# $x_n \rightarrow x$ iff the modified sequence is Cauchy

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?

• 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. – Matt Sep 11 '18 at 2:57
• Agreeing with @Matt. You need to get your $\epsilon$s straight up front. – Randall Sep 11 '18 at 3:00
• @Matt & Randall: So the only problem is to use the same $\epsilon$ ? – user444830 Sep 11 '18 at 3:09
• 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. – Matt Sep 12 '18 at 3:01

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 $x$.