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.