Meaning of being a Cauchy sequence
$begingroup$
A Cauchy sequence in a metric space $(X,d)$ is a sequence for which the distance between two terms can be made as small as we want, provided we look far enough in the sequence.
Let $X subseteq Y$, where $Y$ is a set on which $d$ can be extended. Is it always true that for any Cauchy sequence ${x_n}$ in $X$ one can find $yin Y$ such that $d(x_n,y)rightarrow 0$?
In other words, is it true that every Cauchy sequence "converges" to something? (and if this something is in $X$ we say it is convergent?)
real-analysis convergence metric-spaces cauchy-sequences
asked Dec 28 '18 at 19:34
LeonardoLeonardo
3339
$endgroup$
add a comment |
$begingroup$
A Cauchy sequence in a metric space $(X,d)$ is a sequence for which the distance between two terms can be made as small as we want, provided we look far enough in the sequence.
Let $X subseteq Y$, where $Y$ is a set on which $d$ can be extended. Is it always true that for any Cauchy sequence ${x_n}$ in $X$ one can find $yin Y$ such that $d(x_n,y)rightarrow 0$?
In other words, is it true that every Cauchy sequence "converges" to something? (and if this something is in $X$ we say it is convergent?)
real-analysis convergence metric-spaces cauchy-sequences
asked Dec 28 '18 at 19:34
LeonardoLeonardo
3339
$endgroup$
1
$begingroup$
No. In fact, the property you are describing is called completeness. A metric space is $complete$ if every Cauchy sequence converges.
$endgroup$
– John Douma
Dec 28 '18 at 19:43
add a comment |
$begingroup$
A Cauchy sequence in a metric space $(X,d)$ is a sequence for which the distance between two terms can be made as small as we want, provided we look far enough in the sequence.
Let $X subseteq Y$, where $Y$ is a set on which $d$ can be extended. Is it always true that for any Cauchy sequence ${x_n}$ in $X$ one can find $yin Y$ such that $d(x_n,y)rightarrow 0$?
In other words, is it true that every Cauchy sequence "converges" to something? (and if this something is in $X$ we say it is convergent?)
real-analysis convergence metric-spaces cauchy-sequences
asked Dec 28 '18 at 19:34
LeonardoLeonardo
3339
$endgroup$
A Cauchy sequence in a metric space $(X,d)$ is a sequence for which the distance between two terms can be made as small as we want, provided we look far enough in the sequence.
Let $X subseteq Y$, where $Y$ is a set on which $d$ can be extended. Is it always true that for any Cauchy sequence ${x_n}$ in $X$ one can find $yin Y$ such that $d(x_n,y)rightarrow 0$?
In other words, is it true that every Cauchy sequence "converges" to something? (and if this something is in $X$ we say it is convergent?)
real-analysis convergence metric-spaces cauchy-sequences
real-analysis convergence metric-spaces cauchy-sequences
asked Dec 28 '18 at 19:34
LeonardoLeonardo
3339
asked Dec 28 '18 at 19:34
LeonardoLeonardo
3339
edited Dec 28 '18 at 19:59
Leonardo
asked Dec 28 '18 at 19:34
LeonardoLeonardo
3339
asked Dec 28 '18 at 19:34
LeonardoLeonardo
3339
asked Dec 28 '18 at 19:34
LeonardoLeonardo
3339
3339
1
$begingroup$
No. In fact, the property you are describing is called completeness. A metric space is $complete$ if every Cauchy sequence converges.
$endgroup$
– John Douma
Dec 28 '18 at 19:43
add a comment |
1
$begingroup$
No. In fact, the property you are describing is called completeness. A metric space is $complete$ if every Cauchy sequence converges.
$endgroup$
– John Douma
Dec 28 '18 at 19:43
1
1
$begingroup$
No. In fact, the property you are describing is called completeness. A metric space is $complete$ if every Cauchy sequence converges.
$endgroup$
– John Douma
Dec 28 '18 at 19:43
$begingroup$
No. In fact, the property you are describing is called completeness. A metric space is $complete$ if every Cauchy sequence converges.
$endgroup$
– John Douma
Dec 28 '18 at 19:43
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I think what you're looking for is the concept of the completion of the metric space $X$.
Here you construct $Y$ by adding to $X$ an "artificial" limit element for each Cauchy sequence that doesn't already have one in $X$ -- or rather, for each equivalence class of such Cauchy sequence where sequences $(x_n)$ and $(y_n)$ are related if $d(x_n,y_n)to 0$.
One main example of this is that $mathbb R$ is a metric completion of $mathbb Q$.
answered Dec 28 '18 at 19:58
Henning MakholmHenning Makholm
243k17312556
$endgroup$
$begingroup$
So my question could be rephrased into: does every metric space admit a completion?
$endgroup$
– Leonardo
Dec 28 '18 at 20:02
$begingroup$
@Leonardo: Yes it does. See e.g. Wikipedia for more details.
$endgroup$
– Henning Makholm
Dec 28 '18 at 20:18
add a comment |
$begingroup$
No, in order to show that a Cauchy Sequence converges, a more delicate approach would be needed and that's not always the case. If, though, one proves that for a space $(X,d) subseteq (Y,d)$ it is $d(x_n,y) to 0$ for a Cauchy Sequence ${x_n}_{n in mathbb N} in X$, that means that the space $X$ equipped with the metric $d$ is complete, thus Banach (if the metric $d$ is a norm of course, thus we are dealing with a complete normed space).
answered Dec 28 '18 at 19:49
RebellosRebellos
15.7k31250
$endgroup$
$begingroup$
Banach is terminology for complete normed spaces.
$endgroup$
– user370967
Dec 28 '18 at 19:57
$begingroup$
I didn't specify $y in X$. So what you're saying is that is if $exists y in Y $ such that $ d(x_n,y) $ vanishes then I could prove $y in X$?
$endgroup$
– Leonardo
Dec 28 '18 at 19:58
$begingroup$
@Leonardo I didn't say $y in X$.
$endgroup$
– Rebellos
Dec 28 '18 at 20:02
$begingroup$
@Rebellos But a complete metric space $(X,d)$ is one in which every Cauchy sequence in $X$ is convergent in X, so when you said that $(X,d)$ is complete if $d(x_n,y)$ vanishes you're implying $ y in X$, aren't you?
$endgroup$
– Leonardo
Dec 28 '18 at 20:07
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
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votes
$begingroup$
I think what you're looking for is the concept of the completion of the metric space $X$.
Here you construct $Y$ by adding to $X$ an "artificial" limit element for each Cauchy sequence that doesn't already have one in $X$ -- or rather, for each equivalence class of such Cauchy sequence where sequences $(x_n)$ and $(y_n)$ are related if $d(x_n,y_n)to 0$.
One main example of this is that $mathbb R$ is a metric completion of $mathbb Q$.
answered Dec 28 '18 at 19:58
Henning MakholmHenning Makholm
243k17312556
$endgroup$
$begingroup$
So my question could be rephrased into: does every metric space admit a completion?
$endgroup$
– Leonardo
Dec 28 '18 at 20:02
$begingroup$
@Leonardo: Yes it does. See e.g. Wikipedia for more details.
$endgroup$
– Henning Makholm
Dec 28 '18 at 20:18
add a comment |
$begingroup$
I think what you're looking for is the concept of the completion of the metric space $X$.
Here you construct $Y$ by adding to $X$ an "artificial" limit element for each Cauchy sequence that doesn't already have one in $X$ -- or rather, for each equivalence class of such Cauchy sequence where sequences $(x_n)$ and $(y_n)$ are related if $d(x_n,y_n)to 0$.
One main example of this is that $mathbb R$ is a metric completion of $mathbb Q$.
answered Dec 28 '18 at 19:58
Henning MakholmHenning Makholm
243k17312556
$endgroup$
$begingroup$
So my question could be rephrased into: does every metric space admit a completion?
$endgroup$
– Leonardo
Dec 28 '18 at 20:02
$begingroup$
@Leonardo: Yes it does. See e.g. Wikipedia for more details.
$endgroup$
– Henning Makholm
Dec 28 '18 at 20:18
add a comment |
$begingroup$
I think what you're looking for is the concept of the completion of the metric space $X$.
Here you construct $Y$ by adding to $X$ an "artificial" limit element for each Cauchy sequence that doesn't already have one in $X$ -- or rather, for each equivalence class of such Cauchy sequence where sequences $(x_n)$ and $(y_n)$ are related if $d(x_n,y_n)to 0$.
One main example of this is that $mathbb R$ is a metric completion of $mathbb Q$.
answered Dec 28 '18 at 19:58
Henning MakholmHenning Makholm
243k17312556
$endgroup$
I think what you're looking for is the concept of the completion of the metric space $X$.
Here you construct $Y$ by adding to $X$ an "artificial" limit element for each Cauchy sequence that doesn't already have one in $X$ -- or rather, for each equivalence class of such Cauchy sequence where sequences $(x_n)$ and $(y_n)$ are related if $d(x_n,y_n)to 0$.
One main example of this is that $mathbb R$ is a metric completion of $mathbb Q$.
answered Dec 28 '18 at 19:58
Henning MakholmHenning Makholm
243k17312556
answered Dec 28 '18 at 19:58
Henning MakholmHenning Makholm
243k17312556
answered Dec 28 '18 at 19:58
Henning MakholmHenning Makholm
243k17312556
answered Dec 28 '18 at 19:58
Henning MakholmHenning Makholm
243k17312556
243k17312556
$begingroup$
So my question could be rephrased into: does every metric space admit a completion?
$endgroup$
– Leonardo
Dec 28 '18 at 20:02
$begingroup$
@Leonardo: Yes it does. See e.g. Wikipedia for more details.
$endgroup$
– Henning Makholm
Dec 28 '18 at 20:18
add a comment |
$begingroup$
So my question could be rephrased into: does every metric space admit a completion?
$endgroup$
– Leonardo
Dec 28 '18 at 20:02
$begingroup$
@Leonardo: Yes it does. See e.g. Wikipedia for more details.
$endgroup$
– Henning Makholm
Dec 28 '18 at 20:18
$begingroup$
So my question could be rephrased into: does every metric space admit a completion?
$endgroup$
– Leonardo
Dec 28 '18 at 20:02
$begingroup$
So my question could be rephrased into: does every metric space admit a completion?
$endgroup$
– Leonardo
Dec 28 '18 at 20:02
$begingroup$
@Leonardo: Yes it does. See e.g. Wikipedia for more details.
$endgroup$
– Henning Makholm
Dec 28 '18 at 20:18
$begingroup$
@Leonardo: Yes it does. See e.g. Wikipedia for more details.
$endgroup$
– Henning Makholm
Dec 28 '18 at 20:18
add a comment |
$begingroup$
No, in order to show that a Cauchy Sequence converges, a more delicate approach would be needed and that's not always the case. If, though, one proves that for a space $(X,d) subseteq (Y,d)$ it is $d(x_n,y) to 0$ for a Cauchy Sequence ${x_n}_{n in mathbb N} in X$, that means that the space $X$ equipped with the metric $d$ is complete, thus Banach (if the metric $d$ is a norm of course, thus we are dealing with a complete normed space).
answered Dec 28 '18 at 19:49
RebellosRebellos
15.7k31250
$endgroup$
$begingroup$
Banach is terminology for complete normed spaces.
$endgroup$
– user370967
Dec 28 '18 at 19:57
$begingroup$
I didn't specify $y in X$. So what you're saying is that is if $exists y in Y $ such that $ d(x_n,y) $ vanishes then I could prove $y in X$?
$endgroup$
– Leonardo
Dec 28 '18 at 19:58
$begingroup$
@Leonardo I didn't say $y in X$.
$endgroup$
– Rebellos
Dec 28 '18 at 20:02
$begingroup$
@Rebellos But a complete metric space $(X,d)$ is one in which every Cauchy sequence in $X$ is convergent in X, so when you said that $(X,d)$ is complete if $d(x_n,y)$ vanishes you're implying $ y in X$, aren't you?
$endgroup$
– Leonardo
Dec 28 '18 at 20:07
add a comment |
$begingroup$
No, in order to show that a Cauchy Sequence converges, a more delicate approach would be needed and that's not always the case. If, though, one proves that for a space $(X,d) subseteq (Y,d)$ it is $d(x_n,y) to 0$ for a Cauchy Sequence ${x_n}_{n in mathbb N} in X$, that means that the space $X$ equipped with the metric $d$ is complete, thus Banach (if the metric $d$ is a norm of course, thus we are dealing with a complete normed space).
answered Dec 28 '18 at 19:49
RebellosRebellos
15.7k31250
$endgroup$
$begingroup$
Banach is terminology for complete normed spaces.
$endgroup$
– user370967
Dec 28 '18 at 19:57
$begingroup$
I didn't specify $y in X$. So what you're saying is that is if $exists y in Y $ such that $ d(x_n,y) $ vanishes then I could prove $y in X$?
$endgroup$
– Leonardo
Dec 28 '18 at 19:58
$begingroup$
@Leonardo I didn't say $y in X$.
$endgroup$
– Rebellos
Dec 28 '18 at 20:02
$begingroup$
@Rebellos But a complete metric space $(X,d)$ is one in which every Cauchy sequence in $X$ is convergent in X, so when you said that $(X,d)$ is complete if $d(x_n,y)$ vanishes you're implying $ y in X$, aren't you?
$endgroup$
– Leonardo
Dec 28 '18 at 20:07
add a comment |
$begingroup$
No, in order to show that a Cauchy Sequence converges, a more delicate approach would be needed and that's not always the case. If, though, one proves that for a space $(X,d) subseteq (Y,d)$ it is $d(x_n,y) to 0$ for a Cauchy Sequence ${x_n}_{n in mathbb N} in X$, that means that the space $X$ equipped with the metric $d$ is complete, thus Banach (if the metric $d$ is a norm of course, thus we are dealing with a complete normed space).
answered Dec 28 '18 at 19:49
RebellosRebellos
15.7k31250
$endgroup$
No, in order to show that a Cauchy Sequence converges, a more delicate approach would be needed and that's not always the case. If, though, one proves that for a space $(X,d) subseteq (Y,d)$ it is $d(x_n,y) to 0$ for a Cauchy Sequence ${x_n}_{n in mathbb N} in X$, that means that the space $X$ equipped with the metric $d$ is complete, thus Banach (if the metric $d$ is a norm of course, thus we are dealing with a complete normed space).
answered Dec 28 '18 at 19:49
RebellosRebellos
15.7k31250
edited Dec 28 '18 at 20:03
answered Dec 28 '18 at 19:49
RebellosRebellos
15.7k31250
answered Dec 28 '18 at 19:49
RebellosRebellos
15.7k31250
answered Dec 28 '18 at 19:49
RebellosRebellos
15.7k31250
15.7k31250
$begingroup$
Banach is terminology for complete normed spaces.
$endgroup$
– user370967
Dec 28 '18 at 19:57
$begingroup$
I didn't specify $y in X$. So what you're saying is that is if $exists y in Y $ such that $ d(x_n,y) $ vanishes then I could prove $y in X$?
$endgroup$
– Leonardo
Dec 28 '18 at 19:58
$begingroup$
@Leonardo I didn't say $y in X$.
$endgroup$
– Rebellos
Dec 28 '18 at 20:02
$begingroup$
@Rebellos But a complete metric space $(X,d)$ is one in which every Cauchy sequence in $X$ is convergent in X, so when you said that $(X,d)$ is complete if $d(x_n,y)$ vanishes you're implying $ y in X$, aren't you?
$endgroup$
– Leonardo
Dec 28 '18 at 20:07
add a comment |
$begingroup$
Banach is terminology for complete normed spaces.
$endgroup$
– user370967
Dec 28 '18 at 19:57
$begingroup$
I didn't specify $y in X$. So what you're saying is that is if $exists y in Y $ such that $ d(x_n,y) $ vanishes then I could prove $y in X$?
$endgroup$
– Leonardo
Dec 28 '18 at 19:58
$begingroup$
@Leonardo I didn't say $y in X$.
$endgroup$
– Rebellos
Dec 28 '18 at 20:02
$begingroup$
@Rebellos But a complete metric space $(X,d)$ is one in which every Cauchy sequence in $X$ is convergent in X, so when you said that $(X,d)$ is complete if $d(x_n,y)$ vanishes you're implying $ y in X$, aren't you?
$endgroup$
– Leonardo
Dec 28 '18 at 20:07
$begingroup$
Banach is terminology for complete normed spaces.
$endgroup$
– user370967
Dec 28 '18 at 19:57
$begingroup$
Banach is terminology for complete normed spaces.
$endgroup$
– user370967
Dec 28 '18 at 19:57
$begingroup$
I didn't specify $y in X$. So what you're saying is that is if $exists y in Y $ such that $ d(x_n,y) $ vanishes then I could prove $y in X$?
$endgroup$
– Leonardo
Dec 28 '18 at 19:58
$begingroup$
I didn't specify $y in X$. So what you're saying is that is if $exists y in Y $ such that $ d(x_n,y) $ vanishes then I could prove $y in X$?
$endgroup$
– Leonardo
Dec 28 '18 at 19:58
$begingroup$
@Leonardo I didn't say $y in X$.
$endgroup$
– Rebellos
Dec 28 '18 at 20:02
$begingroup$
@Leonardo I didn't say $y in X$.
$endgroup$
– Rebellos
Dec 28 '18 at 20:02
$begingroup$
@Rebellos But a complete metric space $(X,d)$ is one in which every Cauchy sequence in $X$ is convergent in X, so when you said that $(X,d)$ is complete if $d(x_n,y)$ vanishes you're implying $ y in X$, aren't you?
$endgroup$
– Leonardo
Dec 28 '18 at 20:07
$begingroup$
@Rebellos But a complete metric space $(X,d)$ is one in which every Cauchy sequence in $X$ is convergent in X, so when you said that $(X,d)$ is complete if $d(x_n,y)$ vanishes you're implying $ y in X$, aren't you?
$endgroup$
– Leonardo
Dec 28 '18 at 20:07
add a comment |
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$begingroup$
No. In fact, the property you are describing is called completeness. A metric space is $complete$ if every Cauchy sequence converges.
$endgroup$
– John Douma
Dec 28 '18 at 19:43