Proving convergent sequences are Cauchy sequences












4












$begingroup$



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.






real-analysis limits proof-verification cauchy-sequences







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    "if a sequence is not Cauchy can it converge?" you literally just proved if it converges it is Cauchy....
    $endgroup$
    – mathworker21
    Jan 29 '17 at 23:17






  • 4




    $begingroup$
    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.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Jan 29 '17 at 23:18












  • $begingroup$
    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!
    $endgroup$
    – student_t
    Jan 29 '17 at 23:24












  • $begingroup$
    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.
    $endgroup$
    – DanielWainfleet
    Dec 17 '17 at 5:53










  • $begingroup$
    the title is opposed to your question
    $endgroup$
    – Guy Fsone
    Jan 16 '18 at 12:51
















4












$begingroup$



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.






real-analysis limits proof-verification cauchy-sequences







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    "if a sequence is not Cauchy can it converge?" you literally just proved if it converges it is Cauchy....
    $endgroup$
    – mathworker21
    Jan 29 '17 at 23:17






  • 4




    $begingroup$
    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.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Jan 29 '17 at 23:18












  • $begingroup$
    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!
    $endgroup$
    – student_t
    Jan 29 '17 at 23:24












  • $begingroup$
    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.
    $endgroup$
    – DanielWainfleet
    Dec 17 '17 at 5:53










  • $begingroup$
    the title is opposed to your question
    $endgroup$
    – Guy Fsone
    Jan 16 '18 at 12:51














4












4








4


2



$begingroup$



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.






real-analysis limits proof-verification cauchy-sequences







share|cite|improve this question











$endgroup$





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.






real-analysis limits proof-verification cauchy-sequences




real-analysis limits proof-verification cauchy-sequences






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 30 '17 at 17:51







student_t

















asked Jan 29 '17 at 23:15









student_tstudent_t

59339




59339








  • 3




    $begingroup$
    "if a sequence is not Cauchy can it converge?" you literally just proved if it converges it is Cauchy....
    $endgroup$
    – mathworker21
    Jan 29 '17 at 23:17






  • 4




    $begingroup$
    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.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Jan 29 '17 at 23:18












  • $begingroup$
    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!
    $endgroup$
    – student_t
    Jan 29 '17 at 23:24












  • $begingroup$
    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.
    $endgroup$
    – DanielWainfleet
    Dec 17 '17 at 5:53










  • $begingroup$
    the title is opposed to your question
    $endgroup$
    – Guy Fsone
    Jan 16 '18 at 12:51














  • 3




    $begingroup$
    "if a sequence is not Cauchy can it converge?" you literally just proved if it converges it is Cauchy....
    $endgroup$
    – mathworker21
    Jan 29 '17 at 23:17






  • 4




    $begingroup$
    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.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Jan 29 '17 at 23:18












  • $begingroup$
    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!
    $endgroup$
    – student_t
    Jan 29 '17 at 23:24












  • $begingroup$
    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.
    $endgroup$
    – DanielWainfleet
    Dec 17 '17 at 5:53










  • $begingroup$
    the title is opposed to your question
    $endgroup$
    – Guy Fsone
    Jan 16 '18 at 12:51








3




3




$begingroup$
"if a sequence is not Cauchy can it converge?" you literally just proved if it converges it is Cauchy....
$endgroup$
– mathworker21
Jan 29 '17 at 23:17




$begingroup$
"if a sequence is not Cauchy can it converge?" you literally just proved if it converges it is Cauchy....
$endgroup$
– mathworker21
Jan 29 '17 at 23:17




4




4




$begingroup$
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.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 29 '17 at 23:18






$begingroup$
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.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 29 '17 at 23:18














$begingroup$
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!
$endgroup$
– student_t
Jan 29 '17 at 23:24






$begingroup$
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!
$endgroup$
– student_t
Jan 29 '17 at 23:24














$begingroup$
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.
$endgroup$
– DanielWainfleet
Dec 17 '17 at 5:53




$begingroup$
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.
$endgroup$
– DanielWainfleet
Dec 17 '17 at 5:53












$begingroup$
the title is opposed to your question
$endgroup$
– Guy Fsone
Jan 16 '18 at 12:51




$begingroup$
the title is opposed to your question
$endgroup$
– Guy Fsone
Jan 16 '18 at 12:51










1 Answer
1






active

oldest

votes


















0












$begingroup$

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 $(0, infty)$, $d(x, y) = |x − y|$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I guess I should have clarified. This is for metric spaces.
    $endgroup$
    – student_t
    Dec 16 '17 at 23:00






  • 1




    $begingroup$
    It's not true for all metric spaces. $mathbb Q$ with the normal metric is a metric space where it is not true.
    $endgroup$
    – fleablood
    Feb 20 '18 at 17:04










  • $begingroup$
    Or try $Bbb R-{0}$.
    $endgroup$
    – Ted Shifrin
    Nov 7 '18 at 18:06











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1 Answer
1






active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

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 $(0, infty)$, $d(x, y) = |x − y|$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I guess I should have clarified. This is for metric spaces.
    $endgroup$
    – student_t
    Dec 16 '17 at 23:00






  • 1




    $begingroup$
    It's not true for all metric spaces. $mathbb Q$ with the normal metric is a metric space where it is not true.
    $endgroup$
    – fleablood
    Feb 20 '18 at 17:04










  • $begingroup$
    Or try $Bbb R-{0}$.
    $endgroup$
    – Ted Shifrin
    Nov 7 '18 at 18:06
















0












$begingroup$

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 $(0, infty)$, $d(x, y) = |x − y|$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I guess I should have clarified. This is for metric spaces.
    $endgroup$
    – student_t
    Dec 16 '17 at 23:00






  • 1




    $begingroup$
    It's not true for all metric spaces. $mathbb Q$ with the normal metric is a metric space where it is not true.
    $endgroup$
    – fleablood
    Feb 20 '18 at 17:04










  • $begingroup$
    Or try $Bbb R-{0}$.
    $endgroup$
    – Ted Shifrin
    Nov 7 '18 at 18:06














0












0








0





$begingroup$

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 $(0, infty)$, $d(x, y) = |x − y|$.






share|cite|improve this answer











$endgroup$



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 $(0, infty)$, $d(x, y) = |x − y|$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 16 '17 at 23:25









Xander Henderson

14.8k103555




14.8k103555










answered Dec 16 '17 at 22:59









WronskianaWronskiana

1




1












  • $begingroup$
    I guess I should have clarified. This is for metric spaces.
    $endgroup$
    – student_t
    Dec 16 '17 at 23:00






  • 1




    $begingroup$
    It's not true for all metric spaces. $mathbb Q$ with the normal metric is a metric space where it is not true.
    $endgroup$
    – fleablood
    Feb 20 '18 at 17:04










  • $begingroup$
    Or try $Bbb R-{0}$.
    $endgroup$
    – Ted Shifrin
    Nov 7 '18 at 18:06


















  • $begingroup$
    I guess I should have clarified. This is for metric spaces.
    $endgroup$
    – student_t
    Dec 16 '17 at 23:00






  • 1




    $begingroup$
    It's not true for all metric spaces. $mathbb Q$ with the normal metric is a metric space where it is not true.
    $endgroup$
    – fleablood
    Feb 20 '18 at 17:04










  • $begingroup$
    Or try $Bbb R-{0}$.
    $endgroup$
    – Ted Shifrin
    Nov 7 '18 at 18:06
















$begingroup$
I guess I should have clarified. This is for metric spaces.
$endgroup$
– student_t
Dec 16 '17 at 23:00




$begingroup$
I guess I should have clarified. This is for metric spaces.
$endgroup$
– student_t
Dec 16 '17 at 23:00




1




1




$begingroup$
It's not true for all metric spaces. $mathbb Q$ with the normal metric is a metric space where it is not true.
$endgroup$
– fleablood
Feb 20 '18 at 17:04




$begingroup$
It's not true for all metric spaces. $mathbb Q$ with the normal metric is a metric space where it is not true.
$endgroup$
– fleablood
Feb 20 '18 at 17:04












$begingroup$
Or try $Bbb R-{0}$.
$endgroup$
– Ted Shifrin
Nov 7 '18 at 18:06




$begingroup$
Or try $Bbb R-{0}$.
$endgroup$
– Ted Shifrin
Nov 7 '18 at 18:06


















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Ida-Boy-Ed-Garten

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Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
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