17
$\begingroup$

I am really confused. I will appreciate if somebody can help me to define the difference between Cauchy and convergent sequence.

Many thanks.

$\endgroup$
  • $\begingroup$ If you are only dealing with $\mathbb R$, you SHOULD be confused. $\endgroup$ – user99914 Mar 27 '14 at 8:42
  • 3
    $\begingroup$ One important difference is the way the notion is defined: the notion of Cauchy sequence only refers to the terms of the sequence itself, while the notion of convergent sequence refers to (the existence of) a limit value of the sequence. $\endgroup$ – Marc van Leeuwen Mar 27 '14 at 14:56
17
$\begingroup$

When you have a convergent sequence: $x_n\to\xi$, then for large $n$ all points $x_n$ are near $\xi$, and by the triangle inequality they are then near each other as well – in formal language: The $x_n$ automatically form a Cauchy sequence; no big deal.

But:

In some metric spaces $X$, among them ${\mathbb R}$, ${\mathbb R}^n$, and ${\mathbb C}$, the converse also holds: A Cauchy sequence is automatically convergent. (Such spaces are called complete.) This property of spaces is all-important for the following reason: In order to test a sequence $(x_n)_{\geq1}$ for Cauchy-ness you only have to look at the given $x_n$ themselves, and you don't have to know the limit beforehand. When the sequence passes this "finitary" test you know for sure that it has a limit $\xi\in X$.

In all, the so-called Cauchy criterion (for real sequences, say) is not a little proposition about convergence, but a deep theorem about the fine structure of ${\mathbb R}$.

$\endgroup$
  • 2
    $\begingroup$ "a deep theorem about the fine structure of ${\mathbb R}$" -- or an axiom of ${\mathbb R}$, or a fairly straightforward consequence of the construction of ${\mathbb R}$, depending how you do your foundations :-) $\endgroup$ – Steve Jessop Mar 27 '14 at 11:14
  • $\begingroup$ @Christian Blatter May I ask, does the converse also hold for $\mathbb Q$? $\endgroup$ – Dole Feb 26 '18 at 10:38
  • 2
    $\begingroup$ @Dole: No, it doesn't. The finite decimal approximations to $\sqrt{2}$ converge to $\sqrt{2}$, hence they form a Cauchy sequence of rational numbers. Nevertheless this sequence is divergent in ${\mathbb Q}$. $\endgroup$ – Christian Blatter Feb 26 '18 at 10:54
26
$\begingroup$

Let $(X,d)$ be a metric space.

Definition. A sequence $(x_n)_{n\in\mathbb N}$ with $x_n\in X$ for all $n\in\mathbb N$ is a Cauchy sequence in $X$ if and only if for every $\varepsilon > 0$ there exists $N\in\mathbb N$ such that $d(x_n,x_m)<\varepsilon$ for all $n,m>N$.

Informally speaking, a Cauchy sequence is a sequence where the terms of the sequence are getting closer and closer to each other.

Definition. A sequence $(x_n)_{n\in\mathbb N}$ with $x_n\in X$ for all $n\in\mathbb N$ is convergent if and only if there exists a point $x\in X$ such that for every $\varepsilon > 0$ there exists $N\in\mathbb N$ such that $d(x_n,x)<\varepsilon$. In this situation we say $x$ is a limit of the sequence $(x_n)_{n\in\mathbb N}$, or $(x_n)_{n\in\mathbb N}$ converges to $x$.

Informally speaking, a sequence is convergent if the terms of the sequence are getting closer and closer to some point $x\in X$.

A metric space $X$ is said to be complete if every Cauchy sequence is convergent. This is the case for the spaces $\mathbb R^n$, which is the reason why you might not see the difference of the concepts at first glance.

Let's take a look at a familiar metric space which is not complete, so we have Cauchy sequence that are not convergent. Let $X=\mathbb Q$ be the set of rational numbers with the usual metric given by $d(x,y)=\left|x-y\right|$. Consider the sequence defined by $$ x_1 = 1,\quad x_{n+1} = 1+\frac{1}{1+x_n}. $$ We can show that this sequence is a Cauchy sequence in $\mathbb Q$, but there is no $x\in \mathbb Q$ that is a limit of the sequence.

Why is that? Well, consider the same sequence as a sequence in the complete metric space $X=\mathbb R$, we can now show that $(x_n)_{n\in\mathbb N}$ converges to $\sqrt 2\in\mathbb R$, but $\sqrt 2$ is not in $\mathbb Q$. So while the sequence converges in $\mathbb R$, it doesn't converge in $\mathbb Q$. Nevertheless it is a Cauchy sequence in both spaces.

$\endgroup$
  • $\begingroup$ I know this is an old post , thanks for your answer helped me loads , but just wondering how could one show the limit is root2 I know we could just do limxn=limxn+1 and find L but we cant just assume liman=L because it may not converge , so how do I show it converges in the first place? Thanks $\endgroup$ – Rivaldo Nov 22 '18 at 17:56
  • 2
    $\begingroup$ See for example this thread. $\endgroup$ – Christoph Nov 22 '18 at 19:41
7
$\begingroup$

A convergent sequence is also a Cauchy sequence.

A Cauchy sequence is not necessarily a convergent sequence. For example if our space is $X=\mathbb Q$, then $$ x_n=\frac{\lfloor n\sqrt{2}\rfloor}{n}, $$
is a Cauchy sequence which DOES NOT converge is $\mathbb Q$. It DOES converge is $\mathbb R$ but not in $\mathbb Q$.

A metric space where every Cauchy sequence converges is called complete. The space $\mathbb Q$ with metric the distance $|a-b|$ is not complete, while $\mathbb R$ with the same metric is complete.

$\endgroup$
7
$\begingroup$

Every convergent sequence is a cauchy sequence. The converse may however not hold. For sequences in $\mathbb{R}^k$ the two notions are equal. More generally we call an abstract metric space $X$ such that every cauchy sequence in $X$ converges to a point in $X$ a complete metric space. One can for instance show that the space of continuous functions on a compact set with the uniform metric is complete.

As an example of a space that is not complete, consider the interval $(0,1)$ and the sequence $1/n$, clearly $1/n$ is cauchy, but $1/n$ is not convergent in $(0,1)$ as $0$ is not in $(0,1)$.

In contrast $[0,1]$ is complete, since it is a closed subset of the complete space $\mathbb{R}$.

$\endgroup$
  • $\begingroup$ "Every convergent sequence is a Cauchy sequence." That's what I was looking for! I couldn't think of any counterexamples. $\endgroup$ – Mateen Ulhaq May 8 '17 at 10:19
3
$\begingroup$

Let's start in metric space $\mathbb{R}$ equipped with its normal metric.

A sequence $\left(x_{n}\right)$ in $\mathbb{R}$ is convergent if some $x\in\mathbb{R}$ exists such that for every $\varepsilon>0$ some $n_{0}\in\mathbb{N}$ can be found such that $\left|x-x_{n}\right|<\varepsilon$ for each $n\geq n_{0}$. If that is the case then it can be shown that this $x$ is unique and the sequence is said to converge to $x$.

A sequence $\left(x_{n}\right)$ in $\mathbb{R}$ is a Cauchy-sequence if for every $\varepsilon>0$ some $n_{0}\in\mathbb{N}$ can be found such that $\left|x_{n}-x_{m}\right|<\varepsilon$ for each pair $n,m\in\mathbb{N}$ with $n,m\geq n_{0}$.

It can be shown that a convergent sequence is a Cauchy-sequence. This in any metric space.

In the case we are dealing with (metric space $\mathbb{R}$) the opposite is also true: every Cauchy-sequence in $\mathbb{R}$ is a convergent sequence.

However, there are metric spaces in wich the opposite is not true. An example is $\mathbb{Q}$. Start with a sequence $\left(x_{n}\right)$ in $\mathbb{R}$ that converges to some $x\in\mathbb{R}-\mathbb{Q}$ and let it be that $x_{n}\in\mathbb{Q}$ for each $n$. It is a Cauchy-sequence in $\mathbb{R}$ and is convergent in $\mathbb{R}$. However, if we look at it as a sequence in $\mathbb{Q}$ then it is still a Cauchy-sequence, but it is not convergent, simply because $x\notin\mathbb{Q}$.

Spaces in which the opposite is true are so-called 'complete spaces'.

$\endgroup$

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.