Let $Y$ be the subspace of $B(Bbb N, Bbb F)$ consisting of those sequences tending to zero. Show that $Y$ is...












1














Let $Y$ be the subspace of $B(Bbb N,Bbb F)$ consisting of those sequences tending to zero.
Show that $Y$ is separable.
Here $Bbb F= Bbb C$ or $Bbb R$ (field of complex numbers or real numbers), $Bbb N$ is the field of natural numbers, and $B(Bbb N,Bbb F):= {f : Bbb N → Bbb F : f text{ is bounded}}$.






real-analysis general-topology







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  • A dense subspace is all sequences of rational numbers (or rational real and imaginary part) that are eventually $0$. (What topology do you consider on the space of bounded sequences?)
    – Mirko
    Nov 20 at 3:16












  • Thanks for the coment. this question was in the chapter of my book about metric spaces so, I am not sure what type of topologies you could consider in the space of bounded sequences, but you could turn it into a metric space through the metric: D(f,g) := sup x∈S d(f(x), g(x)).
    – Hypathius
    Nov 20 at 3:31


















1














Let $Y$ be the subspace of $B(Bbb N,Bbb F)$ consisting of those sequences tending to zero.
Show that $Y$ is separable.
Here $Bbb F= Bbb C$ or $Bbb R$ (field of complex numbers or real numbers), $Bbb N$ is the field of natural numbers, and $B(Bbb N,Bbb F):= {f : Bbb N → Bbb F : f text{ is bounded}}$.






real-analysis general-topology







share|cite|improve this question
























  • A dense subspace is all sequences of rational numbers (or rational real and imaginary part) that are eventually $0$. (What topology do you consider on the space of bounded sequences?)
    – Mirko
    Nov 20 at 3:16












  • Thanks for the coment. this question was in the chapter of my book about metric spaces so, I am not sure what type of topologies you could consider in the space of bounded sequences, but you could turn it into a metric space through the metric: D(f,g) := sup x∈S d(f(x), g(x)).
    – Hypathius
    Nov 20 at 3:31
















1












1








1


1





Let $Y$ be the subspace of $B(Bbb N,Bbb F)$ consisting of those sequences tending to zero.
Show that $Y$ is separable.
Here $Bbb F= Bbb C$ or $Bbb R$ (field of complex numbers or real numbers), $Bbb N$ is the field of natural numbers, and $B(Bbb N,Bbb F):= {f : Bbb N → Bbb F : f text{ is bounded}}$.






real-analysis general-topology







share|cite|improve this question















Let $Y$ be the subspace of $B(Bbb N,Bbb F)$ consisting of those sequences tending to zero.
Show that $Y$ is separable.
Here $Bbb F= Bbb C$ or $Bbb R$ (field of complex numbers or real numbers), $Bbb N$ is the field of natural numbers, and $B(Bbb N,Bbb F):= {f : Bbb N → Bbb F : f text{ is bounded}}$.






real-analysis general-topology




real-analysis general-topology






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edited Nov 20 at 3:20









Tianlalu

3,12321038




3,12321038










asked Nov 20 at 3:11









Hypathius

62




62












  • A dense subspace is all sequences of rational numbers (or rational real and imaginary part) that are eventually $0$. (What topology do you consider on the space of bounded sequences?)
    – Mirko
    Nov 20 at 3:16












  • Thanks for the coment. this question was in the chapter of my book about metric spaces so, I am not sure what type of topologies you could consider in the space of bounded sequences, but you could turn it into a metric space through the metric: D(f,g) := sup x∈S d(f(x), g(x)).
    – Hypathius
    Nov 20 at 3:31




















  • A dense subspace is all sequences of rational numbers (or rational real and imaginary part) that are eventually $0$. (What topology do you consider on the space of bounded sequences?)
    – Mirko
    Nov 20 at 3:16












  • Thanks for the coment. this question was in the chapter of my book about metric spaces so, I am not sure what type of topologies you could consider in the space of bounded sequences, but you could turn it into a metric space through the metric: D(f,g) := sup x∈S d(f(x), g(x)).
    – Hypathius
    Nov 20 at 3:31


















A dense subspace is all sequences of rational numbers (or rational real and imaginary part) that are eventually $0$. (What topology do you consider on the space of bounded sequences?)
– Mirko
Nov 20 at 3:16






A dense subspace is all sequences of rational numbers (or rational real and imaginary part) that are eventually $0$. (What topology do you consider on the space of bounded sequences?)
– Mirko
Nov 20 at 3:16














Thanks for the coment. this question was in the chapter of my book about metric spaces so, I am not sure what type of topologies you could consider in the space of bounded sequences, but you could turn it into a metric space through the metric: D(f,g) := sup x∈S d(f(x), g(x)).
– Hypathius
Nov 20 at 3:31






Thanks for the coment. this question was in the chapter of my book about metric spaces so, I am not sure what type of topologies you could consider in the space of bounded sequences, but you could turn it into a metric space through the metric: D(f,g) := sup x∈S d(f(x), g(x)).
– Hypathius
Nov 20 at 3:31












2 Answers
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I assume $B(N,F)$ is a metric space with the metric $$d_infty(x_n,y_n)=sup_n vert x_n-y_n vert$$The set $$S:={y_n in Bbb Q: text{finitely many $y_n$ are non zero }}$$ is a countable dense subset of $Y$




Countable part is easy(?). For the other part, take $x_n=(x_1,x_2,...)in Y$. Then $lim x_n=0$. That is, $$(forall varepsilon>0:exists N in Bbb N): vert x_n vert< varepsilon$$ whenever $n >N$. Choose rational $y_1,y_2,...,y_N$ so that $vert x_i-y_i vert < varepsilon$ for $1 leq i leq N$. Define $y_n=(y_1,y_2,..,y_N,0,0,cdots,0)$, we have $y in S$ and $$d_infty(x_n,y_n) < varepsilon$$






share|cite|improve this answer





















  • The question didn't specify any metrics, so the result should hold for any metric. Now, is S countable because it is a collection of countable sets? Also, the conclusion that $d(x_n,y_n)<ε$, implies that $B_ε(y_n)$ has non empty intersection with $Bbb F$ for all ε>0? (here $B_ε(x)$ is just the open ball centered on x. )
    – Hypathius
    Nov 20 at 3:57












  • Yes for the first one! For the second one, dense means for any $x_n in Y$, there exists $y_n in S$ so that $d(x_n , y_n)< varepsilon$. Thats what I prove!
    – Chinnapparaj R
    Nov 20 at 4:01










  • Oh, I see it now, the sequences from $S$ that need to have nonempty intersection with the sequences from $Y$ for any ε>0. Im insisting on this definition because it is the one used in my book. Thanks for clarifying.
    – Hypathius
    Nov 20 at 4:06





















0














Let $D$ be the set of sequences with rational values that have only at most finitely many non-zero values, This is essentially $bigcup_n mathbb{Q}^n$ so a countable set.



Then for any sequence $(x_n)$ tending to $0$ and any $varepsilon>0$ find $N$ such that all $|x_n| < frac{varepsilon}{2}$ for $n ge N$. Then note that $(x_1,ldots,x_N)$ can be approximated in the sup-metric (on $N$ coordinates) by a rational vector $(q_1,ldots,q_N)$ such that $sup_{i=1,ldots N} |x_i - q_i| < varepsilon$.



Then $(q_1,ldots, q_N,0,0,0,ldots)$ is in $D$ and in $B((x_n),varepsilon)$.



This shows density of $D$ in the set of sequences under consideration.






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    2 Answers
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    2 Answers
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    I assume $B(N,F)$ is a metric space with the metric $$d_infty(x_n,y_n)=sup_n vert x_n-y_n vert$$The set $$S:={y_n in Bbb Q: text{finitely many $y_n$ are non zero }}$$ is a countable dense subset of $Y$




    Countable part is easy(?). For the other part, take $x_n=(x_1,x_2,...)in Y$. Then $lim x_n=0$. That is, $$(forall varepsilon>0:exists N in Bbb N): vert x_n vert< varepsilon$$ whenever $n >N$. Choose rational $y_1,y_2,...,y_N$ so that $vert x_i-y_i vert < varepsilon$ for $1 leq i leq N$. Define $y_n=(y_1,y_2,..,y_N,0,0,cdots,0)$, we have $y in S$ and $$d_infty(x_n,y_n) < varepsilon$$






    share|cite|improve this answer





















    • The question didn't specify any metrics, so the result should hold for any metric. Now, is S countable because it is a collection of countable sets? Also, the conclusion that $d(x_n,y_n)<ε$, implies that $B_ε(y_n)$ has non empty intersection with $Bbb F$ for all ε>0? (here $B_ε(x)$ is just the open ball centered on x. )
      – Hypathius
      Nov 20 at 3:57












    • Yes for the first one! For the second one, dense means for any $x_n in Y$, there exists $y_n in S$ so that $d(x_n , y_n)< varepsilon$. Thats what I prove!
      – Chinnapparaj R
      Nov 20 at 4:01










    • Oh, I see it now, the sequences from $S$ that need to have nonempty intersection with the sequences from $Y$ for any ε>0. Im insisting on this definition because it is the one used in my book. Thanks for clarifying.
      – Hypathius
      Nov 20 at 4:06


















    1














    I assume $B(N,F)$ is a metric space with the metric $$d_infty(x_n,y_n)=sup_n vert x_n-y_n vert$$The set $$S:={y_n in Bbb Q: text{finitely many $y_n$ are non zero }}$$ is a countable dense subset of $Y$




    Countable part is easy(?). For the other part, take $x_n=(x_1,x_2,...)in Y$. Then $lim x_n=0$. That is, $$(forall varepsilon>0:exists N in Bbb N): vert x_n vert< varepsilon$$ whenever $n >N$. Choose rational $y_1,y_2,...,y_N$ so that $vert x_i-y_i vert < varepsilon$ for $1 leq i leq N$. Define $y_n=(y_1,y_2,..,y_N,0,0,cdots,0)$, we have $y in S$ and $$d_infty(x_n,y_n) < varepsilon$$






    share|cite|improve this answer





















    • The question didn't specify any metrics, so the result should hold for any metric. Now, is S countable because it is a collection of countable sets? Also, the conclusion that $d(x_n,y_n)<ε$, implies that $B_ε(y_n)$ has non empty intersection with $Bbb F$ for all ε>0? (here $B_ε(x)$ is just the open ball centered on x. )
      – Hypathius
      Nov 20 at 3:57












    • Yes for the first one! For the second one, dense means for any $x_n in Y$, there exists $y_n in S$ so that $d(x_n , y_n)< varepsilon$. Thats what I prove!
      – Chinnapparaj R
      Nov 20 at 4:01










    • Oh, I see it now, the sequences from $S$ that need to have nonempty intersection with the sequences from $Y$ for any ε>0. Im insisting on this definition because it is the one used in my book. Thanks for clarifying.
      – Hypathius
      Nov 20 at 4:06
















    1












    1








    1






    I assume $B(N,F)$ is a metric space with the metric $$d_infty(x_n,y_n)=sup_n vert x_n-y_n vert$$The set $$S:={y_n in Bbb Q: text{finitely many $y_n$ are non zero }}$$ is a countable dense subset of $Y$




    Countable part is easy(?). For the other part, take $x_n=(x_1,x_2,...)in Y$. Then $lim x_n=0$. That is, $$(forall varepsilon>0:exists N in Bbb N): vert x_n vert< varepsilon$$ whenever $n >N$. Choose rational $y_1,y_2,...,y_N$ so that $vert x_i-y_i vert < varepsilon$ for $1 leq i leq N$. Define $y_n=(y_1,y_2,..,y_N,0,0,cdots,0)$, we have $y in S$ and $$d_infty(x_n,y_n) < varepsilon$$






    share|cite|improve this answer












    I assume $B(N,F)$ is a metric space with the metric $$d_infty(x_n,y_n)=sup_n vert x_n-y_n vert$$The set $$S:={y_n in Bbb Q: text{finitely many $y_n$ are non zero }}$$ is a countable dense subset of $Y$




    Countable part is easy(?). For the other part, take $x_n=(x_1,x_2,...)in Y$. Then $lim x_n=0$. That is, $$(forall varepsilon>0:exists N in Bbb N): vert x_n vert< varepsilon$$ whenever $n >N$. Choose rational $y_1,y_2,...,y_N$ so that $vert x_i-y_i vert < varepsilon$ for $1 leq i leq N$. Define $y_n=(y_1,y_2,..,y_N,0,0,cdots,0)$, we have $y in S$ and $$d_infty(x_n,y_n) < varepsilon$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 20 at 3:35









    Chinnapparaj R

    5,2601826




    5,2601826












    • The question didn't specify any metrics, so the result should hold for any metric. Now, is S countable because it is a collection of countable sets? Also, the conclusion that $d(x_n,y_n)<ε$, implies that $B_ε(y_n)$ has non empty intersection with $Bbb F$ for all ε>0? (here $B_ε(x)$ is just the open ball centered on x. )
      – Hypathius
      Nov 20 at 3:57












    • Yes for the first one! For the second one, dense means for any $x_n in Y$, there exists $y_n in S$ so that $d(x_n , y_n)< varepsilon$. Thats what I prove!
      – Chinnapparaj R
      Nov 20 at 4:01










    • Oh, I see it now, the sequences from $S$ that need to have nonempty intersection with the sequences from $Y$ for any ε>0. Im insisting on this definition because it is the one used in my book. Thanks for clarifying.
      – Hypathius
      Nov 20 at 4:06




















    • The question didn't specify any metrics, so the result should hold for any metric. Now, is S countable because it is a collection of countable sets? Also, the conclusion that $d(x_n,y_n)<ε$, implies that $B_ε(y_n)$ has non empty intersection with $Bbb F$ for all ε>0? (here $B_ε(x)$ is just the open ball centered on x. )
      – Hypathius
      Nov 20 at 3:57












    • Yes for the first one! For the second one, dense means for any $x_n in Y$, there exists $y_n in S$ so that $d(x_n , y_n)< varepsilon$. Thats what I prove!
      – Chinnapparaj R
      Nov 20 at 4:01










    • Oh, I see it now, the sequences from $S$ that need to have nonempty intersection with the sequences from $Y$ for any ε>0. Im insisting on this definition because it is the one used in my book. Thanks for clarifying.
      – Hypathius
      Nov 20 at 4:06


















    The question didn't specify any metrics, so the result should hold for any metric. Now, is S countable because it is a collection of countable sets? Also, the conclusion that $d(x_n,y_n)<ε$, implies that $B_ε(y_n)$ has non empty intersection with $Bbb F$ for all ε>0? (here $B_ε(x)$ is just the open ball centered on x. )
    – Hypathius
    Nov 20 at 3:57






    The question didn't specify any metrics, so the result should hold for any metric. Now, is S countable because it is a collection of countable sets? Also, the conclusion that $d(x_n,y_n)<ε$, implies that $B_ε(y_n)$ has non empty intersection with $Bbb F$ for all ε>0? (here $B_ε(x)$ is just the open ball centered on x. )
    – Hypathius
    Nov 20 at 3:57














    Yes for the first one! For the second one, dense means for any $x_n in Y$, there exists $y_n in S$ so that $d(x_n , y_n)< varepsilon$. Thats what I prove!
    – Chinnapparaj R
    Nov 20 at 4:01




    Yes for the first one! For the second one, dense means for any $x_n in Y$, there exists $y_n in S$ so that $d(x_n , y_n)< varepsilon$. Thats what I prove!
    – Chinnapparaj R
    Nov 20 at 4:01












    Oh, I see it now, the sequences from $S$ that need to have nonempty intersection with the sequences from $Y$ for any ε>0. Im insisting on this definition because it is the one used in my book. Thanks for clarifying.
    – Hypathius
    Nov 20 at 4:06






    Oh, I see it now, the sequences from $S$ that need to have nonempty intersection with the sequences from $Y$ for any ε>0. Im insisting on this definition because it is the one used in my book. Thanks for clarifying.
    – Hypathius
    Nov 20 at 4:06













    0














    Let $D$ be the set of sequences with rational values that have only at most finitely many non-zero values, This is essentially $bigcup_n mathbb{Q}^n$ so a countable set.



    Then for any sequence $(x_n)$ tending to $0$ and any $varepsilon>0$ find $N$ such that all $|x_n| < frac{varepsilon}{2}$ for $n ge N$. Then note that $(x_1,ldots,x_N)$ can be approximated in the sup-metric (on $N$ coordinates) by a rational vector $(q_1,ldots,q_N)$ such that $sup_{i=1,ldots N} |x_i - q_i| < varepsilon$.



    Then $(q_1,ldots, q_N,0,0,0,ldots)$ is in $D$ and in $B((x_n),varepsilon)$.



    This shows density of $D$ in the set of sequences under consideration.






    share|cite|improve this answer


























      0














      Let $D$ be the set of sequences with rational values that have only at most finitely many non-zero values, This is essentially $bigcup_n mathbb{Q}^n$ so a countable set.



      Then for any sequence $(x_n)$ tending to $0$ and any $varepsilon>0$ find $N$ such that all $|x_n| < frac{varepsilon}{2}$ for $n ge N$. Then note that $(x_1,ldots,x_N)$ can be approximated in the sup-metric (on $N$ coordinates) by a rational vector $(q_1,ldots,q_N)$ such that $sup_{i=1,ldots N} |x_i - q_i| < varepsilon$.



      Then $(q_1,ldots, q_N,0,0,0,ldots)$ is in $D$ and in $B((x_n),varepsilon)$.



      This shows density of $D$ in the set of sequences under consideration.






      share|cite|improve this answer
























        0












        0








        0






        Let $D$ be the set of sequences with rational values that have only at most finitely many non-zero values, This is essentially $bigcup_n mathbb{Q}^n$ so a countable set.



        Then for any sequence $(x_n)$ tending to $0$ and any $varepsilon>0$ find $N$ such that all $|x_n| < frac{varepsilon}{2}$ for $n ge N$. Then note that $(x_1,ldots,x_N)$ can be approximated in the sup-metric (on $N$ coordinates) by a rational vector $(q_1,ldots,q_N)$ such that $sup_{i=1,ldots N} |x_i - q_i| < varepsilon$.



        Then $(q_1,ldots, q_N,0,0,0,ldots)$ is in $D$ and in $B((x_n),varepsilon)$.



        This shows density of $D$ in the set of sequences under consideration.






        share|cite|improve this answer












        Let $D$ be the set of sequences with rational values that have only at most finitely many non-zero values, This is essentially $bigcup_n mathbb{Q}^n$ so a countable set.



        Then for any sequence $(x_n)$ tending to $0$ and any $varepsilon>0$ find $N$ such that all $|x_n| < frac{varepsilon}{2}$ for $n ge N$. Then note that $(x_1,ldots,x_N)$ can be approximated in the sup-metric (on $N$ coordinates) by a rational vector $(q_1,ldots,q_N)$ such that $sup_{i=1,ldots N} |x_i - q_i| < varepsilon$.



        Then $(q_1,ldots, q_N,0,0,0,ldots)$ is in $D$ and in $B((x_n),varepsilon)$.



        This shows density of $D$ in the set of sequences under consideration.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 20 at 4:37









        Henno Brandsma

        104k346113




        104k346113






























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