Two disjoint convex closed sets that cannot be separated by a closed hyperplane? When I was reading "Functional Analysis" written by Brezis, I noticed the following counterexample, saying that two disjoint closed convex sets may not be separated by a closed hyperplane.  
Let $E=l^{1}$ (the space of real, absolutely convergent series) and define 
$$X=\{x=(x_{n})_{n\geq1}\in E:x_{2n}=0 \ \forall n\geq 1\}$$
and 
$$Y=\{y=(y_{n})_{n\geq1}\in E:y_{2n}=\frac {1}{2^{n}}y_{2n-1} \ \forall n\geq 1\}$$
(i) Show that $X$ and $Y$ are closed linear spaces and $\overline{X+Y}=E$. 
(ii) Let $c=(c_{n})_{n\geq1}\in E$ be defined by $c_{2n}=\frac{1}{2^{n}}$ and $c_{2n-1}=0$ for all $n\geq 1$. Check that $c\notin X+Y$.
(iii) Set $Z=X-c$ (then $Y\bigcap Z=\emptyset$). Show that $Y$ and $Z$ cannot be separated by a closed hyperplane.
I can prove that (i) and (ii) hold. But I'm stuck in part (iii). How can I prove $Y$ and $Z$ cannot be separated by a closed hyperplane? Thanks!
 A: Suppose that they are separated by a closed hyperplane, so that you have a continuous functional $\varphi:E\to \mathbb{R}$ and some $\alpha\in\mathbb{R}$ such that $\varphi(x-c)>\alpha>\varphi(y)$ for all $x\in X, y\in Y$. Note that since $Y$ is a vector space we must have that $\varphi(Y)=\{0\}$ (because otherwise its image will be $\mathbb{R}$). Similarly, $\varphi(X)=\{0\}$ because otherwise $\varphi(X-c)=\varphi(X)-\varphi(c)=\mathbb{R}-\varphi(c)=\mathbb{R}$. We conclude that $X+Y\subseteq \ker(\varphi)$, and from continuity $E=\overline{X+Y}\subseteq \ker(\varphi)$ and you get a contradiction to the separation.
A: As requested by another user, here is a proof of (i):
Consider the continuous projections $\pi_k:E\to\Bbb R, (c_n)_n\mapsto c_k$. We have $X=\bigcap_{n=1}^\infty \pi_{2n}^{-1}(0)$, so $X$ is closed as a intersection of closed subsets. Similarly $Y=\bigcap_{n=1}^\infty (\pi_{2n}-\frac{1}{2^n}\pi_{2n-1})^{-1}(0)$ is closed.
Now let $\varepsilon>0$ and $(c_n)_n\in E$. Let $N\in\Bbb N$ big enough such that $\sum_{n>N}|c_n|<\varepsilon$. We may assume that $N$ is even. Then define $(y_n)_n\in Y$ as $$(y_n)=(*,c_2,*,c_4,*,\dots,*,c_N,0,0,\dots)$$
Note that the $*$-values are uniquely determined by the condition $(y_n)\in Y$.
Then define $(x_n)_n\in E$ by $x_k=c_k-y_k$ for $k\leq N$ and $x_k=0$ for $k>N$. Note that by definition $(x_n)_n\in X$. We have:
\begin{align*}
\|((x_n)_n+(y_n)_n)-(c_n)_n\|&=\sum_{n=1}^N |x_n+y_n-c_n|+\sum_{n=N+1}^\infty|x_n+y_n-c_n|\\
&=\sum_{n=N+1}^\infty|c_n|\\
&<\varepsilon
\end{align*}
Thus $(c_n)_n\in \overline{X+Y}$.
