1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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
2
$\begingroup$

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?)
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.