Sequence such that every subsequence can have a different real limit [duplicate]











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  • Give an example of a sequence of real numbers with subsequences converging to every real number

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I would like to find a sequence of real numbers $(a_n)_{ninmathbb{N}}$ with this property: for any $Linmathbb{R}$ there is a subsequence $a_{k_n}$ such that $$lim_{ntoinfty} a_{k_n} = L$$ Does such a sequence exist?










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  • Otherwise said, the set ${ a_n | ninmathbb N}$ should be dense in $mathbb R$.
    – Giuseppe Negro
    17 hours ago















up vote
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  • Give an example of a sequence of real numbers with subsequences converging to every real number

    3 answers




I would like to find a sequence of real numbers $(a_n)_{ninmathbb{N}}$ with this property: for any $Linmathbb{R}$ there is a subsequence $a_{k_n}$ such that $$lim_{ntoinfty} a_{k_n} = L$$ Does such a sequence exist?










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marked as duplicate by Martin R, Community 14 hours ago


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.















  • Otherwise said, the set ${ a_n | ninmathbb N}$ should be dense in $mathbb R$.
    – Giuseppe Negro
    17 hours ago













up vote
8
down vote

favorite
1









up vote
8
down vote

favorite
1






1






This question already has an answer here:




  • Give an example of a sequence of real numbers with subsequences converging to every real number

    3 answers




I would like to find a sequence of real numbers $(a_n)_{ninmathbb{N}}$ with this property: for any $Linmathbb{R}$ there is a subsequence $a_{k_n}$ such that $$lim_{ntoinfty} a_{k_n} = L$$ Does such a sequence exist?










share|cite|improve this question
















This question already has an answer here:




  • Give an example of a sequence of real numbers with subsequences converging to every real number

    3 answers




I would like to find a sequence of real numbers $(a_n)_{ninmathbb{N}}$ with this property: for any $Linmathbb{R}$ there is a subsequence $a_{k_n}$ such that $$lim_{ntoinfty} a_{k_n} = L$$ Does such a sequence exist?





This question already has an answer here:




  • Give an example of a sequence of real numbers with subsequences converging to every real number

    3 answers








sequences-and-series limits analysis






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edited 17 hours ago









Especially Lime

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asked 17 hours ago









Riccardo Cazzin

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1905




marked as duplicate by Martin R, Community 14 hours ago


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






marked as duplicate by Martin R, Community 14 hours ago


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • Otherwise said, the set ${ a_n | ninmathbb N}$ should be dense in $mathbb R$.
    – Giuseppe Negro
    17 hours ago


















  • Otherwise said, the set ${ a_n | ninmathbb N}$ should be dense in $mathbb R$.
    – Giuseppe Negro
    17 hours ago
















Otherwise said, the set ${ a_n | ninmathbb N}$ should be dense in $mathbb R$.
– Giuseppe Negro
17 hours ago




Otherwise said, the set ${ a_n | ninmathbb N}$ should be dense in $mathbb R$.
– Giuseppe Negro
17 hours ago










4 Answers
4






active

oldest

votes

















up vote
9
down vote



accepted










Just arrange the set of rational numbers in a sequence ${a_n}$. Given any real number $L$ and any positive integer $n$ there are infinitely many rationals in $(L-frac 1 n, L+frac 1 n)$. Pick $a_{n_1}$ in this interval with $n=1$. Then consider the case $n=2$. You can surely find $n_2 >n_1$ such that $a_{n_2} in (L-frac 1 2, L+frac 1 2)$. Use induction to construct a subsequence $a_{n_k}$ such that $|a_{n_k}-L| <frac 1 k$ for all $k$. Then $a_{n_k} to L$.






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  • Could you please explain better how it would work? Thanks
    – gimusi
    17 hours ago










  • That's nice, we are then using that rationals are countable. Thanks for the clarification. Regards
    – gimusi
    17 hours ago










  • Most welcome, @gimusi !
    – Kavi Rama Murthy
    17 hours ago


















up vote
4
down vote













Take the sequence that sweep the interval $[-1,1]$ by $1/2$ steps, then the interval $[-2,2]$ by $1/2^2$ steps, then the interval $[-n,n]$ by $1/2^n$ steps... and so on.



You’ll be able to prove that every real is a limit point of that sequence.






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  • That's less intuitive at first!
    – gimusi
    17 hours ago


















up vote
3
down vote













Rephrasing:



Consider $a_n$, $n=1,2,3,3,....,$ the sequence of rational numbers. Recall that $mathbb{Q}$ is countable, hence can be written as a sequence $a_n$, $nin mathbb{N}$.



Let $L in mathbb{R}$.



Since $mathbb{Q}$ is dense in $mathbb{R}$, we can construct a subsequence $a_{n_k}$ that converges to $L$.






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    up vote
    1
    down vote













    Consider a general topological space $(X, mathscr{T})$ and a sequence of points $x in X^{mathbb{N}}$. One says that point $t in X$ is adherent to the sequence $x$ (some authors use the terminology 'cluster-point', but I don't fancy it so much) if:



    $$(forall V, n)(V in mathscr{V}_{mathscr{T}}(x) wedge n in mathbb{N} implies (exists m)(m geqslant n wedge x_m in V ))$$



    where $mathscr{V}_{mathscr{T}}(x)$ symbolizes the filter of neighbourhoods of point $x$ induced by the topology $mathscr{T}$. In a more descriptive fashion, $t$ is adherent to sequence $x$ if any neighbourhood of $t$ contains terms of arbitrarily high rank from the sequence $x$. If the filter of neighbourhoods of $t$ admits a countable base, then $t$ can be expressed as the limit of a subsequence of $x$. Therefore, in a space satisfying the First Axiom of Countability (i.e. all points have a countable base of neighbourhoods), the points adherent to a given sequence $x$ can be equivalently characterised as limits of subsequences of $x$.



    Now, if the space $X$ is non-empty and separable, let us fix a certain dense subset $T subseteq X$. As $X$ is non-empty, so must $T$ be. It is not difficult to show that a non-empty countable set $M$ admits a surjection $sigma: mathbb{N} rightarrow M$ such that for each $t in M$ the fibre $sigma^{-1}({t})$ be infinite.



    Consider such a surjection $sigma: mathbb{N} rightarrow T$ and define the sequence $t=(sigma(n))_{n in mathbb{N}}$ (which is actually the graphic of map $sigma$). The condition on the cardinality of the fibers ensures that any element $x in T$ is the limit of a (constant) subsequence of $t$. If we furthermore assume that the space $(X, mathscr{T})$ is $T_1$, then any $x in X setminus T$ will be an accumulation point of $T$ and thus necessarily adherent to sequence $t$.



    To conclude, given a topological space $(X, mathscr{T})$ that is non-empty, separable, first countable and $T_1$, one can always find a sequence $t$ of points within the space such that each element $x in X$ be expressible as the limit of a subsequence of $t$. This applies in particular to the topological space $(mathbb{R}, mathscr{O})$, where $mathscr{O}$ denotes the (usual) order topology.






    share|cite|improve this answer








    New contributor




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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      9
      down vote



      accepted










      Just arrange the set of rational numbers in a sequence ${a_n}$. Given any real number $L$ and any positive integer $n$ there are infinitely many rationals in $(L-frac 1 n, L+frac 1 n)$. Pick $a_{n_1}$ in this interval with $n=1$. Then consider the case $n=2$. You can surely find $n_2 >n_1$ such that $a_{n_2} in (L-frac 1 2, L+frac 1 2)$. Use induction to construct a subsequence $a_{n_k}$ such that $|a_{n_k}-L| <frac 1 k$ for all $k$. Then $a_{n_k} to L$.






      share|cite|improve this answer























      • Could you please explain better how it would work? Thanks
        – gimusi
        17 hours ago










      • That's nice, we are then using that rationals are countable. Thanks for the clarification. Regards
        – gimusi
        17 hours ago










      • Most welcome, @gimusi !
        – Kavi Rama Murthy
        17 hours ago















      up vote
      9
      down vote



      accepted










      Just arrange the set of rational numbers in a sequence ${a_n}$. Given any real number $L$ and any positive integer $n$ there are infinitely many rationals in $(L-frac 1 n, L+frac 1 n)$. Pick $a_{n_1}$ in this interval with $n=1$. Then consider the case $n=2$. You can surely find $n_2 >n_1$ such that $a_{n_2} in (L-frac 1 2, L+frac 1 2)$. Use induction to construct a subsequence $a_{n_k}$ such that $|a_{n_k}-L| <frac 1 k$ for all $k$. Then $a_{n_k} to L$.






      share|cite|improve this answer























      • Could you please explain better how it would work? Thanks
        – gimusi
        17 hours ago










      • That's nice, we are then using that rationals are countable. Thanks for the clarification. Regards
        – gimusi
        17 hours ago










      • Most welcome, @gimusi !
        – Kavi Rama Murthy
        17 hours ago













      up vote
      9
      down vote



      accepted







      up vote
      9
      down vote



      accepted






      Just arrange the set of rational numbers in a sequence ${a_n}$. Given any real number $L$ and any positive integer $n$ there are infinitely many rationals in $(L-frac 1 n, L+frac 1 n)$. Pick $a_{n_1}$ in this interval with $n=1$. Then consider the case $n=2$. You can surely find $n_2 >n_1$ such that $a_{n_2} in (L-frac 1 2, L+frac 1 2)$. Use induction to construct a subsequence $a_{n_k}$ such that $|a_{n_k}-L| <frac 1 k$ for all $k$. Then $a_{n_k} to L$.






      share|cite|improve this answer














      Just arrange the set of rational numbers in a sequence ${a_n}$. Given any real number $L$ and any positive integer $n$ there are infinitely many rationals in $(L-frac 1 n, L+frac 1 n)$. Pick $a_{n_1}$ in this interval with $n=1$. Then consider the case $n=2$. You can surely find $n_2 >n_1$ such that $a_{n_2} in (L-frac 1 2, L+frac 1 2)$. Use induction to construct a subsequence $a_{n_k}$ such that $|a_{n_k}-L| <frac 1 k$ for all $k$. Then $a_{n_k} to L$.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited 17 hours ago

























      answered 17 hours ago









      Kavi Rama Murthy

      45.4k31853




      45.4k31853












      • Could you please explain better how it would work? Thanks
        – gimusi
        17 hours ago










      • That's nice, we are then using that rationals are countable. Thanks for the clarification. Regards
        – gimusi
        17 hours ago










      • Most welcome, @gimusi !
        – Kavi Rama Murthy
        17 hours ago


















      • Could you please explain better how it would work? Thanks
        – gimusi
        17 hours ago










      • That's nice, we are then using that rationals are countable. Thanks for the clarification. Regards
        – gimusi
        17 hours ago










      • Most welcome, @gimusi !
        – Kavi Rama Murthy
        17 hours ago
















      Could you please explain better how it would work? Thanks
      – gimusi
      17 hours ago




      Could you please explain better how it would work? Thanks
      – gimusi
      17 hours ago












      That's nice, we are then using that rationals are countable. Thanks for the clarification. Regards
      – gimusi
      17 hours ago




      That's nice, we are then using that rationals are countable. Thanks for the clarification. Regards
      – gimusi
      17 hours ago












      Most welcome, @gimusi !
      – Kavi Rama Murthy
      17 hours ago




      Most welcome, @gimusi !
      – Kavi Rama Murthy
      17 hours ago










      up vote
      4
      down vote













      Take the sequence that sweep the interval $[-1,1]$ by $1/2$ steps, then the interval $[-2,2]$ by $1/2^2$ steps, then the interval $[-n,n]$ by $1/2^n$ steps... and so on.



      You’ll be able to prove that every real is a limit point of that sequence.






      share|cite|improve this answer





















      • That's less intuitive at first!
        – gimusi
        17 hours ago















      up vote
      4
      down vote













      Take the sequence that sweep the interval $[-1,1]$ by $1/2$ steps, then the interval $[-2,2]$ by $1/2^2$ steps, then the interval $[-n,n]$ by $1/2^n$ steps... and so on.



      You’ll be able to prove that every real is a limit point of that sequence.






      share|cite|improve this answer





















      • That's less intuitive at first!
        – gimusi
        17 hours ago













      up vote
      4
      down vote










      up vote
      4
      down vote









      Take the sequence that sweep the interval $[-1,1]$ by $1/2$ steps, then the interval $[-2,2]$ by $1/2^2$ steps, then the interval $[-n,n]$ by $1/2^n$ steps... and so on.



      You’ll be able to prove that every real is a limit point of that sequence.






      share|cite|improve this answer












      Take the sequence that sweep the interval $[-1,1]$ by $1/2$ steps, then the interval $[-2,2]$ by $1/2^2$ steps, then the interval $[-n,n]$ by $1/2^n$ steps... and so on.



      You’ll be able to prove that every real is a limit point of that sequence.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered 17 hours ago









      mathcounterexamples.net

      23.9k21753




      23.9k21753












      • That's less intuitive at first!
        – gimusi
        17 hours ago


















      • That's less intuitive at first!
        – gimusi
        17 hours ago
















      That's less intuitive at first!
      – gimusi
      17 hours ago




      That's less intuitive at first!
      – gimusi
      17 hours ago










      up vote
      3
      down vote













      Rephrasing:



      Consider $a_n$, $n=1,2,3,3,....,$ the sequence of rational numbers. Recall that $mathbb{Q}$ is countable, hence can be written as a sequence $a_n$, $nin mathbb{N}$.



      Let $L in mathbb{R}$.



      Since $mathbb{Q}$ is dense in $mathbb{R}$, we can construct a subsequence $a_{n_k}$ that converges to $L$.






      share|cite|improve this answer

























        up vote
        3
        down vote













        Rephrasing:



        Consider $a_n$, $n=1,2,3,3,....,$ the sequence of rational numbers. Recall that $mathbb{Q}$ is countable, hence can be written as a sequence $a_n$, $nin mathbb{N}$.



        Let $L in mathbb{R}$.



        Since $mathbb{Q}$ is dense in $mathbb{R}$, we can construct a subsequence $a_{n_k}$ that converges to $L$.






        share|cite|improve this answer























          up vote
          3
          down vote










          up vote
          3
          down vote









          Rephrasing:



          Consider $a_n$, $n=1,2,3,3,....,$ the sequence of rational numbers. Recall that $mathbb{Q}$ is countable, hence can be written as a sequence $a_n$, $nin mathbb{N}$.



          Let $L in mathbb{R}$.



          Since $mathbb{Q}$ is dense in $mathbb{R}$, we can construct a subsequence $a_{n_k}$ that converges to $L$.






          share|cite|improve this answer












          Rephrasing:



          Consider $a_n$, $n=1,2,3,3,....,$ the sequence of rational numbers. Recall that $mathbb{Q}$ is countable, hence can be written as a sequence $a_n$, $nin mathbb{N}$.



          Let $L in mathbb{R}$.



          Since $mathbb{Q}$ is dense in $mathbb{R}$, we can construct a subsequence $a_{n_k}$ that converges to $L$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 17 hours ago









          Peter Szilas

          10.4k2720




          10.4k2720






















              up vote
              1
              down vote













              Consider a general topological space $(X, mathscr{T})$ and a sequence of points $x in X^{mathbb{N}}$. One says that point $t in X$ is adherent to the sequence $x$ (some authors use the terminology 'cluster-point', but I don't fancy it so much) if:



              $$(forall V, n)(V in mathscr{V}_{mathscr{T}}(x) wedge n in mathbb{N} implies (exists m)(m geqslant n wedge x_m in V ))$$



              where $mathscr{V}_{mathscr{T}}(x)$ symbolizes the filter of neighbourhoods of point $x$ induced by the topology $mathscr{T}$. In a more descriptive fashion, $t$ is adherent to sequence $x$ if any neighbourhood of $t$ contains terms of arbitrarily high rank from the sequence $x$. If the filter of neighbourhoods of $t$ admits a countable base, then $t$ can be expressed as the limit of a subsequence of $x$. Therefore, in a space satisfying the First Axiom of Countability (i.e. all points have a countable base of neighbourhoods), the points adherent to a given sequence $x$ can be equivalently characterised as limits of subsequences of $x$.



              Now, if the space $X$ is non-empty and separable, let us fix a certain dense subset $T subseteq X$. As $X$ is non-empty, so must $T$ be. It is not difficult to show that a non-empty countable set $M$ admits a surjection $sigma: mathbb{N} rightarrow M$ such that for each $t in M$ the fibre $sigma^{-1}({t})$ be infinite.



              Consider such a surjection $sigma: mathbb{N} rightarrow T$ and define the sequence $t=(sigma(n))_{n in mathbb{N}}$ (which is actually the graphic of map $sigma$). The condition on the cardinality of the fibers ensures that any element $x in T$ is the limit of a (constant) subsequence of $t$. If we furthermore assume that the space $(X, mathscr{T})$ is $T_1$, then any $x in X setminus T$ will be an accumulation point of $T$ and thus necessarily adherent to sequence $t$.



              To conclude, given a topological space $(X, mathscr{T})$ that is non-empty, separable, first countable and $T_1$, one can always find a sequence $t$ of points within the space such that each element $x in X$ be expressible as the limit of a subsequence of $t$. This applies in particular to the topological space $(mathbb{R}, mathscr{O})$, where $mathscr{O}$ denotes the (usual) order topology.






              share|cite|improve this answer








              New contributor




              ΑΘΩ 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













                Consider a general topological space $(X, mathscr{T})$ and a sequence of points $x in X^{mathbb{N}}$. One says that point $t in X$ is adherent to the sequence $x$ (some authors use the terminology 'cluster-point', but I don't fancy it so much) if:



                $$(forall V, n)(V in mathscr{V}_{mathscr{T}}(x) wedge n in mathbb{N} implies (exists m)(m geqslant n wedge x_m in V ))$$



                where $mathscr{V}_{mathscr{T}}(x)$ symbolizes the filter of neighbourhoods of point $x$ induced by the topology $mathscr{T}$. In a more descriptive fashion, $t$ is adherent to sequence $x$ if any neighbourhood of $t$ contains terms of arbitrarily high rank from the sequence $x$. If the filter of neighbourhoods of $t$ admits a countable base, then $t$ can be expressed as the limit of a subsequence of $x$. Therefore, in a space satisfying the First Axiom of Countability (i.e. all points have a countable base of neighbourhoods), the points adherent to a given sequence $x$ can be equivalently characterised as limits of subsequences of $x$.



                Now, if the space $X$ is non-empty and separable, let us fix a certain dense subset $T subseteq X$. As $X$ is non-empty, so must $T$ be. It is not difficult to show that a non-empty countable set $M$ admits a surjection $sigma: mathbb{N} rightarrow M$ such that for each $t in M$ the fibre $sigma^{-1}({t})$ be infinite.



                Consider such a surjection $sigma: mathbb{N} rightarrow T$ and define the sequence $t=(sigma(n))_{n in mathbb{N}}$ (which is actually the graphic of map $sigma$). The condition on the cardinality of the fibers ensures that any element $x in T$ is the limit of a (constant) subsequence of $t$. If we furthermore assume that the space $(X, mathscr{T})$ is $T_1$, then any $x in X setminus T$ will be an accumulation point of $T$ and thus necessarily adherent to sequence $t$.



                To conclude, given a topological space $(X, mathscr{T})$ that is non-empty, separable, first countable and $T_1$, one can always find a sequence $t$ of points within the space such that each element $x in X$ be expressible as the limit of a subsequence of $t$. This applies in particular to the topological space $(mathbb{R}, mathscr{O})$, where $mathscr{O}$ denotes the (usual) order topology.






                share|cite|improve this answer








                New contributor




                ΑΘΩ 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










                  up vote
                  1
                  down vote









                  Consider a general topological space $(X, mathscr{T})$ and a sequence of points $x in X^{mathbb{N}}$. One says that point $t in X$ is adherent to the sequence $x$ (some authors use the terminology 'cluster-point', but I don't fancy it so much) if:



                  $$(forall V, n)(V in mathscr{V}_{mathscr{T}}(x) wedge n in mathbb{N} implies (exists m)(m geqslant n wedge x_m in V ))$$



                  where $mathscr{V}_{mathscr{T}}(x)$ symbolizes the filter of neighbourhoods of point $x$ induced by the topology $mathscr{T}$. In a more descriptive fashion, $t$ is adherent to sequence $x$ if any neighbourhood of $t$ contains terms of arbitrarily high rank from the sequence $x$. If the filter of neighbourhoods of $t$ admits a countable base, then $t$ can be expressed as the limit of a subsequence of $x$. Therefore, in a space satisfying the First Axiom of Countability (i.e. all points have a countable base of neighbourhoods), the points adherent to a given sequence $x$ can be equivalently characterised as limits of subsequences of $x$.



                  Now, if the space $X$ is non-empty and separable, let us fix a certain dense subset $T subseteq X$. As $X$ is non-empty, so must $T$ be. It is not difficult to show that a non-empty countable set $M$ admits a surjection $sigma: mathbb{N} rightarrow M$ such that for each $t in M$ the fibre $sigma^{-1}({t})$ be infinite.



                  Consider such a surjection $sigma: mathbb{N} rightarrow T$ and define the sequence $t=(sigma(n))_{n in mathbb{N}}$ (which is actually the graphic of map $sigma$). The condition on the cardinality of the fibers ensures that any element $x in T$ is the limit of a (constant) subsequence of $t$. If we furthermore assume that the space $(X, mathscr{T})$ is $T_1$, then any $x in X setminus T$ will be an accumulation point of $T$ and thus necessarily adherent to sequence $t$.



                  To conclude, given a topological space $(X, mathscr{T})$ that is non-empty, separable, first countable and $T_1$, one can always find a sequence $t$ of points within the space such that each element $x in X$ be expressible as the limit of a subsequence of $t$. This applies in particular to the topological space $(mathbb{R}, mathscr{O})$, where $mathscr{O}$ denotes the (usual) order topology.






                  share|cite|improve this answer








                  New contributor




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









                  Consider a general topological space $(X, mathscr{T})$ and a sequence of points $x in X^{mathbb{N}}$. One says that point $t in X$ is adherent to the sequence $x$ (some authors use the terminology 'cluster-point', but I don't fancy it so much) if:



                  $$(forall V, n)(V in mathscr{V}_{mathscr{T}}(x) wedge n in mathbb{N} implies (exists m)(m geqslant n wedge x_m in V ))$$



                  where $mathscr{V}_{mathscr{T}}(x)$ symbolizes the filter of neighbourhoods of point $x$ induced by the topology $mathscr{T}$. In a more descriptive fashion, $t$ is adherent to sequence $x$ if any neighbourhood of $t$ contains terms of arbitrarily high rank from the sequence $x$. If the filter of neighbourhoods of $t$ admits a countable base, then $t$ can be expressed as the limit of a subsequence of $x$. Therefore, in a space satisfying the First Axiom of Countability (i.e. all points have a countable base of neighbourhoods), the points adherent to a given sequence $x$ can be equivalently characterised as limits of subsequences of $x$.



                  Now, if the space $X$ is non-empty and separable, let us fix a certain dense subset $T subseteq X$. As $X$ is non-empty, so must $T$ be. It is not difficult to show that a non-empty countable set $M$ admits a surjection $sigma: mathbb{N} rightarrow M$ such that for each $t in M$ the fibre $sigma^{-1}({t})$ be infinite.



                  Consider such a surjection $sigma: mathbb{N} rightarrow T$ and define the sequence $t=(sigma(n))_{n in mathbb{N}}$ (which is actually the graphic of map $sigma$). The condition on the cardinality of the fibers ensures that any element $x in T$ is the limit of a (constant) subsequence of $t$. If we furthermore assume that the space $(X, mathscr{T})$ is $T_1$, then any $x in X setminus T$ will be an accumulation point of $T$ and thus necessarily adherent to sequence $t$.



                  To conclude, given a topological space $(X, mathscr{T})$ that is non-empty, separable, first countable and $T_1$, one can always find a sequence $t$ of points within the space such that each element $x in X$ be expressible as the limit of a subsequence of $t$. This applies in particular to the topological space $(mathbb{R}, mathscr{O})$, where $mathscr{O}$ denotes the (usual) order topology.







                  share|cite|improve this answer








                  New contributor




                  ΑΘΩ is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                  share|cite|improve this answer



                  share|cite|improve this answer






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                  answered 15 hours ago









                  ΑΘΩ

                  1862




                  1862




                  New contributor




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                  New contributor





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






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