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Let $X$ and $Y$ be metric spaces and let $f : X \to Y$ be a mapping.

Is the following statement true or false ?

If $Y$ is complete and if $f$ is continuous, then the image of every Cauchy sequence in $X$ is a Cauchy sequence in $Y$.

I was taking $f:[0,1]\to[0,1]$ I take $f(x) = x$ then this statement is true.

Am i correct or not? I'm not getting another counter examples.

5xum
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1 Answers1

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Let $f:(0,1)\rightarrow{\bf{R}}$, $f(x)=1/x$. Consider the Cauchy sequence $(x_{n})\subseteq(0,1)$, $x_{n}=1/(n+1)$, the sequence $(f(x_{n}))$ is not Cauchy.

user284331
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