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I have a question regarding the definition of a Cauchy sequence of a sequence in a metric space. The definition I learned and that is consistent with Wikipedia defines a sequence $(x_n)_{n=1}^\infty$ as a Cauchy sequence if $$ \forall\, \varepsilon>0 \;\;\exists\, N\in\mathbb{N}\;\; \forall\, m,n \geq N : d(x_m,x_n)<\varepsilon $$ If I am not mistaken, there is a simpler, but equivalent definition: $$ \forall\, \varepsilon>0 \;\; \exists\, N\in\mathbb{N} \;\; \forall\, m \geq N: d(x_m,x_N)<\varepsilon $$ This is simpler, because it only has two natural numbers in it instead of three. This makes it easier to prove, that a given sequence is a Cauchy sequence.

Note that the equivalence relies on the triangle inequality.

Proof: $(\Rightarrow)$: we simply choose $n=N$.

$(\Leftarrow)$: Let $\varepsilon>0$. Then $$ \exists\, N\in\mathbb{N}\;\; \forall\, m \geq N: d(x_m,x_N)<\frac12\varepsilon $$ This means that for $m,n\geq N$ we have $$ d(x_m,x_n) \leq d(x_m,x_N)+d(x_n,x_N) < \frac12\varepsilon +\frac12\varepsilon = \varepsilon $$

So here is my question: why did I never encounter the more simple definition before? Did I make a mistake somewhere? Are there advantages to the common definition, that I don't see?

Edit: Often the shortest/simplest definition becomes the standard definition. Why not in this case?

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    $\begingroup$ I'm not sure this a good reason but note that your $\Rightarrow$ is shorter than your $\Leftarrow$. So in a statement like: "let $(x_n)$ be Cauchy sequence prove that blabla" You might need the first the first definition and so somehow your proof is shorter. $\endgroup$ – Surb Jun 8 '16 at 14:33
  • $\begingroup$ @Surb ok, maybe it could be useful if you have proofs going in that direction. $\endgroup$ – supinf Jun 8 '16 at 14:54
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Here is what I think. First of all, as @Surb pointed out, $\Rightarrow$ is shorter than $\Leftarrow$ (and in fact is pretty short in its own right). It can then be seen that your second definition is an almost immediate corollary of the original, whereas the first definition is definitely not as immediately deducible from the one you found. So in practice, if you wanted to use your definition instead of the original, you could easily just derive it from the original. On the other hand, suppose you had your definition and in some situation you found that the original would be more useful. It would take a lot more work to derive it from your definition, so that's a little bit inconvenient. But really this is just a question of convenience.

Also, and this is more a statement about intuition than anything, I think the original definition expresses a particular intuitive point more clearly than your second one does. The original basically says a sequence is Cauchy if the terms become arbitrarily close to one another. Your definition essentially conveys the same point, but if you think about it, it isn't as obvious from that definition. They are, as you've shown, equivalent so you could obviously use whichever you want.

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    $\begingroup$ good point with the intuition, that makes sence! $\endgroup$ – supinf Jun 8 '16 at 21:12
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In any context, the "best definition" of a property should be as clear, natural, and intuitive as possible. That's my belief.

Sometimes this comes at a cost. For example, I favor the following definition of linear independence in a vector space: The vectors $v_1,\dots,v_n$ are linearly independent if no $v_k$ is a linear combination of the other $v_j.$ This definition seems to me to get at the heart of the matter more naturally than the usual definition. However, there is a cost to this: the usual definition turns out to be easier to apply most of the time. Fine. In my book, it's best to be kind to your audience and start with the most natural definition, then prove it's the same as the one given in umpteen texts. Then you're off to the races, and the audience, without knowing it, is better off at the expense of an extra "dt" in the time it might take.

In the case of Cauchy sequences, the "usual" definition has one more "variable" than the definition proposed. It is claimed that therefore the proposed definition is simpler. But I don't think it is simpler, nor better, nor more natural. The usual definition tells me that after some point, all terms of the sequence are "close to each other." And I think this is the intuition we're trying to get at. The proposed definition tells me that after the index $N,$ all terms are "close to $x_N$". But this $x_N$ is a moving target! Thus there is a hidden variable in the alternate definition. Why try to hide it? Let's bring it out into the light and deal with it instead of brushing it under the rug like a politician.

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You just rediscovered Cantor's original definition. A good discussion is here. Its wording is slightly different from yours: $$\forall m \in \mathbb{N}\ \lim_{n\rightarrow \infty} \left (x_{n+m}-x_n \right ) = 0 $$

And I think that this definition is more intuitive.

You can give the following justification. You can start with the following exercise:

Theorem 1 If the sequence $\{x_n\}$ is convergent then $\displaystyle{\lim_{n\rightarrow \infty}} \left (x_{n+1}-x_n \right ) = 0 $.

Then you ask the question: is the converse true? Of course not. But you can suggest another exercise
Theorem 2 If the sequence $\{x_n\}$ is convergent then $\displaystyle{\lim_{n\rightarrow \infty}} \left (x_{n+2}-x_n \right ) = 0 $.

Is the converse true here? Again the answer is negative. Does the conclusion of Theorem 1 imply the conclusion of Theorem 2 or other way around? The answer is still negative. Do the conclusions of Theorem 1 and Theorem 2 together imply the convergence? Additional negative answer.

In a similar way you can state Theorem 3, ..., Theorem $m$. Now you can ask: if you take the conclusions of Theorem 1, ..., Theorem $m$ as assumptions, does imply the convergence of the sequence $\{x_n\}$. You can check this for $m=3, 4$ and collect additional negative results. (All these trials can supply a lot of meaningful drill exercises for proving convergence or finding counterexamples)

After all these preparations, as a last resort you can propose: If we assume the conclusions of Theorem $m$ for all $m$, i.e.

if you asssume that $$\displaystyle{\lim_{n\rightarrow \infty}} \left (x_{n+m}-x_n \right ) = 0 $$ for all $m \in \mathbb{N}$ does this imply convergence?

And still you cannot say yes! In the example given above we have
$$\lim\limits_{n\rightarrow \infty} \left (x_{n+m}-x_n \right ) = 0 $$ for each $m\in \mathbb{N}$.

The difference between this condition and Cantor's definition is that the convergence in Cantor's definition is uniform in $m$ (the $\varepsilon$ does not depend on $m$).

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    $\begingroup$ Sorry, I must be daft, but exactly how does that answer the OP's question? It's a nice discussion of Cantor's attempts at formalizing limits, but it's not equivalent to what the OP suggested? $\endgroup$ – Clement C. Sep 5 '17 at 0:46
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The alternative definition is not simpler than the standard one. This is because theorems are complicated by the alternative definition, the alternative definition is not generalizable and it is not intuitive.

Here some cases that will convince you of this.

  1. Conterintuitive. The standard definition translates in "a sequence $x_n$ has elements arbitrarily close to each other for sufficiently large $n$." With the alternate definition it is a bit harder to have this intuition.

  2. Conterintuitive. Form the standard definition it is immediate to deduce the following equivalent definition for a fundamental sequence $$\lim\limits_{n\to\infty\\m\to\infty}d(x_n,x_m)=0$$

  3. Difficult to use. A fundamental sequence $x_n$ having a subsequence $x_{n_k}$ converging to a limit converges to that same limit. Indeed it results that $$\forall n_k>N, d(x_n,x_{n_k})<\varepsilon$$ Taking the limit: $d(x_n, l)<\varepsilon$, that is, $x_n$ converges to $l$. It would be impratical to prove this using the alternate definition.

  4. Conterintuitive. Fundamental sequences are "fundamental" to complete a metric space. Points of the completed space are identified by fundamental sequences of points of the original space. Two fundamental sequences $a_n$ and $b_n$ are equivalent when $$\forall \varepsilon>0, \exists N>0, \forall n>N, d(a_n,b_n)<\varepsilon$$ From the standard definition of fundamental sequence it is immediate to see that all subsequences of a fundamental sequence are equivalent one with each other. The alternative definition is impratical for the purpose.

  5. Difficult to use. Standard definition in metric spaces fits well with definition in other spaces. For instance, in the rationals standard and your alternative definitions translate into:

$$\forall \varepsilon>0,\exists N>0, \forall n,m>N, |x_n-x_m|<\varepsilon$$ $$\forall \varepsilon>0,\exists N>0, \forall n>N, |x_n-x_N|<\varepsilon$$ In $\mathbb{Q}$ given two fundamental sequences $a_n$ and $b_n$ their sum $a_n+b_n$ and product $a_nb_n$ are fundamental. Thanks to these two inequalities:

$$|a_n+b_n-a_m-b_m|\le|a_n-a_m|+|b_n-b_m|$$

$$|a_nb_n-a_mb_m|=|a_nb_n-a_mb_n+a_mb_n-a_mb_m|\le|b_n||a_n-a_m|+|a_m||b_n-b_m|$$

the proof follows straightforwardly from the standard definition of fundamental sequence. The alternative definition is impratical for the purpose.

  1. Not generalizable. The alternative definition cannot be put for fundamental nets in topological linear spaces. Given a net $x_{\alpha}$, this net is fundamental if the net $x_{\alpha'} - x_{\alpha''}$ converges to $0$. This is, in a way, parallel to the standard definition of fundamental sequence of metric space. The alternate definition does not have a counterpart in topological vector space.

The list goes on.

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  • $\begingroup$ Interesting! Can you expand on point #6? (I.e., why the alternative definition is less generalizable than the original in those cases.) $\endgroup$ – Clement C. Sep 5 '17 at 14:51
  • $\begingroup$ @ClementC. Thanks, I expanded on point #6 editing the answer. Please see there, after having considered that in topological vector spaces even though there does not exist in general the concept of distance between elements (needed by the definition of fundamental sequence), but by taking advantage of the algrebaic structure, a similar tool that serves the same role to complete a space as the fundamental sequence is the fundamental net (a net is a kind of generalized sequence where the generalizzation consists in the fact that the domain is not simply $\mathbb{N}$ but a directed set). $\endgroup$ – trying Sep 5 '17 at 15:23

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