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Prove that if $x_n \rightarrow a, n \rightarrow \infty$ then $\{x_n\}$ is a Cauchy sequence.

I believe I have found the proof as follows, wondering if there are any simpler methods or added intuition. For me, it makes sense that if a sequence has a limit, then distances between elements in the sequence must be getting smaller, in order for it to converge.

Given $\epsilon > 0, \exists N_1 \ s.t. \ \forall n \geq N_1:$

$|x_n - a| < \frac{\epsilon}{2} < \epsilon$

and for $m > n \geq N_1$ we also have:

$|x_m -a| < \frac{\epsilon}{2} < \epsilon$

Let $N \geq N_1$, then $\forall n,m \geq N$ we have:

$|x_n-x_m| = |x_n - a -x_m+a| < |x_n - a| + |-(x_m -a)| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$

Therefore $\{x_n\}$ is a Cauchy sequence.

Also, if a sequence is Cauchy does it always converge? In other words, is it sufficient to check if a sequence is Cauchy to check for convergence.

student_t
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    "if a sequence is not Cauchy can it converge?" you literally just proved if it converges it is Cauchy.... – mathworker21 Jan 29 '17 at 23:17
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    Very good proof. Indeed, if a sequence is convergent, then it is Cauchy (it can't be not Cauchy, you have just proved that!). However, the converse is not true: A space where all Cauchy sequences are convergent, is called a complete space. In complete spaces, Cauchy property is equivalent to convergence. However (for example Riemann integrable functions on $\mathbb R$, say) incomplete spaces certainly do exist, but there is an abstract notion of "completion" of a space, which is used time and again to unify these concepts. – Sarvesh Ravichandran Iyer Jan 29 '17 at 23:18
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    Yes haha sorry, what I meant to ask was if a sequence is Cauchy does it always converge, which астон вілла олоф мэллбэрг explained nicely! – student_t Jan 29 '17 at 23:24
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    It is true that a Cauchy sequence in $\Bbb R$ converges to a member of $\Bbb R.$ This is a consequence of the definition of $\Bbb R.$ A Cauchy sequence in $\Bbb Q$ will converge in $\Bbb R$ but it may or may not converge to a member of $\Bbb Q$. So whether a sequence is said to converge depends on the $space$ (e.g. $\Bbb Q$ or $\Bbb R$ ) that is under consideration. – DanielWainfleet Dec 17 '17 at 05:53
  • the title is opposed to your question – Guy Fsone Jan 16 '18 at 12:51
  • " For me, it makes sense that if a sequence has a limit, then distances between elements in the sequence must be getting smaller," One needs to be careful about what you mean "distance between terms". What does that mean? Distances between which terms. Note $a_1 = 1$ and $a_{n} = a_{n-1} + \frac 1n$ is not Cauchy and does not converge although one can say the "distance between terms gets smaller". Oh, and also "getting smaller" doesn't mean approaching zero. $a_{n} = a_n + 1 +\frac 1n$ then distance between terms is getting smaller but its always more than $1$. – fleablood Feb 20 '18 at 17:00
  • In a "complete" space or one with the least upper bound property then Cauchy sequences must converge. But not all spaces are complete. $\mathbb Q$ is not. But $\mathbb R$ is. – fleablood Feb 20 '18 at 17:02
  • Also, if a sequence is Cauchy does it always converge? $\rightarrow$ (see Banach spaces), i.e. in the open interval (0,1) 1/n is a Cauchy sequence but it does not converge. – yugikaiba Dec 30 '20 at 03:24
  • Does this answer your question? Prove that a Cauchy sequence is convergent – Clemens Bartholdy Aug 02 '22 at 10:46

2 Answers2

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Your proof is correct.

Secondly, the property of having every Cauchy sequence converge is very important, and is known as completeness.

As an example, $\mathbb R $ with the usual metric is complete.

Another important area of study is Banach spaces, which roughly are complete metric spaces where the metric comes from a norm.

More generally, there are Hilbert spaces, which are equipped with an inner product.

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With response to the comments above, it is not always true that a Cauchy sequence is convergent. For example, if we think of metric spaces, yes it is true that any Cauchy sequence in $(R, \mathrm{d}E)$ is convergent and any Cauchy sequence in $(X, \mathrm{d}\delta)$ is eventually constant, so convergent. We have the counter example however: Cauchy does not imply convergent in $(0, \infty)$, $d(x, y) = |x − y|$.

Xander Henderson
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Wronskiana
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