Taken from Functional Analysis by Eidelman, Milman and Tsolomitis; Problem 1.6.1
PROBLEM
Consider the linear subspace $\hat{c}$ of double subsequences $x=(x_n)_{-\infty}^\infty$ such that the limits $b_1=\lim_{n\rightarrow\infty} x_n$ and $b_2=\lim_{n\rightarrow-\infty} x_n$ exist. Consider the subspace $\hat{c}_0$ of the sequences $y=(y_n)_{-\infty}^\infty$ such that $\lim_{n\rightarrow\pm\infty} y_n=0$. Find the dimension and a basis of the space $\hat{c}/\hat{c}_0$.
PROOF
For simplicity, we write $\hat{c}=c$ and $\hat{c}_0=c_0$. For any $x\in c$ we can write it as $x=(\dots, b_2+a_{-3}, b_2+a_{-2}, b_2+a_{-1}, b_1+a_{0},b_1+a_{1}, \dots)$ for $(a_n)_{-\infty}^\infty$ in $c$. We will denote this sequence as $a$.
Now consider the vectors $e_1$ which has all 0's before the 0th index, and all 1's after the 0th index, and $e_2$ with all 1's before the 0th index and all 0's after the 0th index.
So $x=b_1e_1+b_2 e_2+a$. If $e_1, e_2$ are linearly independent in $c_0$, then $e_1, e_2$ would form a basis for the quotient space $c/c_0$ and the dimension is then 2.
PROBLEM
Consider the linear subspace $\hat{c}$ of double subsequences $x=(x_n)_{-\infty}^\infty$ such that the limits $b_1=\lim_{n\rightarrow\infty} x_n$ and $b_2=\lim_{n\rightarrow-\infty} x_n$ exist. Consider the subspace $\hat{c}_0$ of the sequences $y=(y_n)_{-\infty}^\infty$ such that $\lim_{n\rightarrow\pm\infty} y_n=0$. Find the dimension and a basis of the space $\hat{c}/\hat{c}_0$.
PROOF
For simplicity, we write $\hat{c}=c$ and $\hat{c}_0=c_0$. For any $x\in c$ we can write it as $x=(\dots, b_2+a_{-3}, b_2+a_{-2}, b_2+a_{-1}, b_1+a_{0},b_1+a_{1}, \dots)$ for $(a_n)_{-\infty}^\infty$ in $c$. We will denote this sequence as $a$.
Now consider the vectors $e_1$ which has all 0's before the 0th index, and all 1's after the 0th index, and $e_2$ with all 1's before the 0th index and all 0's after the 0th index.
So $x=b_1e_1+b_2 e_2+a$. If $e_1, e_2$ are linearly independent in $c_0$, then $e_1, e_2$ would form a basis for the quotient space $c/c_0$ and the dimension is then 2.
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