Proving convergent sequences are Cauchy sequences
$begingroup$
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
asked Jan 29 '17 at 23:15
student_tstudent_t
59339
$endgroup$
|
show 2 more comments
$begingroup$
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
asked Jan 29 '17 at 23:15
student_tstudent_t
59339
$endgroup$
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
|
show 2 more comments
$begingroup$
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
asked Jan 29 '17 at 23:15
student_tstudent_t
59339
$endgroup$
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
asked Jan 29 '17 at 23:15
student_tstudent_t
59339
asked Jan 29 '17 at 23:15
student_tstudent_t
59339
edited Jan 30 '17 at 17:51
student_t
asked Jan 29 '17 at 23:15
student_tstudent_t
59339
asked Jan 29 '17 at 23:15
student_tstudent_t
59339
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
|
show 2 more comments
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
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
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 $(0, infty)$, $d(x, y) = |x − y|$.
edited Dec 16 '17 at 23:25
Xander Henderson
14.8k103555
answered Dec 16 '17 at 22:59
WronskianaWronskiana
1
$endgroup$
$begingroup$
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
add a comment |
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$begingroup$
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 $(0, infty)$, $d(x, y) = |x − y|$.
edited Dec 16 '17 at 23:25
Xander Henderson
14.8k103555
answered Dec 16 '17 at 22:59
WronskianaWronskiana
1
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $(0, infty)$, $d(x, y) = |x − y|$.
edited Dec 16 '17 at 23:25
Xander Henderson
14.8k103555
answered Dec 16 '17 at 22:59
WronskianaWronskiana
1
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $(0, infty)$, $d(x, y) = |x − y|$.
edited Dec 16 '17 at 23:25
Xander Henderson
14.8k103555
answered Dec 16 '17 at 22:59
WronskianaWronskiana
1
$endgroup$
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 $(0, infty)$, $d(x, y) = |x − y|$.
edited Dec 16 '17 at 23:25
Xander Henderson
14.8k103555
answered Dec 16 '17 at 22:59
WronskianaWronskiana
1
edited Dec 16 '17 at 23:25
Xander Henderson
14.8k103555
edited Dec 16 '17 at 23:25
Xander Henderson
14.8k103555
edited Dec 16 '17 at 23:25
Xander Henderson
14.8k103555
14.8k103555
answered Dec 16 '17 at 22:59
WronskianaWronskiana
1
answered Dec 16 '17 at 22:59
WronskianaWronskiana
1
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
add a comment |
$begingroup$
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
add a comment |
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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