Assume $f$ : E -> $\mathbb{R}$ is a continuous function defined on some set E $\subset$ $\mathbb{R}$ and if {$a_j$}$_{j\in\mathbb{N}}$ $\subset$ E is a Cauchy sequence is it true that {$f(a_j)$}$_{j\in\mathbb{N}}$ is also a Cauchy sequence?

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    $\begingroup$ math.stackexchange.com/questions/207559/… $\endgroup$ – Nigel Overmars Aug 13 '16 at 18:01
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    $\begingroup$ If the limit of the $a_i$'s is in $E$, then it is true that the sequence $\big(f\left(a_i\right)\big)_{i=1}^\infty$ is Cauchy. $\endgroup$ – Batominovski Aug 13 '16 at 18:19
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    $\begingroup$ Adding to the previous comment : if the set $E$ is closed, then it is complete and so every cauchy sequence in $E$ is convergent. A continuous map sends convergent sequences to convergent sequences. $\endgroup$ – Dominique R.F. Aug 13 '16 at 18:23

Not necessarily.

A sufficient condition such that $\{f(x_j)\}$ is again Cauchy is that $f$ does not increase distances, as defined in this question: Function doesn't increase distance.

This is because, for any continuous function $f$ such that $\{f(x_j)\}$ is no longer Cauchy, one has that $f$ must increase distances. In fact such an $f$, despite being continuous, would increase distances either just as rapidly or even more rapidly than the distances between terms in the Cauchy sequence decrease -- this would cause the sequence to no longer be Cauchy.

The standard example is the following: let $\{a_j\} = \frac{1}{j}$. Then let $f$ = $\frac{1}{x}$ which is continuous.

But $\{f(a_j)\}$ is the sequence $j$, which is clearly not Cauchy. Clearly the distances between terms for $f(a_j)$ do not decrease to zero -- instead they remain constant at $1$. Thus $f$ in some sense "exactly cancels out" the approach of the inter-term distances to zero which we had for $\{a_j\}$.

(To address how this relates to the comments on the question above -- the limit of the original sequence, $0$, is not in the domain of the function $f(x)=\frac{1}{x}$, hence there is no contradiction.)

  • $\begingroup$ Good answer, but $(f(a_j))$ is not a constant sequence... $\endgroup$ – paf Aug 13 '16 at 18:19
  • $\begingroup$ @paf Thanks for making me aware of the mistake; I think I confused constant with identity for some reason $\endgroup$ – Chill2Macht Aug 13 '16 at 18:28
  • $\begingroup$ $f$ can increase distances and still have the property, for example $f(x) = x^2$ on $\mathbb R.$ It's more a question of the domain not being closed. $\endgroup$ – zhw. Aug 13 '16 at 18:36
  • $\begingroup$ The given property that f preserves Cauchy sequences. $\endgroup$ – zhw. Aug 13 '16 at 18:57
  • $\begingroup$ No, its the opposite: It's sufficient but not necessary. Note $x^2$ increases distances on $[1,\infty)$ but has the property. $\endgroup$ – zhw. Aug 13 '16 at 19:47

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