Cauchy Sequence in subset of a metric space











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True or False:
If E is a subset of a metric space X, then any sequence of points of E that is a Cauchy sequence in X is a Cauchy sequence in E.



I'm having difficulty understanding this language. What does it mean for a sequence to be Cauchy "in" X?










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    True or False:
    If E is a subset of a metric space X, then any sequence of points of E that is a Cauchy sequence in X is a Cauchy sequence in E.



    I'm having difficulty understanding this language. What does it mean for a sequence to be Cauchy "in" X?










    share|cite|improve this question









    New contributor




    51n84d is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






















      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      True or False:
      If E is a subset of a metric space X, then any sequence of points of E that is a Cauchy sequence in X is a Cauchy sequence in E.



      I'm having difficulty understanding this language. What does it mean for a sequence to be Cauchy "in" X?










      share|cite|improve this question









      New contributor




      51n84d is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      True or False:
      If E is a subset of a metric space X, then any sequence of points of E that is a Cauchy sequence in X is a Cauchy sequence in E.



      I'm having difficulty understanding this language. What does it mean for a sequence to be Cauchy "in" X?







      real-analysis sequences-and-series metric-spaces cauchy-sequences






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      edited 16 hours ago









      Henno Brandsma

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      asked 17 hours ago









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          3 Answers
          3






          active

          oldest

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          up vote
          0
          down vote



          accepted










          Since $X$ is a metric space there is a distance $d:Xtimes Xto [0,+infty)$. Then ${x_n}_nsubset X$ is a Cauchy sequence in $X$ (or with respect to this distance $d$) if $$forallepsilon>0,; exists N;:; forall n,m>N, d(x_n,x_m)<epsilon.$$
          If $Esubset X$ and ${x_n}_nsubset E$ then we may say that if ${x_n}_n$ is a Cauchy sequence in $X$ then it also a Cauchy sequence in $E$. $E$ and $X$ share the same distance if nothing else has been specified.






          share|cite|improve this answer























          • Does E being a subset of X imply that E and X share the same distance function?
            – 51n84d
            17 hours ago






          • 1




            Yes, $E$ and $X$ share the same distance.
            – Robert Z
            17 hours ago










          • 51n84d.Given a metric d on X, and $E subset X$, $d_E(x,y) : =d(x,y)$, $x,y in E subset X$.
            – Peter Szilas
            16 hours ago






          • 1




            To the proposer: Strictly speaking, a metric space is a pair $(X,d)$ where $d:Xtimes Xto Bbb R$ is a metric on a set $X$. When speaking of a Cauchy or convergent sequence in some $Esubset X$, it is usually assumed, without saying so, that we mean in the metric space $(E,d|_{Etimes E})$ where $d|_{Etimes E}$ is $d$ restricted to the domain $Etimes E$.
            – DanielWainfleet
            15 hours ago




















          up vote
          1
          down vote













          The definition of Cauchy sequence is$$forall epsilon>0 exists Nin mathbb Ntext{ such that } d(a_n,a_m)<epsilon forall n,m>N$$
          Now, the existence of this N is dependent on the metric space(similar to the fact that limit points of a set may or may not be in a particular metric space).



          Coming back to the question, yes, the given statement is true.



          $because$We know that the sequence is in E, therefore, all of the terms of the sequence is in E. Hence, all the terms after the stage N are also in E and therefore the property satisfies.



          Hope it helps:)






          share|cite|improve this answer






























            up vote
            0
            down vote













            The same sequence may be Cauchy in one metric space but not Cauchy in another. For example, let $M =mathbb{Q}$ with the normal distance function, $d(x, y) = |x-y|$, and let $N = mathbb{Q}$ with $d(x, y) = 1$ if $x not = y$ and $d(x, x) = 0$ (this is called the discrete metric).



            Now notice that $M$ and $N$ have the same set, but they are different metric spaces since a metric space is the set together with the distance function.



            Now consider $x_n = dfrac 1n$. In $M$, this sequence is convergent and therefore Cauchy (the terms bunch up near $0$ and thus get closer and closer to each other).



            Now since the sequence never repeats terms, in $N$, every member of the sequence is at distance $1$ from every other member, so the terms do not get closer and closer.



            Thus $x_n$ is Cauchy in $M$ but not Cauchy in $N$.



            Now your exercise is one of those "so easy that's it's hard". There isn't really anything to do to solve it, but you do have to understand exactly what is going on to write a correct answer.






            share|cite|improve this answer





















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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes








              up vote
              0
              down vote



              accepted










              Since $X$ is a metric space there is a distance $d:Xtimes Xto [0,+infty)$. Then ${x_n}_nsubset X$ is a Cauchy sequence in $X$ (or with respect to this distance $d$) if $$forallepsilon>0,; exists N;:; forall n,m>N, d(x_n,x_m)<epsilon.$$
              If $Esubset X$ and ${x_n}_nsubset E$ then we may say that if ${x_n}_n$ is a Cauchy sequence in $X$ then it also a Cauchy sequence in $E$. $E$ and $X$ share the same distance if nothing else has been specified.






              share|cite|improve this answer























              • Does E being a subset of X imply that E and X share the same distance function?
                – 51n84d
                17 hours ago






              • 1




                Yes, $E$ and $X$ share the same distance.
                – Robert Z
                17 hours ago










              • 51n84d.Given a metric d on X, and $E subset X$, $d_E(x,y) : =d(x,y)$, $x,y in E subset X$.
                – Peter Szilas
                16 hours ago






              • 1




                To the proposer: Strictly speaking, a metric space is a pair $(X,d)$ where $d:Xtimes Xto Bbb R$ is a metric on a set $X$. When speaking of a Cauchy or convergent sequence in some $Esubset X$, it is usually assumed, without saying so, that we mean in the metric space $(E,d|_{Etimes E})$ where $d|_{Etimes E}$ is $d$ restricted to the domain $Etimes E$.
                – DanielWainfleet
                15 hours ago

















              up vote
              0
              down vote



              accepted










              Since $X$ is a metric space there is a distance $d:Xtimes Xto [0,+infty)$. Then ${x_n}_nsubset X$ is a Cauchy sequence in $X$ (or with respect to this distance $d$) if $$forallepsilon>0,; exists N;:; forall n,m>N, d(x_n,x_m)<epsilon.$$
              If $Esubset X$ and ${x_n}_nsubset E$ then we may say that if ${x_n}_n$ is a Cauchy sequence in $X$ then it also a Cauchy sequence in $E$. $E$ and $X$ share the same distance if nothing else has been specified.






              share|cite|improve this answer























              • Does E being a subset of X imply that E and X share the same distance function?
                – 51n84d
                17 hours ago






              • 1




                Yes, $E$ and $X$ share the same distance.
                – Robert Z
                17 hours ago










              • 51n84d.Given a metric d on X, and $E subset X$, $d_E(x,y) : =d(x,y)$, $x,y in E subset X$.
                – Peter Szilas
                16 hours ago






              • 1




                To the proposer: Strictly speaking, a metric space is a pair $(X,d)$ where $d:Xtimes Xto Bbb R$ is a metric on a set $X$. When speaking of a Cauchy or convergent sequence in some $Esubset X$, it is usually assumed, without saying so, that we mean in the metric space $(E,d|_{Etimes E})$ where $d|_{Etimes E}$ is $d$ restricted to the domain $Etimes E$.
                – DanielWainfleet
                15 hours ago















              up vote
              0
              down vote



              accepted







              up vote
              0
              down vote



              accepted






              Since $X$ is a metric space there is a distance $d:Xtimes Xto [0,+infty)$. Then ${x_n}_nsubset X$ is a Cauchy sequence in $X$ (or with respect to this distance $d$) if $$forallepsilon>0,; exists N;:; forall n,m>N, d(x_n,x_m)<epsilon.$$
              If $Esubset X$ and ${x_n}_nsubset E$ then we may say that if ${x_n}_n$ is a Cauchy sequence in $X$ then it also a Cauchy sequence in $E$. $E$ and $X$ share the same distance if nothing else has been specified.






              share|cite|improve this answer














              Since $X$ is a metric space there is a distance $d:Xtimes Xto [0,+infty)$. Then ${x_n}_nsubset X$ is a Cauchy sequence in $X$ (or with respect to this distance $d$) if $$forallepsilon>0,; exists N;:; forall n,m>N, d(x_n,x_m)<epsilon.$$
              If $Esubset X$ and ${x_n}_nsubset E$ then we may say that if ${x_n}_n$ is a Cauchy sequence in $X$ then it also a Cauchy sequence in $E$. $E$ and $X$ share the same distance if nothing else has been specified.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited 17 hours ago

























              answered 17 hours ago









              Robert Z

              89.1k1056128




              89.1k1056128












              • Does E being a subset of X imply that E and X share the same distance function?
                – 51n84d
                17 hours ago






              • 1




                Yes, $E$ and $X$ share the same distance.
                – Robert Z
                17 hours ago










              • 51n84d.Given a metric d on X, and $E subset X$, $d_E(x,y) : =d(x,y)$, $x,y in E subset X$.
                – Peter Szilas
                16 hours ago






              • 1




                To the proposer: Strictly speaking, a metric space is a pair $(X,d)$ where $d:Xtimes Xto Bbb R$ is a metric on a set $X$. When speaking of a Cauchy or convergent sequence in some $Esubset X$, it is usually assumed, without saying so, that we mean in the metric space $(E,d|_{Etimes E})$ where $d|_{Etimes E}$ is $d$ restricted to the domain $Etimes E$.
                – DanielWainfleet
                15 hours ago




















              • Does E being a subset of X imply that E and X share the same distance function?
                – 51n84d
                17 hours ago






              • 1




                Yes, $E$ and $X$ share the same distance.
                – Robert Z
                17 hours ago










              • 51n84d.Given a metric d on X, and $E subset X$, $d_E(x,y) : =d(x,y)$, $x,y in E subset X$.
                – Peter Szilas
                16 hours ago






              • 1




                To the proposer: Strictly speaking, a metric space is a pair $(X,d)$ where $d:Xtimes Xto Bbb R$ is a metric on a set $X$. When speaking of a Cauchy or convergent sequence in some $Esubset X$, it is usually assumed, without saying so, that we mean in the metric space $(E,d|_{Etimes E})$ where $d|_{Etimes E}$ is $d$ restricted to the domain $Etimes E$.
                – DanielWainfleet
                15 hours ago


















              Does E being a subset of X imply that E and X share the same distance function?
              – 51n84d
              17 hours ago




              Does E being a subset of X imply that E and X share the same distance function?
              – 51n84d
              17 hours ago




              1




              1




              Yes, $E$ and $X$ share the same distance.
              – Robert Z
              17 hours ago




              Yes, $E$ and $X$ share the same distance.
              – Robert Z
              17 hours ago












              51n84d.Given a metric d on X, and $E subset X$, $d_E(x,y) : =d(x,y)$, $x,y in E subset X$.
              – Peter Szilas
              16 hours ago




              51n84d.Given a metric d on X, and $E subset X$, $d_E(x,y) : =d(x,y)$, $x,y in E subset X$.
              – Peter Szilas
              16 hours ago




              1




              1




              To the proposer: Strictly speaking, a metric space is a pair $(X,d)$ where $d:Xtimes Xto Bbb R$ is a metric on a set $X$. When speaking of a Cauchy or convergent sequence in some $Esubset X$, it is usually assumed, without saying so, that we mean in the metric space $(E,d|_{Etimes E})$ where $d|_{Etimes E}$ is $d$ restricted to the domain $Etimes E$.
              – DanielWainfleet
              15 hours ago






              To the proposer: Strictly speaking, a metric space is a pair $(X,d)$ where $d:Xtimes Xto Bbb R$ is a metric on a set $X$. When speaking of a Cauchy or convergent sequence in some $Esubset X$, it is usually assumed, without saying so, that we mean in the metric space $(E,d|_{Etimes E})$ where $d|_{Etimes E}$ is $d$ restricted to the domain $Etimes E$.
              – DanielWainfleet
              15 hours ago












              up vote
              1
              down vote













              The definition of Cauchy sequence is$$forall epsilon>0 exists Nin mathbb Ntext{ such that } d(a_n,a_m)<epsilon forall n,m>N$$
              Now, the existence of this N is dependent on the metric space(similar to the fact that limit points of a set may or may not be in a particular metric space).



              Coming back to the question, yes, the given statement is true.



              $because$We know that the sequence is in E, therefore, all of the terms of the sequence is in E. Hence, all the terms after the stage N are also in E and therefore the property satisfies.



              Hope it helps:)






              share|cite|improve this answer



























                up vote
                1
                down vote













                The definition of Cauchy sequence is$$forall epsilon>0 exists Nin mathbb Ntext{ such that } d(a_n,a_m)<epsilon forall n,m>N$$
                Now, the existence of this N is dependent on the metric space(similar to the fact that limit points of a set may or may not be in a particular metric space).



                Coming back to the question, yes, the given statement is true.



                $because$We know that the sequence is in E, therefore, all of the terms of the sequence is in E. Hence, all the terms after the stage N are also in E and therefore the property satisfies.



                Hope it helps:)






                share|cite|improve this answer

























                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  The definition of Cauchy sequence is$$forall epsilon>0 exists Nin mathbb Ntext{ such that } d(a_n,a_m)<epsilon forall n,m>N$$
                  Now, the existence of this N is dependent on the metric space(similar to the fact that limit points of a set may or may not be in a particular metric space).



                  Coming back to the question, yes, the given statement is true.



                  $because$We know that the sequence is in E, therefore, all of the terms of the sequence is in E. Hence, all the terms after the stage N are also in E and therefore the property satisfies.



                  Hope it helps:)






                  share|cite|improve this answer














                  The definition of Cauchy sequence is$$forall epsilon>0 exists Nin mathbb Ntext{ such that } d(a_n,a_m)<epsilon forall n,m>N$$
                  Now, the existence of this N is dependent on the metric space(similar to the fact that limit points of a set may or may not be in a particular metric space).



                  Coming back to the question, yes, the given statement is true.



                  $because$We know that the sequence is in E, therefore, all of the terms of the sequence is in E. Hence, all the terms after the stage N are also in E and therefore the property satisfies.



                  Hope it helps:)







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 16 hours ago









                  Henno Brandsma

                  100k344107




                  100k344107










                  answered 17 hours ago









                  Crazy for maths

                  3305




                  3305






















                      up vote
                      0
                      down vote













                      The same sequence may be Cauchy in one metric space but not Cauchy in another. For example, let $M =mathbb{Q}$ with the normal distance function, $d(x, y) = |x-y|$, and let $N = mathbb{Q}$ with $d(x, y) = 1$ if $x not = y$ and $d(x, x) = 0$ (this is called the discrete metric).



                      Now notice that $M$ and $N$ have the same set, but they are different metric spaces since a metric space is the set together with the distance function.



                      Now consider $x_n = dfrac 1n$. In $M$, this sequence is convergent and therefore Cauchy (the terms bunch up near $0$ and thus get closer and closer to each other).



                      Now since the sequence never repeats terms, in $N$, every member of the sequence is at distance $1$ from every other member, so the terms do not get closer and closer.



                      Thus $x_n$ is Cauchy in $M$ but not Cauchy in $N$.



                      Now your exercise is one of those "so easy that's it's hard". There isn't really anything to do to solve it, but you do have to understand exactly what is going on to write a correct answer.






                      share|cite|improve this answer

























                        up vote
                        0
                        down vote













                        The same sequence may be Cauchy in one metric space but not Cauchy in another. For example, let $M =mathbb{Q}$ with the normal distance function, $d(x, y) = |x-y|$, and let $N = mathbb{Q}$ with $d(x, y) = 1$ if $x not = y$ and $d(x, x) = 0$ (this is called the discrete metric).



                        Now notice that $M$ and $N$ have the same set, but they are different metric spaces since a metric space is the set together with the distance function.



                        Now consider $x_n = dfrac 1n$. In $M$, this sequence is convergent and therefore Cauchy (the terms bunch up near $0$ and thus get closer and closer to each other).



                        Now since the sequence never repeats terms, in $N$, every member of the sequence is at distance $1$ from every other member, so the terms do not get closer and closer.



                        Thus $x_n$ is Cauchy in $M$ but not Cauchy in $N$.



                        Now your exercise is one of those "so easy that's it's hard". There isn't really anything to do to solve it, but you do have to understand exactly what is going on to write a correct answer.






                        share|cite|improve this answer























                          up vote
                          0
                          down vote










                          up vote
                          0
                          down vote









                          The same sequence may be Cauchy in one metric space but not Cauchy in another. For example, let $M =mathbb{Q}$ with the normal distance function, $d(x, y) = |x-y|$, and let $N = mathbb{Q}$ with $d(x, y) = 1$ if $x not = y$ and $d(x, x) = 0$ (this is called the discrete metric).



                          Now notice that $M$ and $N$ have the same set, but they are different metric spaces since a metric space is the set together with the distance function.



                          Now consider $x_n = dfrac 1n$. In $M$, this sequence is convergent and therefore Cauchy (the terms bunch up near $0$ and thus get closer and closer to each other).



                          Now since the sequence never repeats terms, in $N$, every member of the sequence is at distance $1$ from every other member, so the terms do not get closer and closer.



                          Thus $x_n$ is Cauchy in $M$ but not Cauchy in $N$.



                          Now your exercise is one of those "so easy that's it's hard". There isn't really anything to do to solve it, but you do have to understand exactly what is going on to write a correct answer.






                          share|cite|improve this answer












                          The same sequence may be Cauchy in one metric space but not Cauchy in another. For example, let $M =mathbb{Q}$ with the normal distance function, $d(x, y) = |x-y|$, and let $N = mathbb{Q}$ with $d(x, y) = 1$ if $x not = y$ and $d(x, x) = 0$ (this is called the discrete metric).



                          Now notice that $M$ and $N$ have the same set, but they are different metric spaces since a metric space is the set together with the distance function.



                          Now consider $x_n = dfrac 1n$. In $M$, this sequence is convergent and therefore Cauchy (the terms bunch up near $0$ and thus get closer and closer to each other).



                          Now since the sequence never repeats terms, in $N$, every member of the sequence is at distance $1$ from every other member, so the terms do not get closer and closer.



                          Thus $x_n$ is Cauchy in $M$ but not Cauchy in $N$.



                          Now your exercise is one of those "so easy that's it's hard". There isn't really anything to do to solve it, but you do have to understand exactly what is going on to write a correct answer.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 16 hours ago









                          Ovi

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