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?

Question

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..

Answer 1

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

1. If a function takes cauchy sequences into cauchy sequences then its continuous.
2. If $f$ is uniformly continuous then it takes cauchy sequences into cauchy squences
3. If $f$ is a isometry then it takes cauchy sequences into cauchy squences

Answer 2

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!

Answer 3

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.