0
$\begingroup$

A book on topology that I am reading tells me to

Prove that every subsequence of a Cauchy sequence is a Cauchy sequence.

which I do not believe is true. Here is my counterexample:
The sequence $\{x_n\}$ with $x_n=\frac{1}{n}$ for positive integer $n$ is Cauchy (because it converges to $0$, and convergence implies Cauchy). Its subsequence $1,\frac{1}{2},\frac{1}{3}$ is not Cauchy because the distance (with the Euclidean metric) is at least $\frac{1}{6}$ which is not an arbitrarily small $\epsilon$. Is this counterexample valid?

Here are the definitions:

  1. A sequence $\{x_n\}$ of points in a metric space $(X,d)$ is said to be a Cauchy sequence if given any real number $\epsilon>0$, there exists a positive integer $n_0$ such that for all integers $m\geq n_0$ and $n\geq n_0$, $d(x_m,x_n) \leq \epsilon$.
  2. If $\{x_n\}$ is any sequence, then the sequence $x_{n_1}, x_{n_2}, ...$ is said to be a subsequence if $n_1<n_2< ...$
$\endgroup$
2
  • 3
    $\begingroup$ Subsequences are supposed to have infinitely many terms. That is what $'...'$ in the definition indicates. $\endgroup$ Nov 17, 2022 at 9:31
  • $\begingroup$ So $x_n=1/6$ for $n>3$ and this is enough... $\endgroup$
    – dmtri
    Nov 17, 2022 at 9:54

1 Answer 1

1
$\begingroup$

As geetha290krm pointed out in the comments section, the issue with your counterexample is that a subsequence needs to itself be a sequence, and thus have infinitely many terms. With this in mind, we can prove the given statement. Specifically the thing to notice is that if $\{x_{n_i}\}_{i \in \mathbb{N}}$ is a subsequence of a sequence $\{x_n\}_{n \in \mathbb{N}}$, then we must have $n_i \geq i$ for all $i \in \mathbb{N}$. So then given $\epsilon > 0$, we know that there is some $N_\epsilon$ where $i,j \geq N_\epsilon$ implies $|x_i - x_j| < \epsilon$ since the original sequence is Cauchy. Since $n_i \geq i$ and $n_j \geq j$, we have that $n_i, n_j \geq N_\epsilon$, so then $|x_{n_i} - x_{n_j}| < \epsilon$, and we see that the subsequence is Cauchy as well.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .