If $\{x_n\}_{n\in I_m}$ is a Cauchy sequence, then $lim\{x_n-x_{n+c}\}_{n\in I_m}=0$ $\forall c\in \mathbb{N}$

Question

I'm trying to prove the following:

If $\{x_n\}_{n\in I_m}$ is a Cauchy sequence, then: a) $lim\{x_n-x_{n+c}\}_{n\in I_m}=0$ $\forall c\in \mathbb{N}$. b) Given $\varepsilon >0, \exists N \in I_m : x_n \in (x_N-\varepsilon, x_N+\varepsilon)$, except for a finite number of terms.

The definition I have for Cauchy Sequence is "Let $\{a_n\}_{n\in I_m}$ be a sequence, $\{a_n\}_{n\in I_m}$ is a Cauchy sequence if $\forall \varepsilon >0, \exists M_{\varepsilon} \in I_m: \forall n,s \in I_{M_\varepsilon}, |a_n -a_s|<\varepsilon$"

For a), what I've thought is that, using the definition of a Cauchy sequence, then, if $\{x_n\}_{n\in I_m}$ is a Cauchy sequence, given $\varepsilon >0, \exists N \in I_m : \forall n,s \in I_N, |x_n-x_s|<\varepsilon$.

Now, for $c\in \mathbb{N}$, notice that $0<c \Rightarrow n<n+c$ and we can see that $x_n< x_{n+c} $

From this, I'm not sure how to say that $lim\{x_n-x_{n+c}\}_{n\in I_m}=0$ $\forall c\in \mathbb{N}$.

For b), I don't even know how to start.

Answer 1

I'm confused with your indexing notation with the $I_m$'s so let me just write the sequence as $\{x_n\}_{n \in \mathbb{N}}$ instead. I'll leave it to you rewrite it in your notation.

$(a)$ Fix any $c \in \mathbb{N}$. We'll show $lim_{n \rightarrow \infty} \space x_n - x_{n+c} = 0$. So pick any $\epsilon > 0$. By cauchy-ness of $\{x_n\}_{n \in \mathbb{N}}$, there is a $N$ such that for all $n, m \geq N$, $|x_n - x_m| = |0-(x_n-x_m)| < \epsilon$. So in particular for any $n > N$, obviously $n+c > N$ as well, so $|0 - (x_n - x_{n+c})| < \epsilon$. So this shows convergence. You basically got it in your post.

$(b)$ The idea is simply that for cauchy sequences, the points begin to cluster together after some time. So the statement is saying that the $x_n$ cluster in an interval $\epsilon$ small after some point. Of course there can be outliers, but only finitely many.

So pick any $\epsilon > 0$. Then again by definition, there is a $N > 0$ such that for all $n, m \geq N$, $|x_n - x_m| < \epsilon$. Well this is just another way of saying that $x_n \in (x_m - \epsilon, x_m + \epsilon)$ .Well now let $m = N$. This means that for any $n \geq N$, we have that $x_n \in (x_N - \epsilon, x_N + \epsilon)$. Ok so we know that past a certain point, i.e, for all $n \geq N$, we have what we desire: $x_n \in (x_N - \epsilon, x_N + \epsilon)$. The only place left where this property can fail then is in the range of $0 \leq n < N$. But this is a finite range. So even if we have that $x_n \not \in (x_N - \epsilon, x_N + \epsilon)$ for EVERY $n < N$, that's still just finitely ($N$ many) violations. And we know that past $N$, we do have $x_n \in (x_N - \epsilon, x_N + \epsilon)$, so that means we can only really have finitely many violations (in fact we can only have up to $N$ many violations).

Answer 2

There exist $n_0$ such that $|x_i-x_j| <\epsilon$ $\cdots (1)$ if $i,j \geq n_0$ and $i,j \geq m$. Let $N$ be the maximum of $n_0$ and $m$. Then $|x_n-x_{n+c}| <\epsilon$ for $n \geq N$ (and any positive integer $c$).

We already have $x_n \in (x_N-\epsilon,x_N+\epsilon)$ if $n \geq N$ by (1). [Take $i=N$ and $j=n$].