Let $(X,d)$ be a metric space. A function $f:X\rightarrow X$ maps Cauchy sequences to Cauchy sequences. Then if $x_n\rightarrow x$, then $\{x_n\}$ is Cauchy, implying $\{f(x_n)\}$ is Cauchy. But if $X$ is not complete then $\{f(x_n)\}$ may not converge, let alone converging to $f(x)$. So the condition doesn't imply $f$ is continuous.

However what about the converse? If $f$ is continuous does it take Cauchy sequences to Cauchy sequences?

  • $\begingroup$ should the second sentence be fixed somehow because it seems to be asserting the question in the title, the answer to which is no? Maybe "If X is complete, then a continuous function $f:X \to X $ maps Cauchy sequences to Cauchy sequences". I don't know, I'm confused $\endgroup$ – usr0192 Dec 3 '20 at 15:20

No. Take $\iota\colon(0,+\infty)\longrightarrow\mathbb R$ defined by $\iota(x)=\frac1x$ and the sequence $\left(\frac1n\right)_{n\in\mathbb N}$.

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    $\begingroup$ Just out of curiosity, what made you choose the symbol $\iota$ out of all the possible symbols for a function? $\endgroup$ – 5xum Nov 15 '17 at 9:32
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    $\begingroup$ @5xum Because $\iota$ is the greek letter which correponds to our letter i and therefore I often use $\iota$ to denote maps which map each $x$ to its inverse. $\endgroup$ – José Carlos Santos Nov 15 '17 at 9:34
  • $\begingroup$ I see. Makes sense, I guess. $\endgroup$ – 5xum Nov 15 '17 at 9:35

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