Codimension of subspace of sequences that converges to $0$











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Let $bar C =$ be the double converges sequences ($(a_n)^infty_{n=-infty}$), and $bar C_0$ be the double converges sequences where the limit is $0$ for both $-+infty$. What is $codim(bar C_0)$?




I know that in general the codim would be $dim(bar C) - dim(bar C_0)$ but I cannot calculate the dimension in that case.



Any other approaches?










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  • That codimension formula is for finite-dimensional spaces, which is not the case here.
    – Dave
    Nov 13 at 21:56










  • To determine the codimension, can you find a basis for the quotient space $overline{C}/overline{C_0}$?
    – Dave
    Nov 13 at 21:58










  • @Dave So could you give me a hint for that case? Is it possible to find another space which isomorphic to $bar C setminus bar C_0$?
    – ChikChak
    Nov 13 at 21:59










  • My hint would be to try and find a basis for $overline{C}/overline{C_0}$. Try starting with a basis for $C/C_0$ where $C$ and $C_0$ are for regular sequences for $1leq n<infty$.
    – Dave
    Nov 13 at 22:00















up vote
0
down vote

favorite













Let $bar C =$ be the double converges sequences ($(a_n)^infty_{n=-infty}$), and $bar C_0$ be the double converges sequences where the limit is $0$ for both $-+infty$. What is $codim(bar C_0)$?




I know that in general the codim would be $dim(bar C) - dim(bar C_0)$ but I cannot calculate the dimension in that case.



Any other approaches?










share|cite|improve this question
























  • That codimension formula is for finite-dimensional spaces, which is not the case here.
    – Dave
    Nov 13 at 21:56










  • To determine the codimension, can you find a basis for the quotient space $overline{C}/overline{C_0}$?
    – Dave
    Nov 13 at 21:58










  • @Dave So could you give me a hint for that case? Is it possible to find another space which isomorphic to $bar C setminus bar C_0$?
    – ChikChak
    Nov 13 at 21:59










  • My hint would be to try and find a basis for $overline{C}/overline{C_0}$. Try starting with a basis for $C/C_0$ where $C$ and $C_0$ are for regular sequences for $1leq n<infty$.
    – Dave
    Nov 13 at 22:00













up vote
0
down vote

favorite









up vote
0
down vote

favorite












Let $bar C =$ be the double converges sequences ($(a_n)^infty_{n=-infty}$), and $bar C_0$ be the double converges sequences where the limit is $0$ for both $-+infty$. What is $codim(bar C_0)$?




I know that in general the codim would be $dim(bar C) - dim(bar C_0)$ but I cannot calculate the dimension in that case.



Any other approaches?










share|cite|improve this question
















Let $bar C =$ be the double converges sequences ($(a_n)^infty_{n=-infty}$), and $bar C_0$ be the double converges sequences where the limit is $0$ for both $-+infty$. What is $codim(bar C_0)$?




I know that in general the codim would be $dim(bar C) - dim(bar C_0)$ but I cannot calculate the dimension in that case.



Any other approaches?







real-analysis linear-algebra sequences-and-series hilbert-spaces dimension-theory






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edited Nov 13 at 21:56

























asked Nov 13 at 21:45









ChikChak

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  • That codimension formula is for finite-dimensional spaces, which is not the case here.
    – Dave
    Nov 13 at 21:56










  • To determine the codimension, can you find a basis for the quotient space $overline{C}/overline{C_0}$?
    – Dave
    Nov 13 at 21:58










  • @Dave So could you give me a hint for that case? Is it possible to find another space which isomorphic to $bar C setminus bar C_0$?
    – ChikChak
    Nov 13 at 21:59










  • My hint would be to try and find a basis for $overline{C}/overline{C_0}$. Try starting with a basis for $C/C_0$ where $C$ and $C_0$ are for regular sequences for $1leq n<infty$.
    – Dave
    Nov 13 at 22:00


















  • That codimension formula is for finite-dimensional spaces, which is not the case here.
    – Dave
    Nov 13 at 21:56










  • To determine the codimension, can you find a basis for the quotient space $overline{C}/overline{C_0}$?
    – Dave
    Nov 13 at 21:58










  • @Dave So could you give me a hint for that case? Is it possible to find another space which isomorphic to $bar C setminus bar C_0$?
    – ChikChak
    Nov 13 at 21:59










  • My hint would be to try and find a basis for $overline{C}/overline{C_0}$. Try starting with a basis for $C/C_0$ where $C$ and $C_0$ are for regular sequences for $1leq n<infty$.
    – Dave
    Nov 13 at 22:00
















That codimension formula is for finite-dimensional spaces, which is not the case here.
– Dave
Nov 13 at 21:56




That codimension formula is for finite-dimensional spaces, which is not the case here.
– Dave
Nov 13 at 21:56












To determine the codimension, can you find a basis for the quotient space $overline{C}/overline{C_0}$?
– Dave
Nov 13 at 21:58




To determine the codimension, can you find a basis for the quotient space $overline{C}/overline{C_0}$?
– Dave
Nov 13 at 21:58












@Dave So could you give me a hint for that case? Is it possible to find another space which isomorphic to $bar C setminus bar C_0$?
– ChikChak
Nov 13 at 21:59




@Dave So could you give me a hint for that case? Is it possible to find another space which isomorphic to $bar C setminus bar C_0$?
– ChikChak
Nov 13 at 21:59












My hint would be to try and find a basis for $overline{C}/overline{C_0}$. Try starting with a basis for $C/C_0$ where $C$ and $C_0$ are for regular sequences for $1leq n<infty$.
– Dave
Nov 13 at 22:00




My hint would be to try and find a basis for $overline{C}/overline{C_0}$. Try starting with a basis for $C/C_0$ where $C$ and $C_0$ are for regular sequences for $1leq n<infty$.
– Dave
Nov 13 at 22:00










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That is not the general definition of "codimension". The codimension of $X$ in $Y$ is the dimension of $Y/X$. So, we need to find the dimension of $bar{C}/bar{C}_0$.



Let $a = (a_n)_{n=-infty}^infty$ be the sequence such that $a_n = 0$ for all $n leq 0$, and $a_n = 1$ for all $n > 0$.



Let $b = (b_n)_{n=-infty}^infty$ be the sequence such that $b_n = 0$ for all $n geq 0$, and $b_n = 1$ for all $n < 0$.



Now, $a$ and $b$ are linearly independent, and so are their equivalence classes in $bar{C}/bar{C}_0$.



For any $c = (c_n)_{n=-infty}^infty in bar{C}$, let $L(c) = limlimits_{n to infty}c_n$ and $l(c) = limlimits_{nto-infty}c_n$. Then $c = L(c)a + l(c)b + (c - L(c)a - l(c)b)$, and begin{align*}limlimits_{ntoinfty}(c_n - L(c)a_n - l(c)b_n) &= limlimits_{ntoinfty}c_n - L(c)limlimits_{ntoinfty}a_n - l(c)limlimits_{ntoinfty}b_n \&= L(c) - L(c)1 - l(c)0 \&= 0,end{align*}
and
begin{align*}limlimits_{nto-infty}(c_n - L(c)a_n - l(c)b_n) &= limlimits_{nto-infty}c_n - L(c)limlimits_{nto-infty}a_n - l(c)limlimits_{nto-infty}b_n \&= l(c) - L(c)0 - l(c)1 \&= 0,end{align*}
so $c$ differs from a linear combination of $a$ and $b$ by an element of $bar{C}_0$, so in $bar{C}/bar{C}_0$, the equivalence class containing $c$ is a linear combination of the equivalence classes of $a$ and $b$. Thus, ${[a],[b]}$ is a basis for $bar{C}/bar{C}_0$, so $mathop{mathrm{codim}}(bar{C}_0) = dim(bar{C}/bar{C}_0) = 2$.






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    That is not the general definition of "codimension". The codimension of $X$ in $Y$ is the dimension of $Y/X$. So, we need to find the dimension of $bar{C}/bar{C}_0$.



    Let $a = (a_n)_{n=-infty}^infty$ be the sequence such that $a_n = 0$ for all $n leq 0$, and $a_n = 1$ for all $n > 0$.



    Let $b = (b_n)_{n=-infty}^infty$ be the sequence such that $b_n = 0$ for all $n geq 0$, and $b_n = 1$ for all $n < 0$.



    Now, $a$ and $b$ are linearly independent, and so are their equivalence classes in $bar{C}/bar{C}_0$.



    For any $c = (c_n)_{n=-infty}^infty in bar{C}$, let $L(c) = limlimits_{n to infty}c_n$ and $l(c) = limlimits_{nto-infty}c_n$. Then $c = L(c)a + l(c)b + (c - L(c)a - l(c)b)$, and begin{align*}limlimits_{ntoinfty}(c_n - L(c)a_n - l(c)b_n) &= limlimits_{ntoinfty}c_n - L(c)limlimits_{ntoinfty}a_n - l(c)limlimits_{ntoinfty}b_n \&= L(c) - L(c)1 - l(c)0 \&= 0,end{align*}
    and
    begin{align*}limlimits_{nto-infty}(c_n - L(c)a_n - l(c)b_n) &= limlimits_{nto-infty}c_n - L(c)limlimits_{nto-infty}a_n - l(c)limlimits_{nto-infty}b_n \&= l(c) - L(c)0 - l(c)1 \&= 0,end{align*}
    so $c$ differs from a linear combination of $a$ and $b$ by an element of $bar{C}_0$, so in $bar{C}/bar{C}_0$, the equivalence class containing $c$ is a linear combination of the equivalence classes of $a$ and $b$. Thus, ${[a],[b]}$ is a basis for $bar{C}/bar{C}_0$, so $mathop{mathrm{codim}}(bar{C}_0) = dim(bar{C}/bar{C}_0) = 2$.






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      That is not the general definition of "codimension". The codimension of $X$ in $Y$ is the dimension of $Y/X$. So, we need to find the dimension of $bar{C}/bar{C}_0$.



      Let $a = (a_n)_{n=-infty}^infty$ be the sequence such that $a_n = 0$ for all $n leq 0$, and $a_n = 1$ for all $n > 0$.



      Let $b = (b_n)_{n=-infty}^infty$ be the sequence such that $b_n = 0$ for all $n geq 0$, and $b_n = 1$ for all $n < 0$.



      Now, $a$ and $b$ are linearly independent, and so are their equivalence classes in $bar{C}/bar{C}_0$.



      For any $c = (c_n)_{n=-infty}^infty in bar{C}$, let $L(c) = limlimits_{n to infty}c_n$ and $l(c) = limlimits_{nto-infty}c_n$. Then $c = L(c)a + l(c)b + (c - L(c)a - l(c)b)$, and begin{align*}limlimits_{ntoinfty}(c_n - L(c)a_n - l(c)b_n) &= limlimits_{ntoinfty}c_n - L(c)limlimits_{ntoinfty}a_n - l(c)limlimits_{ntoinfty}b_n \&= L(c) - L(c)1 - l(c)0 \&= 0,end{align*}
      and
      begin{align*}limlimits_{nto-infty}(c_n - L(c)a_n - l(c)b_n) &= limlimits_{nto-infty}c_n - L(c)limlimits_{nto-infty}a_n - l(c)limlimits_{nto-infty}b_n \&= l(c) - L(c)0 - l(c)1 \&= 0,end{align*}
      so $c$ differs from a linear combination of $a$ and $b$ by an element of $bar{C}_0$, so in $bar{C}/bar{C}_0$, the equivalence class containing $c$ is a linear combination of the equivalence classes of $a$ and $b$. Thus, ${[a],[b]}$ is a basis for $bar{C}/bar{C}_0$, so $mathop{mathrm{codim}}(bar{C}_0) = dim(bar{C}/bar{C}_0) = 2$.






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        That is not the general definition of "codimension". The codimension of $X$ in $Y$ is the dimension of $Y/X$. So, we need to find the dimension of $bar{C}/bar{C}_0$.



        Let $a = (a_n)_{n=-infty}^infty$ be the sequence such that $a_n = 0$ for all $n leq 0$, and $a_n = 1$ for all $n > 0$.



        Let $b = (b_n)_{n=-infty}^infty$ be the sequence such that $b_n = 0$ for all $n geq 0$, and $b_n = 1$ for all $n < 0$.



        Now, $a$ and $b$ are linearly independent, and so are their equivalence classes in $bar{C}/bar{C}_0$.



        For any $c = (c_n)_{n=-infty}^infty in bar{C}$, let $L(c) = limlimits_{n to infty}c_n$ and $l(c) = limlimits_{nto-infty}c_n$. Then $c = L(c)a + l(c)b + (c - L(c)a - l(c)b)$, and begin{align*}limlimits_{ntoinfty}(c_n - L(c)a_n - l(c)b_n) &= limlimits_{ntoinfty}c_n - L(c)limlimits_{ntoinfty}a_n - l(c)limlimits_{ntoinfty}b_n \&= L(c) - L(c)1 - l(c)0 \&= 0,end{align*}
        and
        begin{align*}limlimits_{nto-infty}(c_n - L(c)a_n - l(c)b_n) &= limlimits_{nto-infty}c_n - L(c)limlimits_{nto-infty}a_n - l(c)limlimits_{nto-infty}b_n \&= l(c) - L(c)0 - l(c)1 \&= 0,end{align*}
        so $c$ differs from a linear combination of $a$ and $b$ by an element of $bar{C}_0$, so in $bar{C}/bar{C}_0$, the equivalence class containing $c$ is a linear combination of the equivalence classes of $a$ and $b$. Thus, ${[a],[b]}$ is a basis for $bar{C}/bar{C}_0$, so $mathop{mathrm{codim}}(bar{C}_0) = dim(bar{C}/bar{C}_0) = 2$.






        share|cite|improve this answer












        That is not the general definition of "codimension". The codimension of $X$ in $Y$ is the dimension of $Y/X$. So, we need to find the dimension of $bar{C}/bar{C}_0$.



        Let $a = (a_n)_{n=-infty}^infty$ be the sequence such that $a_n = 0$ for all $n leq 0$, and $a_n = 1$ for all $n > 0$.



        Let $b = (b_n)_{n=-infty}^infty$ be the sequence such that $b_n = 0$ for all $n geq 0$, and $b_n = 1$ for all $n < 0$.



        Now, $a$ and $b$ are linearly independent, and so are their equivalence classes in $bar{C}/bar{C}_0$.



        For any $c = (c_n)_{n=-infty}^infty in bar{C}$, let $L(c) = limlimits_{n to infty}c_n$ and $l(c) = limlimits_{nto-infty}c_n$. Then $c = L(c)a + l(c)b + (c - L(c)a - l(c)b)$, and begin{align*}limlimits_{ntoinfty}(c_n - L(c)a_n - l(c)b_n) &= limlimits_{ntoinfty}c_n - L(c)limlimits_{ntoinfty}a_n - l(c)limlimits_{ntoinfty}b_n \&= L(c) - L(c)1 - l(c)0 \&= 0,end{align*}
        and
        begin{align*}limlimits_{nto-infty}(c_n - L(c)a_n - l(c)b_n) &= limlimits_{nto-infty}c_n - L(c)limlimits_{nto-infty}a_n - l(c)limlimits_{nto-infty}b_n \&= l(c) - L(c)0 - l(c)1 \&= 0,end{align*}
        so $c$ differs from a linear combination of $a$ and $b$ by an element of $bar{C}_0$, so in $bar{C}/bar{C}_0$, the equivalence class containing $c$ is a linear combination of the equivalence classes of $a$ and $b$. Thus, ${[a],[b]}$ is a basis for $bar{C}/bar{C}_0$, so $mathop{mathrm{codim}}(bar{C}_0) = dim(bar{C}/bar{C}_0) = 2$.







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        answered Nov 13 at 22:03









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