I am having trouble with the following proof:

Prove that a Cauchy sequence in a normed vector space which is known to have a convergent subsequence must itself converge.

I think that the proof should start something like this:

Let there be a convergent subsequence ($\{p_n\}$) of a cauchy sequence ($\{q_n\}$) in $V$. Suppose that $\{q_n\}$ does not converge in $V$. Then there is no $\varepsilon>0$ such that $||q_n-\vec{L}||<\varepsilon$ for all $n>N$ where $N$ is an integer.

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    $\begingroup$ A direct proof would be much simpler $\endgroup$ – Aweygan Oct 13 '16 at 16:13
  • $\begingroup$ @Aweygan Could you provide a suggestion for a direct proof? $\endgroup$ – AzJ Oct 13 '16 at 16:15
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    $\begingroup$ @AzJ: Use the definitions of Cauchy sequence and convergence along with the triangle inequality (also, this works for metric spaces in general; no need to assume a normed linear space) $\endgroup$ – parsiad Oct 13 '16 at 16:15

Hint: Let $(x_n)_n$ be a Cauchy sequence and $(x_{n_k})_k$ be a subsequence converging to $x$.

  1. Pick $N$ such that for all $n,m \geq N$, $\Vert x_n - x_m \Vert < \epsilon / 2$.
  2. Pick $K$ such that for all $k \geq K$, $\Vert x_{n_k} - x \Vert < \epsilon / 2$.
  3. Now, let $M = \max\{N,K\}$. For all $n \geq M$, $\Vert x_n - x \Vert \leq \ldots$ (can you finish the rest?)
  • $\begingroup$ Right idea with the triangle inequality, but you should pick $m$ in a specific way. Hint: you haven't used point (2). Also, please clean up your comments as they are unreadable in their current states. $\endgroup$ – parsiad Oct 13 '16 at 16:38
  • $\begingroup$ Is this correct \begin{align*} \Vert x_n - x \Vert &= \Vert x_n - x \Vert \\ &= \Vert x_n - x_m +x_m -x \Vert \\ &\leq \Vert x_n - x_m\Vert + \Vert x_m -x \Vert \text{ triangle inequality}\\ &\leq \varepsilon /2 + \Vert x_m -x \Vert \\ &\text{ replace $x_m$ with $x_{n_k}$ (as $x_{n_k}$ \supseteq $x_m$), then} &\leq \varepsilon /2 + \Vert x_{n_k} -x \Vert \\ &\leq \varepsilon /2 + \varepsilon /2 \\ \end{align*} Therefore as $\Vert x_n - x \Vert \leq \varepsilon$ the Cauchy sequence converges. $\endgroup$ – AzJ Oct 13 '16 at 17:02

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