sum of two closed subspaces of a Banach space need not be closed. show that the sum of two closed subspaces of a Banach space need not be closed.
I find this example :
Let $M$ and $N$ be the closed subspaces of $c_{0}$ defined by the formulas
$M=\left\{\left(\alpha_{n}\right):\left(\alpha_{n}\right) \in c_{0}, \alpha_{m}=m \alpha_{m-1} \text { for each even } m\right\}$
$N=\left\{\left(\alpha_{n}\right):\left(\alpha_{n}\right) \in c_{0}, \alpha_{m}=0 \text { for each odd } m\right\}$
 .
($c_{0}$ is the Banach space of sequences converging to zero ) 
but i can't prove : $M+N$ is a proper dense subspace of $c_{0}$ ?

To show that $M+N$ is dense, show that $c_{00}\subset M+N$. Let $(a_n)\in c_{00}$, i.e. there exists $n_0$ such that for all $n> 2n_0$ it is $a_n=0$. Now we need to write $(a_n)$ as a sum of $M+N$. We want to find complex numbers $x_i,y_i$ such that
$$(a_1,a_2,\dots,a_{n_0},0,0,0,\dots)=$$ $$=(y_1,2y_1,y_3,4y_3,y_5,6y_5,\dots,2n_0y_{2n_0-1},0,0,0,\dots)+$$ $$(0,x_2,0,x_4,0,\dots,0,x_{2n_0-2},0,x_{2n_0},0,0,0\dots)$$
so we want solutions to the system of equations $a_1=y_1$ and $a_2=x_2+2y_1$, $a_3=y_3$ and $a_4=x_4+4y_3$ and so on. Well, this shows exactly how we can write $(a_n)$ as a sum of $M+N$. To be more specific:
$$(a_n)=$$ $$=(a_1,2a_1,a_3,4a_3,a_5,6a_5,\dots,a_{2n_0-1},2n_0a_{2n_0-1},0,0,0,\dots)+$$ $$(0,a_2-2a_1,0,a_4-4a_3,0,a_6-6a_5,0,\dots,0,a_{2n_0}-2n_0a_{2n_0-1},0,0,0,\dots)$$
The first sequence on this sum belongs to $M$ and the second belongs to $N$.
Since $c_{00}\subset M+N$ as we just showed, we have $\overline{c_{00}}\subset\overline{M+N}$, but $c_{00}$ is dense in $c_0$, so $c_0\subset\overline{M+N}$ and we are done.
To show that it is a proper subspace, consider the sequence $a=(\frac{1}{n})_{n=1}^\infty$. If $a\in M+N$, we have
$$(\tfrac{1}{n})=(y_1,2y_1,y_3,2y_3,\dots)+(0,x_2,0,x_4,0,\dots)$$
so $y_{2k-1}=\frac{1}{2k-1}$ and $2ky_{2k-1}+x_{2k}=\frac{1}{2k}$, so $x_{2k}=\frac{1}{2k}-\frac{2k}{2k-1}=\frac{-4k^2-2k-1}{4k^2-2k}\longrightarrow-1$, which is impossible since $(x_k)\in c_0$, so $x_{2k}\to 0$.
