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Are there Cauchy sequence that are not bounded? [duplicate]
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Proof Check: Every Cauchy Sequence is Bounded
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I was wondering, Is there Cauchy sequence that are not bounded ? Of course, in complete spaces is not possible. I have a theorem that says that if $(A,d)$ is not complete, then $(bar A,d)$ is complete. But are there spaces s.t. indeed for all $varepsilon>0$, there is $N$ s.t. $d(x_n,x_m)<varepsilon$ for all $ngeq N$, and all $rgeq 0$, but the ball also move to $+infty $ ?
cauchy-sequences
asked Jul 15 at 13:56
MathBeginner
707312
marked as duplicate by José Carlos Santos, user223391, Somos, Xander Henderson, Shailesh Jul 16 at 0:09
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |Â
up vote
0
down vote
favorite
This question already has an answer here:
Proof Check: Every Cauchy Sequence is Bounded
2 answers
I was wondering, Is there Cauchy sequence that are not bounded ? Of course, in complete spaces is not possible. I have a theorem that says that if $(A,d)$ is not complete, then $(bar A,d)$ is complete. But are there spaces s.t. indeed for all $varepsilon>0$, there is $N$ s.t. $d(x_n,x_m)<varepsilon$ for all $ngeq N$, and all $rgeq 0$, but the ball also move to $+infty $ ?
cauchy-sequences
asked Jul 15 at 13:56
MathBeginner
707312
marked as duplicate by José Carlos Santos, user223391, Somos, Xander Henderson, Shailesh Jul 16 at 0:09
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
@JoséCarlosSantos: It's not duplicate because I'm not in a normed space but in a metric space.
– MathBeginner
Jul 15 at 14:05
It is a duplicate, since it is the same argument.
– José Carlos Santos
Jul 15 at 14:07
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
This question already has an answer here:
Proof Check: Every Cauchy Sequence is Bounded
2 answers
I was wondering, Is there Cauchy sequence that are not bounded ? Of course, in complete spaces is not possible. I have a theorem that says that if $(A,d)$ is not complete, then $(bar A,d)$ is complete. But are there spaces s.t. indeed for all $varepsilon>0$, there is $N$ s.t. $d(x_n,x_m)<varepsilon$ for all $ngeq N$, and all $rgeq 0$, but the ball also move to $+infty $ ?
cauchy-sequences
asked Jul 15 at 13:56
MathBeginner
707312
This question already has an answer here:
Proof Check: Every Cauchy Sequence is Bounded
2 answers
I was wondering, Is there Cauchy sequence that are not bounded ? Of course, in complete spaces is not possible. I have a theorem that says that if $(A,d)$ is not complete, then $(bar A,d)$ is complete. But are there spaces s.t. indeed for all $varepsilon>0$, there is $N$ s.t. $d(x_n,x_m)<varepsilon$ for all $ngeq N$, and all $rgeq 0$, but the ball also move to $+infty $ ?
This question already has an answer here:
Proof Check: Every Cauchy Sequence is Bounded
2 answers
cauchy-sequences
asked Jul 15 at 13:56
MathBeginner
707312
asked Jul 15 at 13:56
MathBeginner
707312
asked Jul 15 at 13:56
MathBeginner
707312
asked Jul 15 at 13:56
MathBeginner
707312
707312
marked as duplicate by José Carlos Santos, user223391, Somos, Xander Henderson, Shailesh Jul 16 at 0:09
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by José Carlos Santos, user223391, Somos, Xander Henderson, Shailesh Jul 16 at 0:09
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
@JoséCarlosSantos: It's not duplicate because I'm not in a normed space but in a metric space.
– MathBeginner
Jul 15 at 14:05
It is a duplicate, since it is the same argument.
– José Carlos Santos
Jul 15 at 14:07
add a comment |Â
@JoséCarlosSantos: It's not duplicate because I'm not in a normed space but in a metric space.
– MathBeginner
Jul 15 at 14:05
It is a duplicate, since it is the same argument.
– José Carlos Santos
Jul 15 at 14:07
@JoséCarlosSantos: It's not duplicate because I'm not in a normed space but in a metric space.
– MathBeginner
Jul 15 at 14:05
@JoséCarlosSantos: It's not duplicate because I'm not in a normed space but in a metric space.
– MathBeginner
Jul 15 at 14:05
It is a duplicate, since it is the same argument.
– José Carlos Santos
Jul 15 at 14:07
It is a duplicate, since it is the same argument.
– José Carlos Santos
Jul 15 at 14:07
add a comment |Â
2 Answers
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No, that cannot happen. Suppose that $(x_n)_n geq 0$ is a Cauchy sequence in some metric space. By definition, there exists an integer $N geq 0$ such that for all $n,m geq N$ we have $d(x_n,x_m) leq 1$. Thus, all the points $~ n geq N$ are contained in the ball of radius 1 around $x_N$. By adding the finitely many points $x_1,dots,x_N-1$, this sequence cannot become unbounded.
answered Jul 15 at 14:04
Florian R
3458
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1
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Note that $d(x_n,x_N)<epsilon$ for all $nge N$ and that only finitely many $n<N$ are to be considered beyond that.
answered Jul 15 at 14:04
Hagen von Eitzen
265k20258477
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
No, that cannot happen. Suppose that $(x_n)_n geq 0$ is a Cauchy sequence in some metric space. By definition, there exists an integer $N geq 0$ such that for all $n,m geq N$ we have $d(x_n,x_m) leq 1$. Thus, all the points $~ n geq N$ are contained in the ball of radius 1 around $x_N$. By adding the finitely many points $x_1,dots,x_N-1$, this sequence cannot become unbounded.
answered Jul 15 at 14:04
Florian R
3458
add a comment |Â
up vote
4
down vote
No, that cannot happen. Suppose that $(x_n)_n geq 0$ is a Cauchy sequence in some metric space. By definition, there exists an integer $N geq 0$ such that for all $n,m geq N$ we have $d(x_n,x_m) leq 1$. Thus, all the points $~ n geq N$ are contained in the ball of radius 1 around $x_N$. By adding the finitely many points $x_1,dots,x_N-1$, this sequence cannot become unbounded.
answered Jul 15 at 14:04
Florian R
3458
add a comment |Â
up vote
4
down vote
up vote
4
down vote
No, that cannot happen. Suppose that $(x_n)_n geq 0$ is a Cauchy sequence in some metric space. By definition, there exists an integer $N geq 0$ such that for all $n,m geq N$ we have $d(x_n,x_m) leq 1$. Thus, all the points $~ n geq N$ are contained in the ball of radius 1 around $x_N$. By adding the finitely many points $x_1,dots,x_N-1$, this sequence cannot become unbounded.
answered Jul 15 at 14:04
Florian R
3458
No, that cannot happen. Suppose that $(x_n)_n geq 0$ is a Cauchy sequence in some metric space. By definition, there exists an integer $N geq 0$ such that for all $n,m geq N$ we have $d(x_n,x_m) leq 1$. Thus, all the points $~ n geq N$ are contained in the ball of radius 1 around $x_N$. By adding the finitely many points $x_1,dots,x_N-1$, this sequence cannot become unbounded.
answered Jul 15 at 14:04
Florian R
3458
answered Jul 15 at 14:04
Florian R
3458
answered Jul 15 at 14:04
Florian R
3458
answered Jul 15 at 14:04
Florian R
3458
3458
add a comment |Â
add a comment |Â
up vote
1
down vote
Note that $d(x_n,x_N)<epsilon$ for all $nge N$ and that only finitely many $n<N$ are to be considered beyond that.
answered Jul 15 at 14:04
Hagen von Eitzen
265k20258477
add a comment |Â
up vote
1
down vote
Note that $d(x_n,x_N)<epsilon$ for all $nge N$ and that only finitely many $n<N$ are to be considered beyond that.
answered Jul 15 at 14:04
Hagen von Eitzen
265k20258477
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Note that $d(x_n,x_N)<epsilon$ for all $nge N$ and that only finitely many $n<N$ are to be considered beyond that.
answered Jul 15 at 14:04
Hagen von Eitzen
265k20258477
Note that $d(x_n,x_N)<epsilon$ for all $nge N$ and that only finitely many $n<N$ are to be considered beyond that.
answered Jul 15 at 14:04
Hagen von Eitzen
265k20258477
answered Jul 15 at 14:04
Hagen von Eitzen
265k20258477
answered Jul 15 at 14:04
Hagen von Eitzen
265k20258477
answered Jul 15 at 14:04
Hagen von Eitzen
265k20258477
265k20258477
add a comment |Â
add a comment |Â
@JoséCarlosSantos: It's not duplicate because I'm not in a normed space but in a metric space.
– MathBeginner
Jul 15 at 14:05
It is a duplicate, since it is the same argument.
– José Carlos Santos
Jul 15 at 14:07