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Let $(X,d)$ be a metric space.Which of the following statements are true?
(a)A sequence {$x_n$} converges to $x$ in$X$ iff the sequences {$y_n$} is a cauchy sequence in $X$ , where, for $k\ge1$ , $y_{2k-1}=x_k$ and $y_{2k}=x$.
(b)if $f:X \to X$ maps Cauchy sequences into Cauchy sequences,then $f$ is continuous.
(c)If $f:X\to X$ is continuous , then it maps Cauchy sequences into Cauchy sequences.

I know that (c) is not true, it should be uniformly continuous function.But I am stuck on other two.Can somebody help me please.

jadugar
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1 Answers1

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Hints:

(a) The more-complicated direction is solved by noting that a Cauchy sequence having a convergent sub-sequence is convergent in itself (and to the same limit) (why?)

(b) It suffices that $f$ maps convergent sequences to convergent sequences (again, to the proper limit). If $x_n\to x$, then, what could be said about $(f(y_n))$, using the notation of part (a)?

Added

  1. If $x_n\to x$, then clearly $y_n\to x$ as well. In the other direction, if $(y_n)$ is Cauchy, and $y_{2n-1}\equiv x$ (hence $y_{2n-1}\to x$), then $y_n\to x$ as well, and therefore $x_n = y_{2n}\to x$.

  2. It suffices that $f(x_n)\to f(x)$ whenever $x_n\to x$. Take a convergent sequence $x_n\to x$, and by (1) the above-defined $(y_n)$ is Cauchy. Therefore $(f(y_n))$ (which is the result of part (1) construction on the series $(f(x_n))$) is also Cauchy, so again by (1) $f(x_n)\to f(x)$.

  3. As OP noted, this part is false, as $x\mapsto\frac{1}{x}$ over $(0,1)$ demonstrates.

Jonathan Y.
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