If $X = \{x_n:n \in \mathbb N\}$ is a cauchy sequence in a metric space $S$ and $f : S \rightarrow T$ is continuous , is $f(x_n)$ a cauchy sequence? If $X = \{x_n:n \in \mathbb N\}$ is a cauchy sequence in a metric space $S$ and $f : S \rightarrow T$ is a continuous function where $T$ is an another metric space , is $f(x_n)$ a cauchy sequence?
Attempt: $f: S \rightarrow T$ is a continuous function at $x_m \in S$
$\implies \forall \epsilon >0, \exists \delta >0$ such that $d_T(f(x_n),f(x_m) ) < \epsilon$ whenever $d_S(x_n,x_m)< \delta~~~~............(1)$ 
$X$ is a cauchy sequence $\implies  \forall \delta >0, \exists k \in \mathbb N$ such that $d_S( x_n ,  x_m  ) < \delta$ whenever $ m,n \geq k~~~~............(2)$
Substituting $(2) $ in $(1)$, we get : 
$\forall \epsilon >0, \exists  k \in  \mathbb N $ such that $d_T(f(x_n),f(x_m) ) < \epsilon$ whenever $ m,n \geq k~~~~............(2)$
$\implies f(X)$ must also be a cauchy sequence.

Please tell me where I could  have gone wrong in the attempt shown above?

Thank you for your help..
 A: Let $x_n = \{1/n \colon n \in \mathbb{N}\}$ and take $f(x) = 1/x$ over some decent space $S$, then $f(x_n)$ is clearly not cauchy, as it is the set $(1,2,3,\dots)$
Some things you could prove/add


*

*If a function takes cauchy sequences into cauchy sequences then its continuous.

*If $f$ is uniformly continuous then it takes cauchy sequences into cauchy squences

*If $f$ is a isometry then it takes cauchy sequences into cauchy squences

A: No. You really need and extra assumption. $f$ not only has to be continuous if not uniformly continuous. 
Prop: If $f$ is uniformly continuous from metric spaces $M$ to $N$, then  $f$ sends Cauchy sequences in Cauchy sequences.
Pf: Let $(a_n)$ a Cauchy and suppose that $f$ is uniformly continuous. We will show that $f(a_n)$ is Cauchy. Given $\epsilon>0$, there is a $\delta>0$ s.t. $$\forall x, y\in M\,(d_M(x,y)<\delta\Rightarrow d_N(fx,fy)<\epsilon)\tag{1}$$
By definition of uniformly continuous. Now, since $a_n$ is Cauchy, so for a sufficient large $n_0$, we have $d_M(a_n,a_m)<\delta$ for all $n,m\ge n_0$.So by $(1)$, $d_N(f(a_n),f(a_m))<\epsilon$ whenever $n,m\ge n_0$,i.e., $f(a_n)$ is Cauchy. 
In your proof you're wrong because $\delta=\delta(x,\epsilon)$ i.e., for continuous function $\delta$ depends not only of the $\epsilon$, if not also of the point, and this cannot happen in uniformly continuous functions.
A: One more example where a continuous function need not take Cauchy sequence to Cauchy sequence , take $$f:(0,1)\to (1,\infty)\quad \text{as}\quad f(x)=\frac{1}{x}$$ Now $\{x_n\}=\{\frac{1}{n}\}$ is cauchy sequence in $(0,1)$ but $f(x_n)$ is not!
