Cauchy sequence and subsequence

Question

"Let $\{x_n\}\subset U$ be a Cauchy sequence. Give a direct proof that if a subsequence $\{x_{nk}\}\subset S$ has a limit $L$, then the Cauchy sequence $\{x_n\}\subset U$ has $L$ as a limit. Do not assume that general Cauchy sequences are convergent."

Ok I know a lot of people hate helping others with homework when they feel that the person asking the question hasn't worked on the problem. I have spent about 30 minutes going through all my theorems from my textbook on Cauchy sequences and I don't know where to start. Please help me

Answer 1

The actual theorem states that a Cauchy sequence is convergent iff it has a convergent subsequence, and your question only asks us to prove the if part. It follows rather immediately from the triangle inequality. Pick $n_k$ such that $|x_{n_k}-L|\leq\dfrac{\epsilon}{2}$, and $n,m$ such that $|x_n-x_m|\leq \dfrac{\epsilon}{2}$. You should be able to see why these exist for integers sufficiently large. Then

$$|x_n-L|\leq|x_n-x_m|+|x_{n_k}-L|$$.

Answer 2

Fix $\epsilon>0.$ $x_{n_k}\rightarrow L$ implies there is an integer $k_0$ such that $d(x_{n_k},L)<\epsilon/2$ for $k\geq k_0.$ $(x_n)$ is Cauchy implies there is an integer $n_0$ such that $d(x_m,x_n)<\epsilon/2$ for $m,n\geq n_0.$ Take $N=\max\{n_0,n_{k_0}\}.$ Then for $n\geq N,$ $d(x_n,L)\leq d(x_n,x_{n_{k_0}})+d(x_{n_{k_0}},L)<\epsilon.$ Hence $x_n\rightarrow L.$

Answer 3

Let $x_n$ be a Cauchy sequence and $(x_n)_k$ a subsequence. Let $\epsilon >0$. Then for sufficiently large $n$, since $x_n$ is Cauchy, $|x_n-(x_n)_k|<\epsilon /2.$ Since the subsequence converges to $L$ we have $|(x_n)_k-L|<\epsilon /2$ for sufficiently large $k$. Choosing the maximum of $n$ for which the above are satisfied gives $|x_n-L|=|x_n-(x_n)_k+(x_n)_k-L|\leq |x_n-(x_n)_k|+|(x_n)_k-L|<\epsilon$.