# Separation in infinite dimensional normed space

I would like to construct some counterexamples:

• $E$ is an infinite dimensional normed space. Let $C, D$ be nonempty convex subsets in $E$ such that $$C\cap D=\emptyset.$$ There is no vector $f\in E^*\setminus \{0\}$ and $\alpha\in \mathbb{R}$ such that $$f(x)\leq \alpha\leq f(y) \quad \forall x\in C, \quad \forall y\in D.$$

• $E$ is an infinite dimensional normed space. Let $C, D$ be nonempty convex subsets in $E$ such that $$C\cap D=\emptyset$$ and $C$ is compact. There is no vector $f\in E^*\setminus\{0\}$ and $\alpha\in \mathbb{R}$ such that $$f(x)< \alpha< f(y) \quad\forall x\in C, \quad \forall y\in D.$$

Thank you for all comments and helping.

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For the first bullet point see this thread on mathoverflow. For the second start with the closed unit disk $C$ and $D = \{y \gt 1\}$. –  commenter Oct 25 '12 at 11:08