)=\frac{1}{x}$.).
I'm sorry if I wasn't thorough enough with my search but, I've only seen a similar question here with normed vector spaces instead of metric or the mapping was uniformly continuous and I wasn't able to translate it to my problem.   
EDIT:
Forgot to include this attribute of metrics:
$\rho(x_1+x,x_2+x)=\rho(x_1,x_2) \forall x_1,x_2 \in X$
 A: The central theme here doesn't need linearity:

If $(X, \rho), (Y, \xi)$ are metric spaces, $f : X \to Y$ is uniformly continuous, and $\{x_i\}$ is a Cauchy sequence in $X$, then $\{f(x_i)\}$ is a Cauchy sequence in $Y$.



*

*That $\{x_i\}$ is Cauchy means that $\forall \epsilon > 0, \exists N$ such that $\forall m,n > N, \rho(x_n, x_m) < \epsilon$

*That $f$ is uniformly continuous means that $\forall \epsilon > 0, \exists \delta > 0$ such that $\forall x, y\in X,\rho(x,y) <\delta\implies\xi(f(x), f(y)) < \epsilon$. Note the difference between this and ordinary continuity:



$f$ is continuous if $\forall y \in X$ and $\forall \epsilon > 0, \exists \delta > 0$ such that $\forall x\in X,\rho(x,y) <\delta\implies \xi(f(x), f(y)) < \epsilon$.

Regular continuity allows you to choose different $\delta$ values for each $y$, while uniform continuity requires a single $\delta$ to work for every $y$.
To prove the theorem above, we need to show that $\forall \epsilon > 0, \exists N$ such that $\forall m,n > N, \xi(f(x_n), f(x_m)) < \epsilon$. This is straight-forward:
Given $\epsilon$, pick the corresponding $\delta$ that satisfies the uniform continuity condition on $f$. Use this $\delta$ as the "epsilon" for the Cauchy condition on $\{x_i\}$ and consider the corresponding $N$. For $m,n > N$, we have $\rho(x_n, x_m) < \delta$. Now doesn't that look a lot like the uniform continuity condition on $f$?

So why does your problem have the linearity condition? Because you are only given that $T$ is continuous, not uniformly continuous. What you still need to do is to show that because of the linearity, continuity of $T$ implies uniform continuity. This is a simple calculation, as continuity of $T$ at $y = 0$ is sufficient to prove uniform continuity everywhere (by the translation invariance of the two metrics that you added late to your post).
