I'm trying to know a counterexample for separation theorem: If $A$ and $B$ are two disjoint convex set in a topological vector space $X$, one of them has nonempty interior, then there exists $f\in X^*\setminus\{0\}$ such that $$\sup_{x\in A} f(x) \leq \inf_{y\in B} f(y).$$ If $\mbox{int}(A) \cup \mbox{int}(B)=\emptyset$ then $A$ and $B$ might not be separated.
Let $X=\ell^2$ with the complete basis system $\{e_n\}_{n\in \mathbb N}$. Define $$B=\mbox{span}\{e_1\}$$ and $$A=\left\{x\in X: x=\sum x_ne_n, x_1 \geq n^3\left|x_n -\frac{1}{n^2}\right| \mbox{ for all } n\geq 2\right\}.$$ Prove: $A, B$ are convex, $\mbox{int}(A)=\emptyset$, $\mbox{int}(B)=\emptyset$, $\overline{A-B}=X$.
Anyone can help me to show that $\overline{A-B}=X$ ? Thank you very much!
Let $y_k = e_1 + \frac1{k^2}e_k$ for all $k\geqslant 2$. Then $e_k = k^2(y_k - e_1) \in A-B$.Thus $A-B$ contains all finite linear combinations of the basis elements, hence $\overline{A-B} = X$.