Cauchy sequence and a particular subsequence

Question

Let $\{{f_n}\}$ a Cauchy sequence. Then by definition $$\forall k\in\mathbb{N},\quad\exists n(k)\in\mathbb{N}\quad|\quad\forall m,n>n(k)\quad \|f_m-f_n\|<\frac{1}{2^k}$$

I must prove that we can choose a increasing sequence of natural numbers $\{n_k\}$ for which

$$\|f_{n_{k+1}}-f_{n_k}\|< \frac{1}{2^k}\;\text{for all}\;k\in\mathbb{N}$$

If I choose $n_k> n(k)$ for all $k\in\mathbb{N}$ and then I can suppose without loss of generality that the sequence $\{n_k\}$ is increasing.

Is this formally correct or should we add more?

Thanks!

Let's see if I understand: Let $n_1=n(1)+1$ and $n_2=\max\{n(2)+1,n_1+1\}$. Now $n_2> n_1>n(1)$, then $$\|f_{n_2}-f_{n_1}\|<\frac{1}{2}.$$

Now, let $n_3=\max\{n(3)+1,n_2+1\}$, then $n_3>n_2$ and $$\|f_{n_3}-f_{n_2}\|<\frac{1}{4},$$ since $n_3>n_2\ge n(2)+1>n(2)$

We can proceed in this way for every $k$, right?

Yes $n_k$ is actually at least constant, and at most stricly increasing:) — EDX, May 10 '20 at 09:00

Henno Brandsma · Accepted Answer · 2020-05-10T11:07:09.200

2

You should probably explain why you can assume "without loss of generality" that the $n_k$ are decreasing.

If you want to be formal, you can do a "proper" recursive construction:

E.g. let $n_1 = n(1)+1$. Then having defined $n_k$ for some $k$ such that $$\forall 1 \le l < k: \|f_{n_{l+1}} - f_{n_l}\| < \frac{1}{2^l}\tag{k}$$

(which is voidly obeyed for $k=1$ now) we define $n_{k+1}=\max(n(k+1) + 1, n_k+1))$ and so both ensure the "recursion condition" $(k+1)$ from $(k)$ and the definition of $n(k+1)$, and the strict increasingness of the $n_k$.

edited May 10 '20 at 11:07

answered May 10 '20 at 09:06

Henno Brandsma

242,131

I'm sorry but I don't understand what you mean. Could you explain it to me? – Jack J. May 10 '20 at 09:36
@JackJ. I just gave a formal construction... What exact thing about it don't you understand? – Henno Brandsma May 10 '20 at 09:39
This: "and so both ensure the "recursion condition" $(k+1)$ from $(k)$ and the definition of $n(k+1)$ and the strict increasingness of the $n_k$." – Jack J. May 10 '20 at 09:41
What is the inductive step? And what should we show? – Jack J. May 10 '20 at 09:52
Please, help me... – Jack J. May 10 '20 at 10:30
Why the sequence is increasing? – Jack J. May 10 '20 at 11:33
@JackJ. by construction $n_{k+1} > n_k$ for all $k$ as $n_{k+1} = \max(n(k+1)+1, n_k+1) \ge n_{k+1} > n_k$. – Henno Brandsma May 10 '20 at 11:37
@JackJ. $n_k +1$ after the first $\ge$ of course. – Henno Brandsma May 10 '20 at 11:42
I did not understand anything – Jack J. May 10 '20 at 11:45
Let us continue this discussion in chat. – Henno Brandsma May 10 '20 at 11:46
The condition (k) used? – Jack J. May 10 '20 at 13:48
@JackJ. We preserve $(k)$ and so at the end the subsequence does what we want. – Henno Brandsma May 10 '20 at 13:55
But what does this mean? – Jack J. May 10 '20 at 13:56
There is no inductive procedure. Where do we start? We suppose to find $ n_k $ increasing that satisfy $(k)$? What do means "voidly"? – Jack J. May 10 '20 at 13:59
Do we build $n_k$ one by one? – Jack J. May 10 '20 at 14:00
I don't understand what you said in the post. – Jack J. May 10 '20 at 14:06
@JackJ. Yes, you construct $n_k$ one by one, while preserving invariant $(k)$. Based on the recursion theorem; it’s not induction. – Henno Brandsma May 10 '20 at 14:21
When you said: "Then having defined $n_k$ for some k such that", What do you mean? – Jack J. May 10 '20 at 15:01

Cauchy sequence and a particular subsequence

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