Let $(x_n)$ be Cauchy with a subsequence $(x_{n_k})$ such that $\lim_{k\to\infty} (x_{n_k}) = a$. Show that $\lim_{n\to\infty} (x_{n}) = a$

Question

Let $(x_n)$ be Cauchy with a subsequence $(x_{n_k})$ such that $\lim_{k\rightarrow\infty} (x_{n_k}) = a$. Show that $\lim_{k\rightarrow\infty} (x_{n}) = a$.

Proof: Since $(x_n)$ is Cauchy, we know it is convergent to some limit $L$. We know if we have a sequence $x_n \rightarrow L$, then $x_{n_k} \rightarrow L$ for any subsequence. So if $\lim_{k\rightarrow\infty} (x_{n_k}) = a$, we have that $L = a$ and $\lim_{k\rightarrow\infty} (x_{n}) = a$

Unless there's something obvious I missed, this seems pretty clear, want to make sure that it's right. The above facts (Cauchy implying convergence and subsequence converging to same limit are easily proven).

Don't seem to need to use the definition of Cauchy ($(x_n)$ is Cauchy if $\forall\ \epsilon > 0, \exists\ N \in \mathbb{N}$ such that $m, n \geq N \implies |x_{m} - x_n| < \epsilon$) to prove this.

Answer 1

The usual way this goes is by noting that the Cauchy condition drags the rest of the sequence to $a$ along with the subsequence. I.e. there is an $k$ large enough so that $$ |x_{n_j}-a|<\epsilon/2 $$ whenever $j\geq k$ and $$ |x_{n}-x_m|<\epsilon/2 $$ whenever $n,m\geq n_k$ for a given $\epsilon>0$ challenge. Can you conclude from here, without using the Cauchy criterion?

Answer 2

The only thing that you have used is that any Cauchy sequence is convergent. This is true only if the metric space containing the sequence is complete. For example let $$a_n={\lfloor \pi\cdot 10^n\rfloor\over 10^n}$$which is a Cauchy sequence since is convergent to $\pi$ in $\Bbb R$ but not in $\Bbb Q$. A better (and probably safer!) way of proof goes as following:$${\text{Cauchy:}\quad \forall \epsilon>0 \ \ , \ \ \exists N\ \ ,\ \ m,n>N\to |a_m-a_n|<\epsilon\\\text{Convergent subsequence}: \quad \forall \epsilon>0 \ \ , \ \ \exists M\ \ ,\ \ k>N\to |a_{n_k}-L|<\epsilon}$$Now choose $n>\max\{M,N\}$ and $M=n_k$ such that $k>\max\{M,N\}$. Obviously $M=n_k\ge k>\max\{M,N\}$ therefore $$|a_m-L|=|a_{n_k}-L|<\epsilon\\|a_n-a_m|<\epsilon$$therefore $$|a_n-L|=|a_n-a_m+a_m-L|\le |a_n-a_m|+|a_m-L|<2\epsilon$$which completes the proof.