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Let $(X,d)$ be a metric space. How to prove:

A sequence ${x_n} \rightarrow x$ in $X$ if and only if the sequence $\{y_n\}$ is a Cauchy sequence in $X$ where $y_n$ is defined as $y_{2k-1}=x_k$ and $y_{2k}=x$

My try:

Here $(y_n)=\{x_1,x,x_2,x,....\}$

Assume ${x_n} \rightarrow x$ in $X$. Then $d(x_n,x) < \epsilon$ for all $n>N$.

Now to check $d(y_n,y_m) < \epsilon$ for all $m,n>N_1 \in \Bbb{N}$.

Case(i): $m$ and $n$ is even

In this case, $d(y_n,y_m) =d(x,x)=0< \epsilon$

Case(ii): $m$ and $n$ is odd

In this case, $d(y_n,y_m) =d(x_n,x_m)< \epsilon$, since $x_n$ is Cauchy

Case(iii): $m$ is odd and $n$ is even

In this case, $d(y_n,y_m) =d(x,x_m)< \epsilon$

Hence $\{y_n\}$ is Cauchy

Is this correct and how about the other part?

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  • $\begingroup$ How is $N_1$ related to $N$? Your case (ii) seems fishy. Why should $d(x_n,x_m)<\epsilon$? Yes, convergence implies Cauchy, but how do you know that $x_n$ and $x_m$ are that close? All you know is that $x_n$ is close to $x$ and $x_m$ is close to $x$, not necessarily $\epsilon$ close to each other. $\endgroup$ – Matt Sep 11 '18 at 2:57
  • $\begingroup$ Agreeing with @Matt. You need to get your $\epsilon$s straight up front. $\endgroup$ – Randall Sep 11 '18 at 3:00
  • $\begingroup$ @Matt & Randall: So the only problem is to use the same $\epsilon$ ? $\endgroup$ – user444830 Sep 11 '18 at 3:09
  • $\begingroup$ I mean, the $N$ and the $\epsilon$ you use in the first line $d(x_n,x)<\epsilon$ for all $n\geq N$, cannot tell you that $d(x_n,x_m)<\epsilon$ for all $n,m\geq N$. Yes, convergence implies Cauchy, but the $N$ you have in each setting may be different. So you cannot use the same $\epsilon$ in both settings here. $\endgroup$ – Matt Sep 12 '18 at 3:01
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The direct side of the theorem is easy. If $x_n$ is convergent so is $y_n$ therefore $y_n$ is also Cauchy. For proving the converse side, let $y_n$ be Cauchy, then we have:$$\forall\epsilon>0\quad,\quad\exists N\quad,\quad\forall m,n>N\quad,\quad|y_m-y_n|<\epsilon$$For each such $N$ choose $n$ such that $2n,2n-1>N$. This choice leads to$$|y_{2n}-y_{2n-1}|<\epsilon$$because $y_n$ is Cauchy or equivalently$$|x_n-x|<\epsilon$$which means that $x_n$ is convergent to $x$.

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  • $\begingroup$ Thanks!........ $\endgroup$ – user444830 Sep 15 '18 at 9:44
  • $\begingroup$ You're welcome. Wish you luck! $\endgroup$ – Mostafa Ayaz Sep 16 '18 at 8:41

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