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












8















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















marked as duplicate by Martin R, Community Dec 9 '18 at 14:29


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
    Dec 9 '18 at 11:49
















8















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















marked as duplicate by Martin R, Community Dec 9 '18 at 14:29


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
    Dec 9 '18 at 11:49














8












8








8


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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 '18 at 11:46









Especially Lime

21.6k22858




21.6k22858










asked Dec 9 '18 at 11:31









Riccardo Cazzin

1905




1905




marked as duplicate by Martin R, Community Dec 9 '18 at 14:29


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 Dec 9 '18 at 14:29


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
    Dec 9 '18 at 11:49


















  • Otherwise said, the set ${ a_n | ninmathbb N}$ should be dense in $mathbb R$.
    – Giuseppe Negro
    Dec 9 '18 at 11:49
















Otherwise said, the set ${ a_n | ninmathbb N}$ should be dense in $mathbb R$.
– Giuseppe Negro
Dec 9 '18 at 11:49




Otherwise said, the set ${ a_n | ninmathbb N}$ should be dense in $mathbb R$.
– Giuseppe Negro
Dec 9 '18 at 11:49










4 Answers
4






active

oldest

votes


















9














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
    Dec 9 '18 at 11:40










  • That's nice, we are then using that rationals are countable. Thanks for the clarification. Regards
    – gimusi
    Dec 9 '18 at 12:00










  • Most welcome, @gimusi !
    – Kavi Rama Murthy
    Dec 9 '18 at 12:04



















4














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
    Dec 9 '18 at 12:01



















3














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





























    1














    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




























      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      9














      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
        Dec 9 '18 at 11:40










      • That's nice, we are then using that rationals are countable. Thanks for the clarification. Regards
        – gimusi
        Dec 9 '18 at 12:00










      • Most welcome, @gimusi !
        – Kavi Rama Murthy
        Dec 9 '18 at 12:04
















      9














      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
        Dec 9 '18 at 11:40










      • That's nice, we are then using that rationals are countable. Thanks for the clarification. Regards
        – gimusi
        Dec 9 '18 at 12:00










      • Most welcome, @gimusi !
        – Kavi Rama Murthy
        Dec 9 '18 at 12:04














      9












      9








      9






      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 Dec 9 '18 at 11:45

























      answered Dec 9 '18 at 11:37









      Kavi Rama Murthy

      50.1k31854




      50.1k31854












      • Could you please explain better how it would work? Thanks
        – gimusi
        Dec 9 '18 at 11:40










      • That's nice, we are then using that rationals are countable. Thanks for the clarification. Regards
        – gimusi
        Dec 9 '18 at 12:00










      • Most welcome, @gimusi !
        – Kavi Rama Murthy
        Dec 9 '18 at 12:04


















      • Could you please explain better how it would work? Thanks
        – gimusi
        Dec 9 '18 at 11:40










      • That's nice, we are then using that rationals are countable. Thanks for the clarification. Regards
        – gimusi
        Dec 9 '18 at 12:00










      • Most welcome, @gimusi !
        – Kavi Rama Murthy
        Dec 9 '18 at 12:04
















      Could you please explain better how it would work? Thanks
      – gimusi
      Dec 9 '18 at 11:40




      Could you please explain better how it would work? Thanks
      – gimusi
      Dec 9 '18 at 11:40












      That's nice, we are then using that rationals are countable. Thanks for the clarification. Regards
      – gimusi
      Dec 9 '18 at 12:00




      That's nice, we are then using that rationals are countable. Thanks for the clarification. Regards
      – gimusi
      Dec 9 '18 at 12:00












      Most welcome, @gimusi !
      – Kavi Rama Murthy
      Dec 9 '18 at 12:04




      Most welcome, @gimusi !
      – Kavi Rama Murthy
      Dec 9 '18 at 12:04











      4














      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
        Dec 9 '18 at 12:01
















      4














      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
        Dec 9 '18 at 12:01














      4












      4








      4






      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 Dec 9 '18 at 11:42









      mathcounterexamples.net

      24.4k21753




      24.4k21753












      • That's less intuitive at first!
        – gimusi
        Dec 9 '18 at 12:01


















      • That's less intuitive at first!
        – gimusi
        Dec 9 '18 at 12:01
















      That's less intuitive at first!
      – gimusi
      Dec 9 '18 at 12:01




      That's less intuitive at first!
      – gimusi
      Dec 9 '18 at 12:01











      3














      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


























        3














        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
























          3












          3








          3






          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 Dec 9 '18 at 11:51









          Peter Szilas

          10.7k2720




          10.7k2720























              1














              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


























                1














                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
























                  1












                  1








                  1






                  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












                  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












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 9 '18 at 14:09









                  ΑΘΩ

                  2363




                  2363















                      Popular posts from this blog

                      Plaza Victoria

                      Puebla de Zaragoza

                      Musa