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..


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
  • $\begingroup$ Thank you .. Could you please tell me where I could have possibly gone wrong in the attempt ? $\endgroup$ – MathMan Oct 7 '14 at 23:48
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    $\begingroup$ @VHP I think you're using that they're uniformly continuous in (1), every $\delta$ depends on $x_n$ or $x_m$, if you exchanged the words continuous by uniformly continuous your proof would work :). $\endgroup$ – Aram Oct 7 '14 at 23:52
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    $\begingroup$ @VHP remember that a function is continuous at $x$ if given $\varepsilon > 0$ there exists $\delta (x,\varepsilon)$ (depending on $\varepsilon$ and $x$) such that $d(f(x),f(y)) < \varepsilon$ if $d(x,y) < \delta$. And a function is uniformly continuous if given $\varepsilon$ there exists some $\delta(\varepsilon)$ (depends solely on epsilon) such that $d(f(x),f(y)) < \varepsilon$ whenever any $x,y$ are at distance less than $\delta$ $\endgroup$ – Aram Oct 7 '14 at 23:55
  • $\begingroup$ Corresponding to every $x_n,x_m$ we would have a $\delta_0$ ( even if $\delta$ in general is a function of $x_i,x_n$). Now, corresponding to this $\delta_0$, we find the value of required natural number $k$. I am not sure if we used that $f$ is uniformly continuous . Could you please explain a bit more? $\endgroup$ – MathMan Oct 8 '14 at 0:08
  • $\begingroup$ I think i got it. Thank you very much $\endgroup$ – MathMan Oct 8 '14 at 0:17

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.


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!

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    $\begingroup$ And by "one more example" you mean "the same example as Aram's answer"? $\endgroup$ – Najib Idrissi Oct 10 '14 at 13:41
  • $\begingroup$ Frankly speaking I didn't saw his answer otherwise I'd have given $\displaystyle\sin \frac{1}{x}, x\in (0,1)$ $\endgroup$ – Mathronaut Oct 10 '14 at 13:56

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