Complementary subspaces of $c_{00} (\mathbb{N})$.

Question

Let $c_{00} (\mathbb{N})$ denote the space of finitely non-zero sequences, and let $(\beta_n)_{n \in \mathbb{N}} \subset \mathbb{F}$ be a sequence of scalars. Then the subsets $$X := \{(x_n)_{n \in \mathbb{N}} \in c_{00} (\mathbb{N}) \; | \; x_{2n} = 0, \; \forall n \in \mathbb{N} \}, \quad Y := \{(x_n)_{n \in \mathbb{N}} \in c_{00} (\mathbb{N}) \; | \; x_{2n-1} + \beta_n x_{2n} = 0, \; \forall n \in \mathbb{N} \}$$ are complementary subspaces of $c_{00} (\mathbb{N})$, that is, the subsets $X, Y$ are closed subspaces and $c_{00} (\mathbb{N})$ is the internal direct sum of $X$ and $Y$.

It follows readily that $X$ and $Y$ are closed subspaces of $c_{00} (\mathbb{N})$. However, I do not succeed in showing that $c_{00} (\mathbb{N})$ is the internal direct sum of $X$ and $Y$.

Answer 1

In the meantime, I have found the following solution myself:

Consider the linear operator $P : c_{00}(\mathbb{N}) \to c_{00} (\mathbb{N})$ defined by $$ P[(x_n)_{n \in \mathbb{N}}] = (x_1 + \beta_1 x_2, 0, x_3 + \beta_2 x_4, 0,...). $$ Then it follows readily that $P^2 = P$, and hence that $P$ is a projection on $c_{00} (\mathbb{N})$. Note that $P(c_{00} (\mathbb{N})) = X$ and $\text{ker}(P) = Y$. Thus $c_{00} (\mathbb{N})$ is the internal direct sum of $X$ and $Y$.

Answer 2

Consider the space $\Bbb F^2$ of pairs. You want to express $(x_1, x_2)$ as a sum $(y_1,y_2)+(z_1,z_2)$ with $(y_1,y_2)\in X$ (and so $y_2=0$) and $(z_1,z_2)\in Y$ (and so $z_1+\beta_1z_2 = 0$, i.e. $z_1=\beta_1 z_2$). You therefore have $(x_1,x_2)=(y_1,0)+(-\beta_1x_2,x_2)$. Solving for $y_1$ (which is the only unknown variable remaining) yields $(x_1, x_2)=(x_1+\beta_1 x_2,0)+(-\beta_1 x_2,x_2)$.

Now that you can do this for pairs, you can do it for sequence of pairs. You have $(x_1,x_2,x_3,\dots)$ and see it as $((x_1,x_2),(x_3,x_4),\dots)$. Doing the above transformation on all pairs (and checking that $y$ and $z$ are still in $c_{00}$, given that $x$ was in $c_{00}$) gives you that $c_{00}(\Bbb N)=X+Y$.

Then you just need to show that $X\cap Y = \{0\}$ and you have the result $c_{00}(\Bbb N)=X\oplus Y$.