Cauchy sequences and convergent sequences












3














I am very confused with a little problem.



What is the difference between a Cauchy sequence and a convergent sequence?










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  • Every convergent sequence is a Cauchy sequence. The converse is not always true. If it is, we call such metric space complete.
    – APC89
    Dec 22 at 20:08












  • Can you Give me an example which is Cauchy but not convergent an any space?
    – Masaud Khan
    Dec 22 at 20:10










  • Consider the metric space $M := (0,1]subsettextbf{R}$. The sequence $a_{n} = 1/n$ is a Cauchy sequence, but it does not converge in $M$, since $lim a_{n} = 0$. Therefore $M$ is not complete.
    – APC89
    Dec 22 at 20:12












  • Thanks. I was reading my real analysis book and get confused with the problem. And now i realized that real R is complete..
    – Masaud Khan
    Dec 22 at 20:16










  • I think what makes it initially confusing is that we're already used to convergent sequences of real numbers before starting analysis, so it's instinctive to visualise Cauchy sequences as being exactly the same thing.
    – timtfj
    Dec 22 at 23:54
















3














I am very confused with a little problem.



What is the difference between a Cauchy sequence and a convergent sequence?










share|cite|improve this question









New contributor




Masaud Khan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • Every convergent sequence is a Cauchy sequence. The converse is not always true. If it is, we call such metric space complete.
    – APC89
    Dec 22 at 20:08












  • Can you Give me an example which is Cauchy but not convergent an any space?
    – Masaud Khan
    Dec 22 at 20:10










  • Consider the metric space $M := (0,1]subsettextbf{R}$. The sequence $a_{n} = 1/n$ is a Cauchy sequence, but it does not converge in $M$, since $lim a_{n} = 0$. Therefore $M$ is not complete.
    – APC89
    Dec 22 at 20:12












  • Thanks. I was reading my real analysis book and get confused with the problem. And now i realized that real R is complete..
    – Masaud Khan
    Dec 22 at 20:16










  • I think what makes it initially confusing is that we're already used to convergent sequences of real numbers before starting analysis, so it's instinctive to visualise Cauchy sequences as being exactly the same thing.
    – timtfj
    Dec 22 at 23:54














3












3








3







I am very confused with a little problem.



What is the difference between a Cauchy sequence and a convergent sequence?










share|cite|improve this question









New contributor




Masaud Khan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











I am very confused with a little problem.



What is the difference between a Cauchy sequence and a convergent sequence?







functional-analysis






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New contributor




Masaud Khan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









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Masaud Khan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




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edited Dec 22 at 20:21









dantopa

6,41132042




6,41132042






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asked Dec 22 at 20:06









Masaud Khan

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161




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New contributor





Masaud Khan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Check out our Code of Conduct.












  • Every convergent sequence is a Cauchy sequence. The converse is not always true. If it is, we call such metric space complete.
    – APC89
    Dec 22 at 20:08












  • Can you Give me an example which is Cauchy but not convergent an any space?
    – Masaud Khan
    Dec 22 at 20:10










  • Consider the metric space $M := (0,1]subsettextbf{R}$. The sequence $a_{n} = 1/n$ is a Cauchy sequence, but it does not converge in $M$, since $lim a_{n} = 0$. Therefore $M$ is not complete.
    – APC89
    Dec 22 at 20:12












  • Thanks. I was reading my real analysis book and get confused with the problem. And now i realized that real R is complete..
    – Masaud Khan
    Dec 22 at 20:16










  • I think what makes it initially confusing is that we're already used to convergent sequences of real numbers before starting analysis, so it's instinctive to visualise Cauchy sequences as being exactly the same thing.
    – timtfj
    Dec 22 at 23:54


















  • Every convergent sequence is a Cauchy sequence. The converse is not always true. If it is, we call such metric space complete.
    – APC89
    Dec 22 at 20:08












  • Can you Give me an example which is Cauchy but not convergent an any space?
    – Masaud Khan
    Dec 22 at 20:10










  • Consider the metric space $M := (0,1]subsettextbf{R}$. The sequence $a_{n} = 1/n$ is a Cauchy sequence, but it does not converge in $M$, since $lim a_{n} = 0$. Therefore $M$ is not complete.
    – APC89
    Dec 22 at 20:12












  • Thanks. I was reading my real analysis book and get confused with the problem. And now i realized that real R is complete..
    – Masaud Khan
    Dec 22 at 20:16










  • I think what makes it initially confusing is that we're already used to convergent sequences of real numbers before starting analysis, so it's instinctive to visualise Cauchy sequences as being exactly the same thing.
    – timtfj
    Dec 22 at 23:54
















Every convergent sequence is a Cauchy sequence. The converse is not always true. If it is, we call such metric space complete.
– APC89
Dec 22 at 20:08






Every convergent sequence is a Cauchy sequence. The converse is not always true. If it is, we call such metric space complete.
– APC89
Dec 22 at 20:08














Can you Give me an example which is Cauchy but not convergent an any space?
– Masaud Khan
Dec 22 at 20:10




Can you Give me an example which is Cauchy but not convergent an any space?
– Masaud Khan
Dec 22 at 20:10












Consider the metric space $M := (0,1]subsettextbf{R}$. The sequence $a_{n} = 1/n$ is a Cauchy sequence, but it does not converge in $M$, since $lim a_{n} = 0$. Therefore $M$ is not complete.
– APC89
Dec 22 at 20:12






Consider the metric space $M := (0,1]subsettextbf{R}$. The sequence $a_{n} = 1/n$ is a Cauchy sequence, but it does not converge in $M$, since $lim a_{n} = 0$. Therefore $M$ is not complete.
– APC89
Dec 22 at 20:12














Thanks. I was reading my real analysis book and get confused with the problem. And now i realized that real R is complete..
– Masaud Khan
Dec 22 at 20:16




Thanks. I was reading my real analysis book and get confused with the problem. And now i realized that real R is complete..
– Masaud Khan
Dec 22 at 20:16












I think what makes it initially confusing is that we're already used to convergent sequences of real numbers before starting analysis, so it's instinctive to visualise Cauchy sequences as being exactly the same thing.
– timtfj
Dec 22 at 23:54




I think what makes it initially confusing is that we're already used to convergent sequences of real numbers before starting analysis, so it's instinctive to visualise Cauchy sequences as being exactly the same thing.
– timtfj
Dec 22 at 23:54










3 Answers
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5














Every convergent sequence is Cauchy but not every Cauchy sequence is convergent depending on which space you are considering.



A Cauchy sequence is a sequence where the terms of the sequence get arbitrarily close to each other after a while.



A convergent sequence is a sequence where the terms get arbitrarily close to a specific point.



Formally a convergent sequence ${x_n}_{n}$ converging to $x$ satisfies:



$$forall varepsilon>0, exists N>0, n>NRightarrow |x_n-x|<varepsilon.$$



A Cauchy sequence ${x_n}_n$ satisfies:



$$forall varepsilon>0, exists N>0, n,m>NRightarrow |x_n-x_m|<varepsilon.$$



For real numbers with the euclidean metric the properties are equivalent but not for every metric space. Consider for instance in $mathbf{Q}$ the sequence 



$$x_1 = 3, x_2 = 3.1, x_3 = 3.14, x_4 = 3.141, x_5 = 3.1415 dotsc$$



whic gets arbitrarily close to the real number $pi$. Now $pinotin mathbf{Q}$ so the sequence does not converge in $mathbf{Q}$. It is however a Cauchy sequence






share|cite|improve this answer































    2














    You can easily find the definitions. Every convergent sequence is necessarily Cauchy but not every Cauchy sequence converges. The notions can be defined in any metric space. The notions are tied to the notion of completeness: A space is complete if, and only if, a sequence converges precisely when it is Cauchy. Also, a sequence is Cauchy if, and only if, it converges in the completion of the space.






    share|cite|improve this answer





























      0














      Let $(X,d)$ be a metric space (or a normed vector space, if you want. In this case we define $d(x,y) := ||x-y||$). We call $(x_n)$ convergent with limit $x$ iff $forall varepsilon>0;exists Ninmathbb{N};forall ngeq N: d(x_n,x) < varepsilon$ (or, intuitively speaking, if the distance between $x_n$ and $x$ will eventually approach zero as $n rightarrow infty$). We say $(x_n)$ is a cauchy sequence iff $forall varepsilon > 0;exists Ninmathbb{N};forall m,ngeq N: d(x_m,x_n) < varepsilon$ (intuitively, if the distance between two arbitrary sequence elements will eventually approach zero as $n rightarrow infty$). One can show that any convergent sequence is a cauchy sequence (you can try to prove this, it's a very easy exercise). If for any metric space or normed vector space $X$ the converse holds true (i.e. every cauchy sequence converges to some limit $x$), we call that space complete. We can show that any normed vector space with finite dimension is complete (though this requires some work, you could start with proving the Bolzano Weierstraß theorem, first for $mathbb{R}$, then for $mathbb{R}^n$ via induction and then for $mathbb{C}^n$ via $mathbb{C}^n cong mathbb{R}^{2n}$, then by showing that cauchy sequences are bounded and that every cauchy sequence with a convergent subsequence is in fact convergent you can show that $mathbb{R}^n$ and $mathbb{C}^n$ are complete, then you can generalize this result for all normed vector spaces of finite dimension because all vector spaces of the same finite dimension are isomorphic to $mathbb{R}^n$ or $mathbb{C}^n$ and all norms are equivalent on $mathbb{R}^n$ and $mathbb{C}^n$ (the equivalence of norms part is really difficult to prove and requires some additional work so you should skip that if you really want to prove everything else).






      share|cite|improve this answer





















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        3 Answers
        3






        active

        oldest

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        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        5














        Every convergent sequence is Cauchy but not every Cauchy sequence is convergent depending on which space you are considering.



        A Cauchy sequence is a sequence where the terms of the sequence get arbitrarily close to each other after a while.



        A convergent sequence is a sequence where the terms get arbitrarily close to a specific point.



        Formally a convergent sequence ${x_n}_{n}$ converging to $x$ satisfies:



        $$forall varepsilon>0, exists N>0, n>NRightarrow |x_n-x|<varepsilon.$$



        A Cauchy sequence ${x_n}_n$ satisfies:



        $$forall varepsilon>0, exists N>0, n,m>NRightarrow |x_n-x_m|<varepsilon.$$



        For real numbers with the euclidean metric the properties are equivalent but not for every metric space. Consider for instance in $mathbf{Q}$ the sequence 



        $$x_1 = 3, x_2 = 3.1, x_3 = 3.14, x_4 = 3.141, x_5 = 3.1415 dotsc$$



        whic gets arbitrarily close to the real number $pi$. Now $pinotin mathbf{Q}$ so the sequence does not converge in $mathbf{Q}$. It is however a Cauchy sequence






        share|cite|improve this answer




























          5














          Every convergent sequence is Cauchy but not every Cauchy sequence is convergent depending on which space you are considering.



          A Cauchy sequence is a sequence where the terms of the sequence get arbitrarily close to each other after a while.



          A convergent sequence is a sequence where the terms get arbitrarily close to a specific point.



          Formally a convergent sequence ${x_n}_{n}$ converging to $x$ satisfies:



          $$forall varepsilon>0, exists N>0, n>NRightarrow |x_n-x|<varepsilon.$$



          A Cauchy sequence ${x_n}_n$ satisfies:



          $$forall varepsilon>0, exists N>0, n,m>NRightarrow |x_n-x_m|<varepsilon.$$



          For real numbers with the euclidean metric the properties are equivalent but not for every metric space. Consider for instance in $mathbf{Q}$ the sequence 



          $$x_1 = 3, x_2 = 3.1, x_3 = 3.14, x_4 = 3.141, x_5 = 3.1415 dotsc$$



          whic gets arbitrarily close to the real number $pi$. Now $pinotin mathbf{Q}$ so the sequence does not converge in $mathbf{Q}$. It is however a Cauchy sequence






          share|cite|improve this answer


























            5












            5








            5






            Every convergent sequence is Cauchy but not every Cauchy sequence is convergent depending on which space you are considering.



            A Cauchy sequence is a sequence where the terms of the sequence get arbitrarily close to each other after a while.



            A convergent sequence is a sequence where the terms get arbitrarily close to a specific point.



            Formally a convergent sequence ${x_n}_{n}$ converging to $x$ satisfies:



            $$forall varepsilon>0, exists N>0, n>NRightarrow |x_n-x|<varepsilon.$$



            A Cauchy sequence ${x_n}_n$ satisfies:



            $$forall varepsilon>0, exists N>0, n,m>NRightarrow |x_n-x_m|<varepsilon.$$



            For real numbers with the euclidean metric the properties are equivalent but not for every metric space. Consider for instance in $mathbf{Q}$ the sequence 



            $$x_1 = 3, x_2 = 3.1, x_3 = 3.14, x_4 = 3.141, x_5 = 3.1415 dotsc$$



            whic gets arbitrarily close to the real number $pi$. Now $pinotin mathbf{Q}$ so the sequence does not converge in $mathbf{Q}$. It is however a Cauchy sequence






            share|cite|improve this answer














            Every convergent sequence is Cauchy but not every Cauchy sequence is convergent depending on which space you are considering.



            A Cauchy sequence is a sequence where the terms of the sequence get arbitrarily close to each other after a while.



            A convergent sequence is a sequence where the terms get arbitrarily close to a specific point.



            Formally a convergent sequence ${x_n}_{n}$ converging to $x$ satisfies:



            $$forall varepsilon>0, exists N>0, n>NRightarrow |x_n-x|<varepsilon.$$



            A Cauchy sequence ${x_n}_n$ satisfies:



            $$forall varepsilon>0, exists N>0, n,m>NRightarrow |x_n-x_m|<varepsilon.$$



            For real numbers with the euclidean metric the properties are equivalent but not for every metric space. Consider for instance in $mathbf{Q}$ the sequence 



            $$x_1 = 3, x_2 = 3.1, x_3 = 3.14, x_4 = 3.141, x_5 = 3.1415 dotsc$$



            whic gets arbitrarily close to the real number $pi$. Now $pinotin mathbf{Q}$ so the sequence does not converge in $mathbf{Q}$. It is however a Cauchy sequence







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 22 at 20:17

























            answered Dec 22 at 20:12









            Olof Rubin

            980315




            980315























                2














                You can easily find the definitions. Every convergent sequence is necessarily Cauchy but not every Cauchy sequence converges. The notions can be defined in any metric space. The notions are tied to the notion of completeness: A space is complete if, and only if, a sequence converges precisely when it is Cauchy. Also, a sequence is Cauchy if, and only if, it converges in the completion of the space.






                share|cite|improve this answer


























                  2














                  You can easily find the definitions. Every convergent sequence is necessarily Cauchy but not every Cauchy sequence converges. The notions can be defined in any metric space. The notions are tied to the notion of completeness: A space is complete if, and only if, a sequence converges precisely when it is Cauchy. Also, a sequence is Cauchy if, and only if, it converges in the completion of the space.






                  share|cite|improve this answer
























                    2












                    2








                    2






                    You can easily find the definitions. Every convergent sequence is necessarily Cauchy but not every Cauchy sequence converges. The notions can be defined in any metric space. The notions are tied to the notion of completeness: A space is complete if, and only if, a sequence converges precisely when it is Cauchy. Also, a sequence is Cauchy if, and only if, it converges in the completion of the space.






                    share|cite|improve this answer












                    You can easily find the definitions. Every convergent sequence is necessarily Cauchy but not every Cauchy sequence converges. The notions can be defined in any metric space. The notions are tied to the notion of completeness: A space is complete if, and only if, a sequence converges precisely when it is Cauchy. Also, a sequence is Cauchy if, and only if, it converges in the completion of the space.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 22 at 20:09









                    Ittay Weiss

                    63.4k6101183




                    63.4k6101183























                        0














                        Let $(X,d)$ be a metric space (or a normed vector space, if you want. In this case we define $d(x,y) := ||x-y||$). We call $(x_n)$ convergent with limit $x$ iff $forall varepsilon>0;exists Ninmathbb{N};forall ngeq N: d(x_n,x) < varepsilon$ (or, intuitively speaking, if the distance between $x_n$ and $x$ will eventually approach zero as $n rightarrow infty$). We say $(x_n)$ is a cauchy sequence iff $forall varepsilon > 0;exists Ninmathbb{N};forall m,ngeq N: d(x_m,x_n) < varepsilon$ (intuitively, if the distance between two arbitrary sequence elements will eventually approach zero as $n rightarrow infty$). One can show that any convergent sequence is a cauchy sequence (you can try to prove this, it's a very easy exercise). If for any metric space or normed vector space $X$ the converse holds true (i.e. every cauchy sequence converges to some limit $x$), we call that space complete. We can show that any normed vector space with finite dimension is complete (though this requires some work, you could start with proving the Bolzano Weierstraß theorem, first for $mathbb{R}$, then for $mathbb{R}^n$ via induction and then for $mathbb{C}^n$ via $mathbb{C}^n cong mathbb{R}^{2n}$, then by showing that cauchy sequences are bounded and that every cauchy sequence with a convergent subsequence is in fact convergent you can show that $mathbb{R}^n$ and $mathbb{C}^n$ are complete, then you can generalize this result for all normed vector spaces of finite dimension because all vector spaces of the same finite dimension are isomorphic to $mathbb{R}^n$ or $mathbb{C}^n$ and all norms are equivalent on $mathbb{R}^n$ and $mathbb{C}^n$ (the equivalence of norms part is really difficult to prove and requires some additional work so you should skip that if you really want to prove everything else).






                        share|cite|improve this answer


























                          0














                          Let $(X,d)$ be a metric space (or a normed vector space, if you want. In this case we define $d(x,y) := ||x-y||$). We call $(x_n)$ convergent with limit $x$ iff $forall varepsilon>0;exists Ninmathbb{N};forall ngeq N: d(x_n,x) < varepsilon$ (or, intuitively speaking, if the distance between $x_n$ and $x$ will eventually approach zero as $n rightarrow infty$). We say $(x_n)$ is a cauchy sequence iff $forall varepsilon > 0;exists Ninmathbb{N};forall m,ngeq N: d(x_m,x_n) < varepsilon$ (intuitively, if the distance between two arbitrary sequence elements will eventually approach zero as $n rightarrow infty$). One can show that any convergent sequence is a cauchy sequence (you can try to prove this, it's a very easy exercise). If for any metric space or normed vector space $X$ the converse holds true (i.e. every cauchy sequence converges to some limit $x$), we call that space complete. We can show that any normed vector space with finite dimension is complete (though this requires some work, you could start with proving the Bolzano Weierstraß theorem, first for $mathbb{R}$, then for $mathbb{R}^n$ via induction and then for $mathbb{C}^n$ via $mathbb{C}^n cong mathbb{R}^{2n}$, then by showing that cauchy sequences are bounded and that every cauchy sequence with a convergent subsequence is in fact convergent you can show that $mathbb{R}^n$ and $mathbb{C}^n$ are complete, then you can generalize this result for all normed vector spaces of finite dimension because all vector spaces of the same finite dimension are isomorphic to $mathbb{R}^n$ or $mathbb{C}^n$ and all norms are equivalent on $mathbb{R}^n$ and $mathbb{C}^n$ (the equivalence of norms part is really difficult to prove and requires some additional work so you should skip that if you really want to prove everything else).






                          share|cite|improve this answer
























                            0












                            0








                            0






                            Let $(X,d)$ be a metric space (or a normed vector space, if you want. In this case we define $d(x,y) := ||x-y||$). We call $(x_n)$ convergent with limit $x$ iff $forall varepsilon>0;exists Ninmathbb{N};forall ngeq N: d(x_n,x) < varepsilon$ (or, intuitively speaking, if the distance between $x_n$ and $x$ will eventually approach zero as $n rightarrow infty$). We say $(x_n)$ is a cauchy sequence iff $forall varepsilon > 0;exists Ninmathbb{N};forall m,ngeq N: d(x_m,x_n) < varepsilon$ (intuitively, if the distance between two arbitrary sequence elements will eventually approach zero as $n rightarrow infty$). One can show that any convergent sequence is a cauchy sequence (you can try to prove this, it's a very easy exercise). If for any metric space or normed vector space $X$ the converse holds true (i.e. every cauchy sequence converges to some limit $x$), we call that space complete. We can show that any normed vector space with finite dimension is complete (though this requires some work, you could start with proving the Bolzano Weierstraß theorem, first for $mathbb{R}$, then for $mathbb{R}^n$ via induction and then for $mathbb{C}^n$ via $mathbb{C}^n cong mathbb{R}^{2n}$, then by showing that cauchy sequences are bounded and that every cauchy sequence with a convergent subsequence is in fact convergent you can show that $mathbb{R}^n$ and $mathbb{C}^n$ are complete, then you can generalize this result for all normed vector spaces of finite dimension because all vector spaces of the same finite dimension are isomorphic to $mathbb{R}^n$ or $mathbb{C}^n$ and all norms are equivalent on $mathbb{R}^n$ and $mathbb{C}^n$ (the equivalence of norms part is really difficult to prove and requires some additional work so you should skip that if you really want to prove everything else).






                            share|cite|improve this answer












                            Let $(X,d)$ be a metric space (or a normed vector space, if you want. In this case we define $d(x,y) := ||x-y||$). We call $(x_n)$ convergent with limit $x$ iff $forall varepsilon>0;exists Ninmathbb{N};forall ngeq N: d(x_n,x) < varepsilon$ (or, intuitively speaking, if the distance between $x_n$ and $x$ will eventually approach zero as $n rightarrow infty$). We say $(x_n)$ is a cauchy sequence iff $forall varepsilon > 0;exists Ninmathbb{N};forall m,ngeq N: d(x_m,x_n) < varepsilon$ (intuitively, if the distance between two arbitrary sequence elements will eventually approach zero as $n rightarrow infty$). One can show that any convergent sequence is a cauchy sequence (you can try to prove this, it's a very easy exercise). If for any metric space or normed vector space $X$ the converse holds true (i.e. every cauchy sequence converges to some limit $x$), we call that space complete. We can show that any normed vector space with finite dimension is complete (though this requires some work, you could start with proving the Bolzano Weierstraß theorem, first for $mathbb{R}$, then for $mathbb{R}^n$ via induction and then for $mathbb{C}^n$ via $mathbb{C}^n cong mathbb{R}^{2n}$, then by showing that cauchy sequences are bounded and that every cauchy sequence with a convergent subsequence is in fact convergent you can show that $mathbb{R}^n$ and $mathbb{C}^n$ are complete, then you can generalize this result for all normed vector spaces of finite dimension because all vector spaces of the same finite dimension are isomorphic to $mathbb{R}^n$ or $mathbb{C}^n$ and all norms are equivalent on $mathbb{R}^n$ and $mathbb{C}^n$ (the equivalence of norms part is really difficult to prove and requires some additional work so you should skip that if you really want to prove everything else).







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 22 at 23:09









                            Nicolas

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