Vtyjkyuk

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 $m,n>N_1 in BbbN$.

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?

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:$$forallepsilon>0quad,quadexists Nquad,quadforall m,n>Nquad,quad|y_m-y_n|<epsilon$$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 $x$.