# show that if a subsequence of a cauchy sequence converges, then the whole sequence converges

I am trying to prove that If a subsequence of a Cauchy sequence converges, then the whole sequence converges

proof:

Let {$a_n$} be a Cauchy sequence such that it has a convergent subsequence {$a_{n_k}$} that converges to L

Let $\epsilon$ > 0 then

$\exists$ $N_1$ $\in$ $\Bbb{N}$ : $\forall$ $m ,n \geq$ $N_1$ |$a_n$ - $a_m$| < $\epsilon$/2

$\exists$ $N_2$ $\in$ $\Bbb{N}$ : $\forall$ $k \geq$ $N_2$ |$a_{n_k}$ - L| < $\epsilon$/2

Let $N_3$ = max{$N_1$,$N_2$}

Using the strictly increasing sequence {$n_k$}

If $k$ $\geq$ $N_3$ $\Rightarrow$ $n_k$ $\geq$ $k$ $\geq$ $N_3$ $\Rightarrow$ $n_k$ $\geq$ $N_3$

so |$a_{n_k}$ - L|<$\epsilon$/2 and |$a_k$ - $a_{n_k}$| < $\epsilon$/2

so |$a_k$ - L| = |$a_k$ - $a_{n_k}$ + $a_{n_k}$ - L| $\leq$ |$a_k$ - $a_{n_k}$| + |$a_{n_k}$ - L| < $\epsilon$ $\Rightarrow$ |$a_k$ - L| < $\epsilon$

Therefore {$a_n$} converges to L

Is my proof correct?

• Why are you using $|a_n - a_m|$? Is this suppose to be a general metric space or $\mathbb{C}$? If it is the latter then this is automatically true. Mar 29, 2016 at 0:48

You have the right idea, but to write it a bit more cleanly; let $\{a_n\}$ be a Cauchy sequence in a metric space $(X,d)$ with a convergent subsequence $a_{n_k}\stackrel{k\to\infty}\longrightarrow L$. Choose $N$ so that $n,m\geqslant N$ implies $$d\left(a_n,a_m\right)<\frac\varepsilon2$$ and $N'$ so that $n_k\geqslant N'$ implies $$d\left(a_{n_k},L\right)<\frac\varepsilon2.$$ Fix $j$ such that $n_j>\max\{N,N'\}$. Then for $n\geqslant\max\{N,N'\}$ we have $$d\left(a_n,L\right)\leqslant d\left(a_n,a_{n_j}\right) + d\left(a_{n_j},L\right) < \frac\varepsilon2+\frac\varepsilon2=\varepsilon,$$ and so $\lim_{n\to\infty} a_n=L$.
• @JoseVega A metric space $(X,d)$ is simply a nonempty set $X$ with a distance function (or metric) $d$ which assigns to every pair of points in $X$ a nonnegative real number and satisfies the following: 1. $d(x,y)=0$ if and only if $x=y$. 2. $d(x,y)=d(y,x)$. 3. $d(x,z)\leqslant d(x,y)+d(y,z)$. If that is too abstract, then you can simply think of the set of real number $\mathbb R$ with the metric $d(x,y)=|x-y|$. Mar 30, 2016 at 19:33