Show $0$ is the only cluster point of the set $A = $$frac1n : ninmathbb N$

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Need a check of working, I'm new to cluster points.



Define $a_n = frac1n$ for all $ninmathbb N$. Clearly $A =$ $frac1n:ninmathbb N$ $=$ $(a_n)$. Now $a_n$ converges to $0$, thus every subsequence of $a_n$ also converges to $0$. Hence $0$ is the only cluster point of the set A.



Does that suffice, or is there something to add?










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  • 1




    Whether or not your argument is correct depends on what you know about cluster points. What is the definition? Is the definition given in terms of sequences?
    – user99914
    Sep 10 at 1:30







  • 1




    The definition I've become familiar with; Let $A$ be a subset of $mathbb R$ and let $x_0 in mathbb R$, $x_0$ is a cluster point pf $A$ if for all $epsilon > 0$ the set $(x_0-epsilon,x_0+epsilon) cap Asetminus$$x_0$ $ne$ the empty set.
    – Florian Suess
    Sep 10 at 1:34







  • 1




    I've also come familiar with the synonymity of $x_0 $ is a cluster point AND there exists a sequence $(a_n)_ninmathbb N$ in $Asetminus$$x_0$ such that $lim_nrightarrowinftya_n = x_0$
    – Florian Suess
    Sep 10 at 1:36







  • 1




    Moreover I use the fact that if a sequence converges to say $L$, then every subsequence (of that sequence) must also converge to $L$
    – Florian Suess
    Sep 10 at 1:37










  • How are you defining "cluster point"? Typical definition: a point $x$ is a cluster point of $A$ if every open neighborhood of $x$ contains a point of $A$ other than $x$ itself, So, the fact that a sequence of points in $A$ converges to $x$ doesn't automatically make $x$ a cluster point (consider the constant sequence $a_n = x$).
    – Bungo
    Sep 10 at 3:07















up vote
0
down vote

favorite
1












Need a check of working, I'm new to cluster points.



Define $a_n = frac1n$ for all $ninmathbb N$. Clearly $A =$ $frac1n:ninmathbb N$ $=$ $(a_n)$. Now $a_n$ converges to $0$, thus every subsequence of $a_n$ also converges to $0$. Hence $0$ is the only cluster point of the set A.



Does that suffice, or is there something to add?










share|cite|improve this question



















  • 1




    Whether or not your argument is correct depends on what you know about cluster points. What is the definition? Is the definition given in terms of sequences?
    – user99914
    Sep 10 at 1:30







  • 1




    The definition I've become familiar with; Let $A$ be a subset of $mathbb R$ and let $x_0 in mathbb R$, $x_0$ is a cluster point pf $A$ if for all $epsilon > 0$ the set $(x_0-epsilon,x_0+epsilon) cap Asetminus$$x_0$ $ne$ the empty set.
    – Florian Suess
    Sep 10 at 1:34







  • 1




    I've also come familiar with the synonymity of $x_0 $ is a cluster point AND there exists a sequence $(a_n)_ninmathbb N$ in $Asetminus$$x_0$ such that $lim_nrightarrowinftya_n = x_0$
    – Florian Suess
    Sep 10 at 1:36







  • 1




    Moreover I use the fact that if a sequence converges to say $L$, then every subsequence (of that sequence) must also converge to $L$
    – Florian Suess
    Sep 10 at 1:37










  • How are you defining "cluster point"? Typical definition: a point $x$ is a cluster point of $A$ if every open neighborhood of $x$ contains a point of $A$ other than $x$ itself, So, the fact that a sequence of points in $A$ converges to $x$ doesn't automatically make $x$ a cluster point (consider the constant sequence $a_n = x$).
    – Bungo
    Sep 10 at 3:07













up vote
0
down vote

favorite
1









up vote
0
down vote

favorite
1






1





Need a check of working, I'm new to cluster points.



Define $a_n = frac1n$ for all $ninmathbb N$. Clearly $A =$ $frac1n:ninmathbb N$ $=$ $(a_n)$. Now $a_n$ converges to $0$, thus every subsequence of $a_n$ also converges to $0$. Hence $0$ is the only cluster point of the set A.



Does that suffice, or is there something to add?










share|cite|improve this question















Need a check of working, I'm new to cluster points.



Define $a_n = frac1n$ for all $ninmathbb N$. Clearly $A =$ $frac1n:ninmathbb N$ $=$ $(a_n)$. Now $a_n$ converges to $0$, thus every subsequence of $a_n$ also converges to $0$. Hence $0$ is the only cluster point of the set A.



Does that suffice, or is there something to add?







real-analysis sequences-and-series






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edited Sep 10 at 3:40









Harambe

5,91121843




5,91121843










asked Sep 10 at 1:18









Florian Suess

329110




329110







  • 1




    Whether or not your argument is correct depends on what you know about cluster points. What is the definition? Is the definition given in terms of sequences?
    – user99914
    Sep 10 at 1:30







  • 1




    The definition I've become familiar with; Let $A$ be a subset of $mathbb R$ and let $x_0 in mathbb R$, $x_0$ is a cluster point pf $A$ if for all $epsilon > 0$ the set $(x_0-epsilon,x_0+epsilon) cap Asetminus$$x_0$ $ne$ the empty set.
    – Florian Suess
    Sep 10 at 1:34







  • 1




    I've also come familiar with the synonymity of $x_0 $ is a cluster point AND there exists a sequence $(a_n)_ninmathbb N$ in $Asetminus$$x_0$ such that $lim_nrightarrowinftya_n = x_0$
    – Florian Suess
    Sep 10 at 1:36







  • 1




    Moreover I use the fact that if a sequence converges to say $L$, then every subsequence (of that sequence) must also converge to $L$
    – Florian Suess
    Sep 10 at 1:37










  • How are you defining "cluster point"? Typical definition: a point $x$ is a cluster point of $A$ if every open neighborhood of $x$ contains a point of $A$ other than $x$ itself, So, the fact that a sequence of points in $A$ converges to $x$ doesn't automatically make $x$ a cluster point (consider the constant sequence $a_n = x$).
    – Bungo
    Sep 10 at 3:07













  • 1




    Whether or not your argument is correct depends on what you know about cluster points. What is the definition? Is the definition given in terms of sequences?
    – user99914
    Sep 10 at 1:30







  • 1




    The definition I've become familiar with; Let $A$ be a subset of $mathbb R$ and let $x_0 in mathbb R$, $x_0$ is a cluster point pf $A$ if for all $epsilon > 0$ the set $(x_0-epsilon,x_0+epsilon) cap Asetminus$$x_0$ $ne$ the empty set.
    – Florian Suess
    Sep 10 at 1:34







  • 1




    I've also come familiar with the synonymity of $x_0 $ is a cluster point AND there exists a sequence $(a_n)_ninmathbb N$ in $Asetminus$$x_0$ such that $lim_nrightarrowinftya_n = x_0$
    – Florian Suess
    Sep 10 at 1:36







  • 1




    Moreover I use the fact that if a sequence converges to say $L$, then every subsequence (of that sequence) must also converge to $L$
    – Florian Suess
    Sep 10 at 1:37










  • How are you defining "cluster point"? Typical definition: a point $x$ is a cluster point of $A$ if every open neighborhood of $x$ contains a point of $A$ other than $x$ itself, So, the fact that a sequence of points in $A$ converges to $x$ doesn't automatically make $x$ a cluster point (consider the constant sequence $a_n = x$).
    – Bungo
    Sep 10 at 3:07








1




1




Whether or not your argument is correct depends on what you know about cluster points. What is the definition? Is the definition given in terms of sequences?
– user99914
Sep 10 at 1:30





Whether or not your argument is correct depends on what you know about cluster points. What is the definition? Is the definition given in terms of sequences?
– user99914
Sep 10 at 1:30





1




1




The definition I've become familiar with; Let $A$ be a subset of $mathbb R$ and let $x_0 in mathbb R$, $x_0$ is a cluster point pf $A$ if for all $epsilon > 0$ the set $(x_0-epsilon,x_0+epsilon) cap Asetminus$$x_0$ $ne$ the empty set.
– Florian Suess
Sep 10 at 1:34





The definition I've become familiar with; Let $A$ be a subset of $mathbb R$ and let $x_0 in mathbb R$, $x_0$ is a cluster point pf $A$ if for all $epsilon > 0$ the set $(x_0-epsilon,x_0+epsilon) cap Asetminus$$x_0$ $ne$ the empty set.
– Florian Suess
Sep 10 at 1:34





1




1




I've also come familiar with the synonymity of $x_0 $ is a cluster point AND there exists a sequence $(a_n)_ninmathbb N$ in $Asetminus$$x_0$ such that $lim_nrightarrowinftya_n = x_0$
– Florian Suess
Sep 10 at 1:36





I've also come familiar with the synonymity of $x_0 $ is a cluster point AND there exists a sequence $(a_n)_ninmathbb N$ in $Asetminus$$x_0$ such that $lim_nrightarrowinftya_n = x_0$
– Florian Suess
Sep 10 at 1:36





1




1




Moreover I use the fact that if a sequence converges to say $L$, then every subsequence (of that sequence) must also converge to $L$
– Florian Suess
Sep 10 at 1:37




Moreover I use the fact that if a sequence converges to say $L$, then every subsequence (of that sequence) must also converge to $L$
– Florian Suess
Sep 10 at 1:37












How are you defining "cluster point"? Typical definition: a point $x$ is a cluster point of $A$ if every open neighborhood of $x$ contains a point of $A$ other than $x$ itself, So, the fact that a sequence of points in $A$ converges to $x$ doesn't automatically make $x$ a cluster point (consider the constant sequence $a_n = x$).
– Bungo
Sep 10 at 3:07





How are you defining "cluster point"? Typical definition: a point $x$ is a cluster point of $A$ if every open neighborhood of $x$ contains a point of $A$ other than $x$ itself, So, the fact that a sequence of points in $A$ converges to $x$ doesn't automatically make $x$ a cluster point (consider the constant sequence $a_n = x$).
– Bungo
Sep 10 at 3:07











3 Answers
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Your argument shows that $0$ is a cluster point, since $(a_n)$ is a sequence in $Asetminus 0$ that converges to $A$. However, it doesn't prove that there are no other cluster points, since not all sequences in $Asetminus 0$ are subsequences of $(a_n)$.



To prove that no other cluster points exist, you can use the definition. That is, show that given any other $x_0 in mathbb R$, there exists $varepsilon > 0$ such that
$$[(x_0 - varepsilon, x_0 + varepsilon)cap A]setminus x_0 = varnothing.$$
You can do this easily by splitting into cases:



  1. $x_0 < 0$

  2. $x_0 > 1$

  3. $x_0 in (0,1]cap A$

  4. $x_0 in (0,1]setminus A$

I'll prove one case for you, and by drawing pictures, you can figure out how to do the other three.



Suppose $x_0 > 1$. Let $varepsilon = fracx_0-12$. Then $[(x_0 - varepsilon, x_0 + varepsilon)cap A]setminus x_0 = varnothing$, so $A$ has no cluster points greater than $1$.






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    up vote
    1
    down vote













    If this is for a class, I think that $0$ being the limit of the sequence should suffice as an argument for $0$ being a cluster point of $A$ (if you discussed this in class yet). However I would suggest being more explicit in your proof that this is the only cluster point of the set $A$. Something like,



    Suppose that $p in mathbbR$ is a cluster point of the set $A$. Then $forall epsilon >0$, the set $a_n in A, $ contains an infinite number of points. By the least upper bound property of $mathbbN$, there is necessarily an index $i in mathbbN$ such that $a_i=min a_n in A, $. You can continue this argument to show that the subsequence starting with $a_1,a_2,...,a_i$, and continuing with the points of $a_n in A, $, ordered by their indices, is a subsequence of $a_n$ with limit $p$. Then by the uniqueness of the limit of a sequence, $p=0$.






    share|cite|improve this answer




















    • Just in case you are not aware, you can edit your answer instead of posting a new one. Just click "edit" below the answer.
      – user99914
      Sep 10 at 3:50










    • Thanks, I just made my first stack exchange account, I didnt realize anyone could see a deleted answer
      – clmundergrad
      Sep 10 at 7:48

















    up vote
    0
    down vote













    There is indeed a flaw I've just realised. A subsequence of $a_n$ does not represent the entire set of subsequences in $A$. Hence I have not discovered what every subsequence in $A$ actually looks like.






    share|cite|improve this answer






















    • For example the constant sequence $frac12$ and so on. It is not a direct subsequence of $a_n$ hence I have not considered it.
      – Florian Suess
      Sep 10 at 1:47










    • I'm just hoping for confirmation from some professionals, because as I said, I'm new to cluster points.
      – Florian Suess
      Sep 10 at 1:48










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    3 Answers
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    3 Answers
    3






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    active

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    up vote
    2
    down vote



    accepted










    Your argument shows that $0$ is a cluster point, since $(a_n)$ is a sequence in $Asetminus 0$ that converges to $A$. However, it doesn't prove that there are no other cluster points, since not all sequences in $Asetminus 0$ are subsequences of $(a_n)$.



    To prove that no other cluster points exist, you can use the definition. That is, show that given any other $x_0 in mathbb R$, there exists $varepsilon > 0$ such that
    $$[(x_0 - varepsilon, x_0 + varepsilon)cap A]setminus x_0 = varnothing.$$
    You can do this easily by splitting into cases:



    1. $x_0 < 0$

    2. $x_0 > 1$

    3. $x_0 in (0,1]cap A$

    4. $x_0 in (0,1]setminus A$

    I'll prove one case for you, and by drawing pictures, you can figure out how to do the other three.



    Suppose $x_0 > 1$. Let $varepsilon = fracx_0-12$. Then $[(x_0 - varepsilon, x_0 + varepsilon)cap A]setminus x_0 = varnothing$, so $A$ has no cluster points greater than $1$.






    share|cite|improve this answer


























      up vote
      2
      down vote



      accepted










      Your argument shows that $0$ is a cluster point, since $(a_n)$ is a sequence in $Asetminus 0$ that converges to $A$. However, it doesn't prove that there are no other cluster points, since not all sequences in $Asetminus 0$ are subsequences of $(a_n)$.



      To prove that no other cluster points exist, you can use the definition. That is, show that given any other $x_0 in mathbb R$, there exists $varepsilon > 0$ such that
      $$[(x_0 - varepsilon, x_0 + varepsilon)cap A]setminus x_0 = varnothing.$$
      You can do this easily by splitting into cases:



      1. $x_0 < 0$

      2. $x_0 > 1$

      3. $x_0 in (0,1]cap A$

      4. $x_0 in (0,1]setminus A$

      I'll prove one case for you, and by drawing pictures, you can figure out how to do the other three.



      Suppose $x_0 > 1$. Let $varepsilon = fracx_0-12$. Then $[(x_0 - varepsilon, x_0 + varepsilon)cap A]setminus x_0 = varnothing$, so $A$ has no cluster points greater than $1$.






      share|cite|improve this answer
























        up vote
        2
        down vote



        accepted







        up vote
        2
        down vote



        accepted






        Your argument shows that $0$ is a cluster point, since $(a_n)$ is a sequence in $Asetminus 0$ that converges to $A$. However, it doesn't prove that there are no other cluster points, since not all sequences in $Asetminus 0$ are subsequences of $(a_n)$.



        To prove that no other cluster points exist, you can use the definition. That is, show that given any other $x_0 in mathbb R$, there exists $varepsilon > 0$ such that
        $$[(x_0 - varepsilon, x_0 + varepsilon)cap A]setminus x_0 = varnothing.$$
        You can do this easily by splitting into cases:



        1. $x_0 < 0$

        2. $x_0 > 1$

        3. $x_0 in (0,1]cap A$

        4. $x_0 in (0,1]setminus A$

        I'll prove one case for you, and by drawing pictures, you can figure out how to do the other three.



        Suppose $x_0 > 1$. Let $varepsilon = fracx_0-12$. Then $[(x_0 - varepsilon, x_0 + varepsilon)cap A]setminus x_0 = varnothing$, so $A$ has no cluster points greater than $1$.






        share|cite|improve this answer














        Your argument shows that $0$ is a cluster point, since $(a_n)$ is a sequence in $Asetminus 0$ that converges to $A$. However, it doesn't prove that there are no other cluster points, since not all sequences in $Asetminus 0$ are subsequences of $(a_n)$.



        To prove that no other cluster points exist, you can use the definition. That is, show that given any other $x_0 in mathbb R$, there exists $varepsilon > 0$ such that
        $$[(x_0 - varepsilon, x_0 + varepsilon)cap A]setminus x_0 = varnothing.$$
        You can do this easily by splitting into cases:



        1. $x_0 < 0$

        2. $x_0 > 1$

        3. $x_0 in (0,1]cap A$

        4. $x_0 in (0,1]setminus A$

        I'll prove one case for you, and by drawing pictures, you can figure out how to do the other three.



        Suppose $x_0 > 1$. Let $varepsilon = fracx_0-12$. Then $[(x_0 - varepsilon, x_0 + varepsilon)cap A]setminus x_0 = varnothing$, so $A$ has no cluster points greater than $1$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Sep 10 at 6:31

























        answered Sep 10 at 2:56









        Harambe

        5,91121843




        5,91121843




















            up vote
            1
            down vote













            If this is for a class, I think that $0$ being the limit of the sequence should suffice as an argument for $0$ being a cluster point of $A$ (if you discussed this in class yet). However I would suggest being more explicit in your proof that this is the only cluster point of the set $A$. Something like,



            Suppose that $p in mathbbR$ is a cluster point of the set $A$. Then $forall epsilon >0$, the set $a_n in A, $ contains an infinite number of points. By the least upper bound property of $mathbbN$, there is necessarily an index $i in mathbbN$ such that $a_i=min a_n in A, $. You can continue this argument to show that the subsequence starting with $a_1,a_2,...,a_i$, and continuing with the points of $a_n in A, $, ordered by their indices, is a subsequence of $a_n$ with limit $p$. Then by the uniqueness of the limit of a sequence, $p=0$.






            share|cite|improve this answer




















            • Just in case you are not aware, you can edit your answer instead of posting a new one. Just click "edit" below the answer.
              – user99914
              Sep 10 at 3:50










            • Thanks, I just made my first stack exchange account, I didnt realize anyone could see a deleted answer
              – clmundergrad
              Sep 10 at 7:48














            up vote
            1
            down vote













            If this is for a class, I think that $0$ being the limit of the sequence should suffice as an argument for $0$ being a cluster point of $A$ (if you discussed this in class yet). However I would suggest being more explicit in your proof that this is the only cluster point of the set $A$. Something like,



            Suppose that $p in mathbbR$ is a cluster point of the set $A$. Then $forall epsilon >0$, the set $a_n in A, $ contains an infinite number of points. By the least upper bound property of $mathbbN$, there is necessarily an index $i in mathbbN$ such that $a_i=min a_n in A, $. You can continue this argument to show that the subsequence starting with $a_1,a_2,...,a_i$, and continuing with the points of $a_n in A, $, ordered by their indices, is a subsequence of $a_n$ with limit $p$. Then by the uniqueness of the limit of a sequence, $p=0$.






            share|cite|improve this answer




















            • Just in case you are not aware, you can edit your answer instead of posting a new one. Just click "edit" below the answer.
              – user99914
              Sep 10 at 3:50










            • Thanks, I just made my first stack exchange account, I didnt realize anyone could see a deleted answer
              – clmundergrad
              Sep 10 at 7:48












            up vote
            1
            down vote










            up vote
            1
            down vote









            If this is for a class, I think that $0$ being the limit of the sequence should suffice as an argument for $0$ being a cluster point of $A$ (if you discussed this in class yet). However I would suggest being more explicit in your proof that this is the only cluster point of the set $A$. Something like,



            Suppose that $p in mathbbR$ is a cluster point of the set $A$. Then $forall epsilon >0$, the set $a_n in A, $ contains an infinite number of points. By the least upper bound property of $mathbbN$, there is necessarily an index $i in mathbbN$ such that $a_i=min a_n in A, $. You can continue this argument to show that the subsequence starting with $a_1,a_2,...,a_i$, and continuing with the points of $a_n in A, $, ordered by their indices, is a subsequence of $a_n$ with limit $p$. Then by the uniqueness of the limit of a sequence, $p=0$.






            share|cite|improve this answer












            If this is for a class, I think that $0$ being the limit of the sequence should suffice as an argument for $0$ being a cluster point of $A$ (if you discussed this in class yet). However I would suggest being more explicit in your proof that this is the only cluster point of the set $A$. Something like,



            Suppose that $p in mathbbR$ is a cluster point of the set $A$. Then $forall epsilon >0$, the set $a_n in A, $ contains an infinite number of points. By the least upper bound property of $mathbbN$, there is necessarily an index $i in mathbbN$ such that $a_i=min a_n in A, $. You can continue this argument to show that the subsequence starting with $a_1,a_2,...,a_i$, and continuing with the points of $a_n in A, $, ordered by their indices, is a subsequence of $a_n$ with limit $p$. Then by the uniqueness of the limit of a sequence, $p=0$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Sep 10 at 1:48









            clmundergrad

            464




            464











            • Just in case you are not aware, you can edit your answer instead of posting a new one. Just click "edit" below the answer.
              – user99914
              Sep 10 at 3:50










            • Thanks, I just made my first stack exchange account, I didnt realize anyone could see a deleted answer
              – clmundergrad
              Sep 10 at 7:48
















            • Just in case you are not aware, you can edit your answer instead of posting a new one. Just click "edit" below the answer.
              – user99914
              Sep 10 at 3:50










            • Thanks, I just made my first stack exchange account, I didnt realize anyone could see a deleted answer
              – clmundergrad
              Sep 10 at 7:48















            Just in case you are not aware, you can edit your answer instead of posting a new one. Just click "edit" below the answer.
            – user99914
            Sep 10 at 3:50




            Just in case you are not aware, you can edit your answer instead of posting a new one. Just click "edit" below the answer.
            – user99914
            Sep 10 at 3:50












            Thanks, I just made my first stack exchange account, I didnt realize anyone could see a deleted answer
            – clmundergrad
            Sep 10 at 7:48




            Thanks, I just made my first stack exchange account, I didnt realize anyone could see a deleted answer
            – clmundergrad
            Sep 10 at 7:48










            up vote
            0
            down vote













            There is indeed a flaw I've just realised. A subsequence of $a_n$ does not represent the entire set of subsequences in $A$. Hence I have not discovered what every subsequence in $A$ actually looks like.






            share|cite|improve this answer






















            • For example the constant sequence $frac12$ and so on. It is not a direct subsequence of $a_n$ hence I have not considered it.
              – Florian Suess
              Sep 10 at 1:47










            • I'm just hoping for confirmation from some professionals, because as I said, I'm new to cluster points.
              – Florian Suess
              Sep 10 at 1:48














            up vote
            0
            down vote













            There is indeed a flaw I've just realised. A subsequence of $a_n$ does not represent the entire set of subsequences in $A$. Hence I have not discovered what every subsequence in $A$ actually looks like.






            share|cite|improve this answer






















            • For example the constant sequence $frac12$ and so on. It is not a direct subsequence of $a_n$ hence I have not considered it.
              – Florian Suess
              Sep 10 at 1:47










            • I'm just hoping for confirmation from some professionals, because as I said, I'm new to cluster points.
              – Florian Suess
              Sep 10 at 1:48












            up vote
            0
            down vote










            up vote
            0
            down vote









            There is indeed a flaw I've just realised. A subsequence of $a_n$ does not represent the entire set of subsequences in $A$. Hence I have not discovered what every subsequence in $A$ actually looks like.






            share|cite|improve this answer














            There is indeed a flaw I've just realised. A subsequence of $a_n$ does not represent the entire set of subsequences in $A$. Hence I have not discovered what every subsequence in $A$ actually looks like.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Sep 10 at 1:53

























            answered Sep 10 at 1:46









            Florian Suess

            329110




            329110











            • For example the constant sequence $frac12$ and so on. It is not a direct subsequence of $a_n$ hence I have not considered it.
              – Florian Suess
              Sep 10 at 1:47










            • I'm just hoping for confirmation from some professionals, because as I said, I'm new to cluster points.
              – Florian Suess
              Sep 10 at 1:48
















            • For example the constant sequence $frac12$ and so on. It is not a direct subsequence of $a_n$ hence I have not considered it.
              – Florian Suess
              Sep 10 at 1:47










            • I'm just hoping for confirmation from some professionals, because as I said, I'm new to cluster points.
              – Florian Suess
              Sep 10 at 1:48















            For example the constant sequence $frac12$ and so on. It is not a direct subsequence of $a_n$ hence I have not considered it.
            – Florian Suess
            Sep 10 at 1:47




            For example the constant sequence $frac12$ and so on. It is not a direct subsequence of $a_n$ hence I have not considered it.
            – Florian Suess
            Sep 10 at 1:47












            I'm just hoping for confirmation from some professionals, because as I said, I'm new to cluster points.
            – Florian Suess
            Sep 10 at 1:48




            I'm just hoping for confirmation from some professionals, because as I said, I'm new to cluster points.
            – Florian Suess
            Sep 10 at 1:48

















             

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