I'm going over an old homework question in my Analysis class that I missed. My professor's solution was confusing so I am looking for additional help.
The question is: If $\{p_n\}_{n \in \mathbb{N}} $ is a cauchy sequence in a metric space, show that for any $\epsilon > 0$ there exists a subsequence $p_{n_i}$ such that $d(p_{n_i}, p_{n_{i+1}}) < \frac{\epsilon}{2^{i+1}}$.
My solution was to choose $p_{n_1}$ and $p_{n_2}$ such that $d(p_{n_1}, p_{n_2}) < \frac{\epsilon}{4}$ and then choose $p_{n_j}$ for $j \geq 2$ such that $d(p_{n_j}, p_{n_{j+1}}) < \frac{\epsilon}{2^{j+1}}$ which we can do since the sequence is Cauchy.
If you could explain how to do it and why my solution doesn't work, that would be much appreciated. Thanks in advance.
What does it mean for a sequence to be Cauchy? It means for each $\epsilon > 0$, there is some $N \in \mathbb{N}$ such that $n, m \geq N$ implies $d(x_{n}, x_{m}) < \epsilon$.
Since this is true for every $\epsilon > 0$, it is true if $\epsilon = \frac{1}{2}$, for example. That is, there is some $N_{1} \in \mathbb{N}$ such that $n, m \geq N_{1}$ implies $d(x_{n}, x_{m}) < \frac{1}{2}$. Let $n_{1} = N_{1}$.
Similarly, we know if $\epsilon= \frac{1}{2^{2}}$, there is some $N_{2} \in \mathbb{N}$ such that $n, m \geq N_{2}$ implies $d(x_{n}, x_{m}) < \frac{1}{2^{2}}$. But if this is true for all $n, m \geq N_{2}$, then it is true for all $n, m \geq N_{2} + 1$. And for all $n, m \geq N_{2} + 2$. And for all $n, m \geq p > N_{2}$ where $p$ is any integer greater than $N_{2}$. In particular, this is true for $\max \{n_{1}, N_{2} \} + 1$. So we can let $n_{2} = \max \{n_{1}, N_{2} \} + 1$, and clearly, $n_{2} > n_{1}$, and if $n, m \geq n_{2}$, $d(x_{n}, x_{m}) < \frac{1}{2^{2}}$.
Similarly, we know if $\epsilon = \frac{1}{2^{3}}$, there is some $N_{3} \in \mathbb{N}$ such that $n, m \geq N_{3}$ implies $d(x_{n}, x_{m}) < \frac{1}{2^{3}}$. If we let $n_{3} = \max \{ n_{2}, N_{3} \} + 1$, then clearly $n_{3} > n_{2}$ and if $n, m \geq n_{3}$, then $d(x_{n}, x_{m}) < \frac{1}{2^{3}}$.
Note that because $n_{1} < n_{2} < n_{3} < \dots$ by our construction, and since we picked $n_{i}$ such that $d(x_{n}, x_{m}) < \frac{1}{2^{i}}$ if $n, m \geq n_{i}$, AND $n_{i + 1}, n_{i} \geq n_{i}$, it follows that $d(x_{n_{i + 1}}, x_{n_{i}}) < \frac{1}{2^{i}}$.