3
$\begingroup$

I recently read the proof of why every convergent sequence is a Cauchy sequence and the mechanics of the proof make total sense to me. However, the proof is rather different from what my intuitive brain pictured. Let me explain. For a convergent sequence we know the sequence $\{ x_n \}_n$ is converging to a specific point say $x$. What this means to me intuitively is the following picture:

enter image description here

In other words, since the distance from $x_n$ to $x_m$ to $x$ is bounded then its "obvious" that the distance between any $x_m$,$x_n$ should also be bounded. But not only that it seems that the distance between any $x_n$ and $x_m$ should be bounded by some $\epsilon'$ less than the original $ \epsilon$ in the convergent definition. However, the proof seems to imply that it twice as big as the original $\epsilon$. Why is the relation between the epsilons the other way round from what I expect if the distance between $x_n$ to $x_m$ is included in the distance between $x_n$ and $x$. Essentially since in the picture $ \| x_n - x_m \| \leq \| x_n - x \| $, then its unexpected to have the constant bounding the Cauchy sequence be twice as large.

Is the constant to loose or do I have a fundamental misconception of how the proof should have turned out?


Maybe I'll try to explain the proof in my own words and reveal my misunderstanding?

Consider a sequence $ \{ x_n\}_n$ that converges to $x$. This means that $\forall \epsilon_{convergent} > 0$, $\exists N \in \mathbb{N}$ s.t. $ \| x_n - x\| < \epsilon_{convergent} $. Fix a $\epsilon_{convergent}$. This corresponds to some $N$. Let it be $N_{convergent}$. Then any two points in the sequence after $N_{convergent}$ have the property $ \| x_n - x\| < \epsilon_{convergent}$, $ \| x_m - x\| < \epsilon_{convergent}$. Therefore consider the quantity of interest and apply the definition of norm (using triangle inequality):

$$ \| x_n - x_m \| = \| x_n - x + x - x_m \| \leq \| x_n - x\| + \| x_m - x \|$$

but since we know both are bounded by $\epsilon_{convergent}$ we get that for that same $N_{convergent}$ the above difference is bounded by $\epsilon_{cauchy} = 2 \epsilon_{convergent}$. Which we used all the rules and axioms correctly so mechanically it looks fine but the constant $\epsilon_{cauchy}$ seems to be larger than I expected (when I thought it should have obviously been smaller), which could mean my intuition is wrong. I just hope to correct it or maybe I don't understand the proof as well as I thought.

$\endgroup$
  • 2
    $\begingroup$ In general you do not have $\Vert x_n - x_m \Vert \leq \Vert x_n -x \Vert$. Just draw your picture slightly different, namely let $(x_n)_{n\geq 1}$ approach $x$ not only from the left, but from both sides. This is exactly where the factor 2 is coming from. $\endgroup$ – Severin Schraven Sep 24 '16 at 22:12
  • $\begingroup$ @SeverinSchraven I see that you have an argument that explains why I'm wrong but I don't understand what I'm suppose to draw. I know you described it but I don't get it I guess. $\endgroup$ – Charlie Parker Sep 24 '16 at 22:17
  • 1
    $\begingroup$ @SeverinSchraven: good answer. I suggest you post it as an answer and not as a comment. $\endgroup$ – Martin Argerami Sep 24 '16 at 22:18
2
$\begingroup$

In general you do not have $\Vert x_n−x_m\Vert \leq \Vert x_n−x \Vert$. Just draw your picture slightly different, namely let $(x_n)_{n\geq 1}$ approach x not only from the left, but from both sides. This is exactly where the factor 2 is coming from.

Look for example at the sequence $-1, 1, -\frac{1}{2}, \frac{1}{2}, -\frac{1}{3}, \frac{1}{3}\dots$ (or formally correct $x_{2n+1}:= -\frac{1}{n}$ and $x_{2n+2}:=\frac{1}{n}$).

This sequence converges to $x=0$. We have

$$ \Vert x_{2n+ 1} - x_{2n+2} \Vert = \Vert -\frac{1}{n} - \frac{1}{n} \Vert = \frac{2}{n},$$

but

$$\Vert x_{2n+1} - x \Vert = \Vert -\frac{1}{n} - 0 \Vert = \frac{1}{n}.$$

$\endgroup$
  • $\begingroup$ a picture would make this answer perfect. Just a suggestion. Thanks btw. :) $\endgroup$ – Charlie Parker Sep 25 '16 at 16:50
1
$\begingroup$

take the example of a sequence which converges to x alternatively from the left and the right sides like

$$x_n=sin(n)e^{-n}$$

then you could have

$$|x_n-x_m|>|x_n-x|$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.