Does a continuous function take Cauchy sequences to Cauchy sequences?

Question

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?

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 — usr0192, Dec 03 '20 at 15:20

score 18 · Accepted Answer · answered Nov 15 '17 at 09:29

18

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

answered Nov 15 '17 at 09:29

José Carlos Santos

440,053

10

Just out of curiosity, what made you choose the symbol $\iota$ out of all the possible symbols for a function? – 5xum Nov 15 '17 at 09:32
9

@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. – José Carlos Santos Nov 15 '17 at 09:34
1

I see. Makes sense, I guess. – 5xum Nov 15 '17 at 09:35

Does a continuous function take Cauchy sequences to Cauchy sequences?

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