Two disjoint convex closed sets that cannot be separated by a closed hyperplane?

Question

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!

Answer 1

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.