If $(x^{(n)})^∞_{n=m}$ is Cauchy and if some subsequence of $(x^{(n)})^∞_{n=m}$ converges then so does $(x^{(n)})^∞_{n=m}$

Question

Let $(x^{(n)})^∞_{n=m}$ be a Cauchy sequence in $(X,d)$. Suppose that there is some subsequence $(x^{(n_j)})^∞_{j=1}$ of this sequence which converges to a limit $x_0$ in $X$. Then the original sequence $(x^{(n)})^∞_{n=m}$ also converges to $x_0$.

My thoughts so far:

$(x^{(n)})^∞_{n=m}$ is a Cauchy sequence. Then $\forall$ $\epsilon >0$, $\exists $ $N\ge m$ such that $d(x^{(j)},x^{(k)}) <\epsilon$ for all $j,k\ge N$

Some subsequence $(x^{(n_j)})^∞_{j=1}$ converges to limit $x_0$ in X.

This means that $\forall$ $\epsilon>0$ , $\exists$ $N\ge1$ such that $|(x^{(n_j)})^∞_{j=1}-x_0|<\epsilon$

This also means that $d((x^{(n_j)})^∞_{j=1},x_0) <\epsilon$, $\forall$ $j\ge N$

I pretty much just listed the definitions, but I don't how to show much more than this. Some help in proving this would be greatly appreciated!

Answer 1

Let $(x_n)_{n\in \Bbb N}$ be Cauchy and let $f:\Bbb N\to \Bbb N$ be strictly increasing, such that $(x_{f(n)})_n$ converges to $x.$ Given $r>0,$ take $m$ such that $$(1)\quad n'\geq n>m\implies d(x_{n'},x_n)<r/2$$ and take $m'$ such that $$(2)\quad n>m'\; \implies d(x,x_{f(n)})<r/2.$$

Let $m''=\max (m,m').$

Now $f$ is strictly increasing so $f(n)\geq n$ for every $n\in \Bbb N.$

Consider any $n>m''$ with $n'=f(n).$ ( So $n'\geq n$.)

We have $d(x_{f(n)},x_n)=d(x_{n'},x_n)<r/2$ by (1) because $n'\geq n>m''\geq m.$ And we also have $d(x,x_{f(n)})<r/2$ by (2) because $n> m''\geq m'.$

Therefore $d(x,x_n)\leq d(x,x_{f(n)})+d(x_{f(n)},x_n)<r$ for all $n>m''.$ QED.

Answer 2

Hint

$$d(x_n, a) \leq d(x_n, x_{n_j})+d(x_{n_j}, a)$$

You need to show that there exists some $N$ such that, for all $n >N$ you can make the RHS smaller than $\epsilon$.

Try to show that you can find $N$ such that, for each $n >N$ you can find some $n_j$, such that each term on the RHS is smaller than $\epsilon/2$.