Meaning of being a Cauchy sequence












1












$begingroup$


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







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$endgroup$








  • 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
















1












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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














1












1








1





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 28 '18 at 19:59







Leonardo

















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














  • 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










2 Answers
2






active

oldest

votes


















1












$begingroup$

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 $mathbb Q$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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



















1












$begingroup$

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).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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












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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

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 $mathbb Q$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















1












$begingroup$

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 $mathbb Q$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














1












1








1





$begingroup$

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 $mathbb Q$.






share|cite|improve this answer









$endgroup$



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 $mathbb Q$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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











1












$begingroup$

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).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















1












$begingroup$

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).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














1












1








1





$begingroup$

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).






share|cite|improve this answer











$endgroup$



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).







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 28 '18 at 20:03

























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


















  • $begingroup$
    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


















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