# Talk:Cauchy sequence

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Cauchy net redirects here, yet there seems to be nothing about the concept here.... Vivacissamamente

## Untitled

Roughly speaking, the terms of the sequence are getting closer and closer together in a way that suggests that the sequence ought to converge. Nonetheless, Cauchy sequences do not always converge.

Some example please --Taw 11 December 2001

added an example, it's kind of kludgy though -- RAE

I'd like to put something along the following lines: Cauchy Seqs are initially useful in spaces such as the Reals because they are a test of convergence which doesn't require a value for the potential limit. -- the flip side is that IF all CSs converge then a space is complete.

There's a sort of switch in perception as things move up a level of abstraction which as a mathematician I find self-evident (and interesting), but I suspect non-mathematicians find baffling or even terrifying:

• theorem: cauchy seqs converge on the Reals
• abstraction: cauchy seqs on other space, where they might not converge
• axiom: part of the defn of complete space

Has this been general idea been coverered anywhere in the maths section? -- Tarquin

I don't think it has been covered; it would fit either here or in complete space. AxelBoldt, Wednesday, June 12, 2002

All Cauchy sequences of real or complex numbers converge, hence testing that a sequence is Cauchy is a test of convergence. This is more useful than using the definition of convergence, since that requires the possible limit to be known. With this idea in mind, a metric space in which all Cauchy sequences converge is called complete.
Thus R and C are complete; but Q is not. The standard construction of the real numbers involves Cauchy sequences of rational numbers; (something about R being the completion of Q...)

...and something on Mathematical abstraction in general somewhere else. I'll see if I can dig up or remember the proof outlines for "Every convergent sequence is a Cauchy sequence" and "every Cauchy sequence is bounded" -- Tarquin June 12 2002

## Definition

How about some sort of formal definition? My elementary analysis textbook states: A sequence $(s_n)$ of real numbers is called a Cauchy sequence if $\forall\epsilon>0\exists N\mbox{ such that }m,n>N\mbox{ implies }\vert s_n-s_m\vert<\epsilon.$66.71.96.78 17:01, 3 October 2005 (UTC)

The formal definition in the article is more general than yours, applying to metric spaces in general rather than specifically to the real line. —Caesura(t) 17:07, 6 December 2005 (UTC)
It would also be nice to have Cauchy sequences defined for other absolute values, in particular for p-adic absolute values. Would this be a problem? Gene Ward Smith 09:05, 6 May 2006 (UTC)
I guess that could go in the generalization section, as is not really central to the concept of Cauchy sequences as used in analysis. Oleg Alexandrov (talk) 15:48, 6 May 2006 (UTC)

## p-adic material out of place

The p-adic material just after the heading Cauchy sequence in a metric space doesn't seem to belong there. McKay 11:18, 16 June 2006 (UTC)

I cut it out, together with other fluff. The whole article was a mumbo jumbo of things without clear connections. Oleg Alexandrov (talk) 16:42, 16 June 2006 (UTC)

## 'All two' or 'any two'?

The first paragraph has again been changed to "all two remaining elements ... ". I don't want to start a revert war here, so I would appreciate other views. My view is that it has to be "any two", as "all two" is both mathematically wrong and grammatically wrong. Madmath789 06:38, 21 June 2006 (UTC)

I agree, but "any two" is not very precise. It is the maximum distance between two of the remaining elements that has to be small. I changed it. McKay 07:57, 21 June 2006 (UTC)

## Reference List

The reference list includes two on algebra and one of constructive mathematics. How about a reference to a good analysis text since after all, Cauchy sequences typically are learned as part of analysis, not algebra.

Just any analysis text, even if it isn't apparently used as a reference? Why not little Rudin or something? But is that right? There isn't anything, offhand, I can think to add to this article, from a source or otherwise. —vivacissamamente 03:13, 21 October 2006 (UTC)
I feel that there should be an analysis text in the references:

I suggest M.Spivak's 'Calculus' as an option. It gives a good treatment of real Cauchy sequences, and of the construction of the reals using Cauchy sequences, as well as other constructions (it's definitely an Analysis text, rather than a Calculus text, despite the title). I'll try to look ISBN, etc, if someone doesn't beat me to it. Messagetolove 13:56, 26 May 2007 (UTC)

Have put in Spivak reference ( it's even in Wiki in its own right).

Messagetolove 19:15, 26 May 2007 (UTC)

## Recent edits

Here's my revert. Here are the issues.

• The sequence
0, 1, 1.5, 2, 2.25, 2.5, 2.75, 3, 3.125, 3.25, ..
appears convergent to me, in spite of what the example claims
• There is no need to emphasize that
$d(x_m, x_n) < r$

implies not only consecutive terms but all remaining terms are getting closer and closer together. It is obvious that we are talking about all terms, since we use different indeces for m and n.

• Why remove the text about completeness from the def of cauchy sequence? That's the best place to make that point! It is obvious to anybody there that a Cauchy sequence looks as if it is convergent. Then make the reader pay attention right there, rather than moving that text half an article down. Oleg Alexandrov (talk) 15:55, 2 June 2007 (UTC)
Completeness has its own section. Especially with the added issue of counterexamples the two issues should not be mixed up.--Patrick 08:45, 3 June 2007 (UTC)
It looks to me as if Patrick may have intended to be taking the partial sums of the series obtained by adding 1 once, 1/2 2 times, 1/4 4 times, 1/8 8 times, etc that is, essentially the argument used to prove that the harmonic series diverges (shifted a bit). Clearly, if that is the intended sequence, it is not convergent, but it isn't entirely made clear that this is really what is intended, and I haven't heard that referred to as "harmonic" before. I agree with your other points.
Messagetolove 19:06, 2 June 2007 (UTC)
Yes, that is what I mean. However, I wrote that this sequence and the sequence of harmonic numbers are counterexamples, I did not write that this sequence is called harmonic. This sequence has easier numbers than the harmonic numbers, so it is easier to see that it diverges, with all natural numbers occurring in it. If the sequence is not clear enough we can add more terms.--Patrick 22:37, 2 June 2007 (UTC)
But it is much harder to see the pattern in this sequence, even with an explanation. The harmonic series look simpler to me, one just adds 1/n each time. Also, I am not sure this counterexample is relevant here. I'll add a picture these days, that should make things clearer. Oleg Alexandrov (talk) 23:07, 2 June 2007 (UTC)

I reverted the counterexample of consecutive terms, that one is kind of pointless, and poorly written too (it is not the difference of consecutive terms which goes to zero, it is the distance between them, per the previous paragraph). The interesting counterexample is a Cauchy sequence that is not convergent, and that is below. Oleg Alexandrov (talk) 11:44, 3 June 2007 (UTC)

(1) That is not at all an argument for deletion, that is just changing one word.
(2) That is not a counterexample, it is a different (equally interesting) issue.
Patrick 12:15, 3 June 2007 (UTC)
I agree with (1), of course. I don't agree that the thing about consecutive terms is interesting. It is clear enough that the terms are not consecutive from the def. The big deal about Cauchy sequence is relation to completeness. Oleg Alexandrov (talk) 12:24, 3 June 2007 (UTC)

## question

Oleg said

> this is clear enough, n is on the horisontal axis, and x_n is on the vertical one. This is a standard way of graphing functions.

yes, i see this is true. but, the graph ploted by blue points lies on the "plane", not on the axis. so, the blue points do not show the sequence of x_n, do that? sorry for my poor english, thank you. --218.42.230.29 21:48, 5 August 2007 (UTC)

I clarified things a bit saying that what is graphed is not the sequence itself, but its plot. Are things clearer now? Oleg Alexandrov (talk) 00:37, 6 August 2007 (UTC)
i thank your kindness. :-) --218.42.230.61 01:41, 6 August 2007 (UTC)

## A squiggle I dont know how to say or from what alphabet it comes

I would simply like to point out that there is a squiggle on this page that assumes a fair bit of pre-knowledge that is not documented by a re-link into Wikipedia explaining the character symbol and associated concepts

Generalizing, it would be nice if every mathematical symbol used an article had a link to all that the reader should know about it. Could Wikipedia make this some kind of standard?

Maybe this could be further generalized to some kind of overall standard that any relevant concept with sufficient complexity and obscurity be re-linked back into Wikipedia, at least with a requested article. —Preceding unsigned comment added by Jjalexand (talkcontribs) 12:50, 2 September 2008 (UTC)

## Values of exp, sin and cos always irrational?

I think this is not true:

The values of the exponential, sine and cosine functions, exp(x), sin(x), cos(x), are known to be irrational for any rational value of x≠0, [...]

Counter-example: $\cos(60^{\circ}) = \frac{1}{2} \,$.

(Francisco Albani (talk) 00:31, 22 September 2008 (UTC)).

The statement you quoted above is correct in "natural units" which in the case of the trigonometric functions means radians instead of degrees: 60° = 60•2π/360 = π/3, and a rational multiple of an irrational number (π) is still irrational. — Tobias Bergemann (talk) 11:29, 22 September 2008 (UTC)

## "m,n > N" vs. "m,n >= N"

Is there any reason "m,n > N" is better than "m,n >= N"? With ">", it seems to me that the first element is always discarded -- which doesn't matter, I guess, but made me wonder if that is necessary or not. 88.65.186.193 (talk) 14:08, 25 April 2009 (UTC)

Since the choice of N is arbitrary so is whether or not you use a strict inequality (or not). For example, if you require that m,n > 199 then m,n >= 200 is just as good. Some authors like it one way, others another. Similarly, when doing a delta-epsilon proof, if two elements {f(y) and f(x)} in a sequence are less than epsilon whenever {x and y} are closer than "delta", then the sequence is convergent... but the same is true if they are less than 5*epsilon or epsilon^2. or sqrt(epsilon) or 1000*epsilon, because the choice is ARBITRARY! Hope that helps.... Brydustin (talk) 03:20, 25 February 2012 (UTC)

In short: it means the same but takes one character less... --CiaPan (talk) 07:50, 1 March 2012 (UTC)

## Convergence

Why don't we just simply state:

a Cauchy sequence convergence in a metric space, though the limit might not in the metric space.

Jackzhp (talk) 22:56, 17 June 2009 (UTC)

Because when we say something converges to something else, we implicitly suggest that the limit is in the space? I mean, if the limit is not in the space, where is it? It's not easy to talk about things that are outside the space. -- Taku (talk) 21:39, 18 June 2009 (UTC)
That's right. I'd even add some emphasis and say 'because by definition a sequence is convergent if and only if it has a limit,' and a limit of a sequence is defined as an element of the same space to which the sequence's items belong. That's why 'the same' sequence of $(1/n)_{n\in\Bbb N}$ is convergent in real numbers (with a limit equal zero) and is divergent in $\Bbb R^+=(0,\infty)$ --CiaPan (talk) 12:05, 2 July 2009 (UTC)

## General topology

What about generalization into general topology? I suggest a definition: (Cn) is Cauchy iff there is a set such that any its open covering contains an open set S such that there exists an integer N such that for any n>N, Cn is in S.

Is it good? Does it appear somewhere? —Preceding unsigned comment added by 84.229.68.163 (talk) 06:56, 2 July 2009 (UTC)

This sounds like the definition of convergence. By definition, a sequence $x_n$ converges to x if every open set containing x contains infinitely many $x_n$. The problem with generalizing Cauhy sequence is that we need some way to measure the distance between two points; this can be done in a metric space, of course. But in general, not possible. So, we need uniform structure or something else. -- Taku (talk) 11:09, 2 July 2009 (UTC)
'Contains infinitely many' seems too weak for me. Consider the real sequence $a_n = \sin n$ It has no limit in the $(0,1)$ interval although any open subset of $(0,1)$ contains infinitely many terms of $(a_n)$. May be we can say 'any open set containing x contains ALL $x_n$ for n greater than some k,' that is 'all except at most a finite number ot initial terms.' --CiaPan (talk) 07:14, 3 July 2009 (UTC)
You're absolutely right. A simpler counterexample would be "1, 0, 1, 0, 1, 0, ...". It should be "x_n -> x if every neighborhood of x contains all x_n except some finite many terms." -- Taku (talk) 13:43, 3 July 2009 (UTC)

My definition is intended precisely to circumvent this obstacle. The idea is to confine the sequence into an open set "as small as we want" (for sufficiently large n). The difficulty was to define what is "an open set as small as we want" without limit point or metric. What I found appears an elegant solution for this: open covering can be as fine as we want. My definition doesn't require anything undefined in general topology and clearly does generalize topological convergent sequences, but I don't see whether it generalizes metric Cauchy sequences, too. If not, can it be fixed by replacing "any its open covering" with "any its finite open covering"?

By the way, how can one add comment to an arbitrary section or to a comment? (one ugly way to do it is certainly editing) --87.68.41.183 (talk) 15:29, 2 July 2009 (UTC)

## English pronunciation of "Cauchy"

In my experience, the name "Cauchy" is pronounced in English as /ˈkoʊʃi/. Should I add this pronunciation to the article, or wait for someone to come along with a reliable source, or at least say, "yeah, I pronounce it that way, too"? —Tanner Swett (talk) 19:28, 21 June 2011 (UTC)

I wouldn't know, although one could in principle search YouTube for evidence that people really pronounce it this way, but I've trusted your observation, been bold and added the pronunciation you are suggesting to the article. This way, it is more visible, and the likelihood of people noticing it and complaining or changing it if it is not the most frequent pronunciation is higher. --Florian Blaschke (talk) 15:46, 6 March 2014 (UTC)

## Confusing Example

The example of a Cauchy sequence says "The customary acceptance of the fact that any real number x has a decimal expansion is an implicit acknowledgment that a particular Cauchy sequence of rational numbers (whose terms are the successive truncations of the decimal expansion of x) has the real limit x."

But this is followed by "counter example: rational numbers" that says "The rational numbers Q are not complete (for the usual distance): There are sequences of rationals that converge (in R) to irrational numbers; these are Cauchy sequences having no limit in Q."

There is no contradiction in saying that people intuitively understand that a Cauchy sequence of rationals converges to a real number, but a casual reading of the two paragraphs invites confusion. — Preceding unsigned comment added by 208.86.181.160 (talk) 15:10, 19 July 2012 (UTC)

## Generalizations to topological groups

I think there is a problem with the definition of the completion of a topological group G when G does not have a countable basis of open neighbourhood of 0. In particular, if K is a field with a Krull valuation, whose value group does not have countable cofinality, then, according to the definition given here, K will be complete (since the only Cauchy sequences in K, as defined here, are eventually constant), contrary to usual terminology. 193.206.101.2 (talk) 14:11, 4 July 2013 (UTC)

Another example of the same problem: let G be the additive group of the Banach space $\ell^1$, equipped with the weak topology. Then every weakly Cauchy SEQUENCE converges in norm, by Schur's property and the fact that $L^1$-spaces are weakly sequentially complete. However the group (the Banach space) is certainly not weakly complete. In order to fix this, one should go to the much more complex setting of uniform spaces and Cauchy filters. Bdmy (talk) 18:32, 4 July 2013 (UTC)