If $(x_n)_{n = 1}^{\infty}$ is Cauchy, show subsequence $(x_{n_{k}})$ such that $\sum_{k = 1}^{\infty}|x_{n_{k}} - x_{n_{k+1}}| < \infty$

Question

If $(x_n)_{n = 1}^{\infty}$ is Cauchy, show that it has a subsequence $(x_{n_{k}})$ such that $\sum_{k = 1}^{\infty}|x_{n_{k}} - x_{n_{k+1}}| < \infty$

Attempt:

Since $x_n$ is Cauchy and since $\mathbb{R}$ is complete $\Rightarrow$ then $(x_n)$ converges to a point $x$ in the set $\mathbb{R}$. This means that for all $\epsilon > 0$ there exists an $N_{1} \in \mathbb{N}$ s.t. $\forall \ m,n > N_{1} \ |x_{m} - x_{n}| < \epsilon$.

Since $(x_n)$ converges $\Rightarrow$ $(x_n)$ is bounded. So by the Bolzano Weirstrauss theorem, there exists a subsequence $(x_{n_{k}})$ that converges to $x$. That is: $\forall \epsilon > 0$ there exists an $N_{2} \in \mathbb{N}$ s.t. $\forall \ n_{k} > N_{2} \ |x_{n_{k}} - x| < \epsilon$.

Therefore:

$$\sum_{k = 1}^{\infty} |x_{n_{k}} - x_{n_{k+1}}| = \sum_{k = 1}^{\infty} |x_{n_{k}} - x_{m} + x_{m} - x + x - x_{n_{k+1}}| \leq \sum_{k = 1}^{\infty}(|x_{n_{k}} - x_{m}| + |x_{m} - x| + |x - x_{n_{k+1}}|) = \sum_{k = 1}^{\infty}|x_{n_{k}} - x_{m}| + \sum_{k = 1}^{\infty}|x_{m} - x| + \sum_{k = 1}^{\infty}|x - x_{n_{k+1}}| < \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon $$

Final Thought: Two things:

1) Am I interpreting that $\sum_{k = 1}^{\infty}|x_{n_{k}} - x_{n_{k+1}}| < \infty$ means the series is convergent?

2) I don't know if I needed to include the $x_{m}$ cauchy term or if this is the right approach all together. My reasoning for it was that since the other sequences are convergent then their series would also be convergent so I applied that. Any advice would be appreciated.

Answer 1

Your answer is wrong. If you know that all terms are smaller than $\frac{\epsilon}{3}$ it doesn't mean their infinite sum is smaller than $\frac{\epsilon}{3}$ as well.

Here is a possible solution: since $(x_n)$ is Cauchy we can conclude that there is $n_1$ such that for all $m,p\geq n_1$ we have $|x_m-x_p|<\frac{1}{2}$. Next, there is $n_2>n_1$ such that for all $m,p\geq n_2$ we have $|x_m-x_p|<\frac{1}{2^2}$. Continue by induction, every time choose $n_k>n_{k-1}$ such that for all $m,p\geq n_k$ we have $|x_m-x_p|<\frac{1}{2^k}$. And now $(x_{n_k})$ is the required subsequence. Since $n_{k+1}, n_k\geq n_k$ we know that $|x_{n_k}-x_{n_{k+1}}|<\frac{1}{2^k}$, this is true for all $k\in\mathbb{N}$. And then:

$\sum_{k=1}^\infty |x_{n_k}-x_{n_{k+1}}|\leq\sum_{k=1}^\infty\frac{1}{2^k}<\infty$

Answer 2

There is an error in your proof: for example, $\sum\limits_{k=1}^\infty |x_m - x|$ doesn't converge unless $x_m = x$ already (and it can be impossible if sequence doesn't reach limit). You also don't need to take convergent subsequence - original sequence was Cauchy, thus already convergent.

The main idea is correct - we need to use Cauchy criteria. Take some $M_0$ s.t. if $m \geqslant M_0$ and $n \geqslant M_0$ then $|x_m - x_n| < 1$, some $M_1 > M_0$ s.t. if $m \geqslant M_1$ and $n \geqslant M_1$ then $|x_m - x_n| < \frac{1}{2}$, ... - for any $i$ take $M_i > M_{i - 1}$ s.t. if $m \geqslant M_i$ and $n \geqslant M_i$ then $|x_m - x_n| < \frac{1}{2^i}$. Now take $x_{M_0}, x_{M_1}, \ldots$ as subsequence: $\sum_{k=1}^\infty |x_{M_k} - x_{M_{k + 1}}| \leqslant \sum_{k=1}^\infty \frac{1}{2^k}$ that converges.

Answer 3

Your attempt doesn't work at all. You're adding infinitely many terms, but all you know so far is that each is less than $\epsilon$. How are you getting the sum to be less than $\epsilon$?

Hint: Take $N_k$ so that for all $m, n \le N_k$, $|x_m - x_n| < 1/2^k$. Then in particular $|x_{N_k} - x_{N_{k+1}}| < 1/2^k$.

Answer 4

Since the sequence $(x_n)_{n\in\mathbb N}$ is a Cauchy sequence, there is a $n_1\in\mathbb N$ such that$$m,n\geqslant n_1\implies\lvert x_m-x_n\rvert<\frac12.$$And there is a $n_2>n_1$ such that$$m,n\geqslant n_2\implies\lvert x_m-x_n\rvert<\frac1{2^2}.$$And there is a $n_3>n_2$ such that$$m,n\geqslant n_3\implies\lvert x_m-x_n\rvert<\frac1{2^3}.$$And so on. Therefore$$\sum_{k=1}^\infty\lvert x_{n_k}-x_{n_{k+1}}\rvert<\sum_{k=1}^\infty\frac1{2^k}=1.$$