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?
Best Answer
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 $\lim_{n\to\infty} a_n=L$.