Can any two disjoint nonempty convex sets in a vector space be separated by a hyperplane?

Question

Let $V$ be a normed vector space over $\mathbb{R}$, and let $A$ and $B$ be two disjoint nonempty convex subsets of $V$. A geometric form of Hahn-Banach Theorem states that $A$ and $B$ can be separated by a closed hyperplane (i.e. there is $f \in V^\ast$ and $\alpha \in \mathbb{R}$ such that $f(a) \le \alpha, \forall a \in A$ and $\alpha \le f(b), \forall b \in B$) if either $A$ or $B$ is open, or $A$ is closed and $B$ is compact. (This statement is not in the full generality.) There are examples of two disjoint nonempty convex sets which cannot be separated by a closed hyperplane.(These convex sets don't satisfy the condition of the previous statement.)

My question is: If the separating hyperplane need not be closed, can any pair of disjoint nonempty convex sets be separated by a hyperplane? More precisely, for any vector space $V$ over $\mathbb{R}$ and two disjoint nonempty convex subsets $A$ and $B$ of $V$, does there exist a linear functional $f:V\to\mathbb{R}$ and a real number $\alpha\in\mathbb{R}$ such that $f(a) \le \alpha, \forall a \in A$ and $\alpha \le f(b), \forall b \in B$?

This question doesn't involve any topological concepts. For the finite dimensional case, it is known that the separation is possible. What would happen if the underlying space is infinite dimensional?

Answer 1

I edited this post, because I misunderstood the problem. However the solution is almost the same. Please check it.

Let us prove that there is a vector space $V$ and two convex subsets of $V$, $A$ and $B$, such that the only functional $f\in V^*$ with the following property is the zero functional. The property is for every $a\in A$ and $b\in B$, there is $\alpha\in \mathbb{R}$ such that $f(a)\leq\alpha\leq f(b)$.

Consider the vector space of real sequences $\mathbb{R}^{\mathbb{N}}$.

Let $A$ be the subspace of $\mathbb{R}^{\mathbb{N}}$ formed by sequences with finitely many non zero coordinates. Notice that $A$ is convex since it is a vector space.

Let $B$ be the subset of $\mathbb{R}^{\mathbb{N}}$ formed by sequences of non negative numbers converging to $0$ with infinitely many non zero coordinates. Notice that $A\cap B=\emptyset$ and any convex combination of elements in $B$ still belongs to $B$, thus $B$ is also convex.

Now, let $V=\text{span } A\cup B$. Let $f\in V^*$ be any linear functional such that $f(a)\leq\alpha\in\mathbb{R}$, for every $a\in A$. Since $A$ is a subspace, the only possibility is $f(a)=0$, for every $a\in A$.

Suppose that $f(b)\geq 0$ for every $b\in B$.

Let $(b_n)_{n\in\mathbb{N}}\in B$ and notice that $(\sqrt{b_n})_{n\in\mathbb{N}}\in B$. Let $k>0$.

Notice that $\dfrac{\sqrt{b_n}}{k}-b_n=\sqrt{b_n}(\dfrac{1}{k}-\sqrt{b_n})$ and since $(\sqrt{b_n})_{n\in\mathbb{N}}$ converges to $0$, there is $N\in\mathbb{N}$ such that $\dfrac{1}{k}-\sqrt{b_n}> 0$ for $n>N$. Therefore, $\dfrac{\sqrt{b_n}}{k}-b_n\geq 0$ for $n>N$.

Define $c_n=0$ for $n\leq N$ and $c_n=\dfrac{\sqrt{b_n}}{k}-b_n$ for $n>N$. Thus, $(c_n)_{n\in\mathbb{N}}\in B$.

Next, define $d_n=\dfrac{\sqrt{b_n}}{k}-b_n$ for $n\leq N$ and $d_n=0$ for $n>N$. Thus, $(d_n)_{n\in\mathbb{N}}\in A$.

Therefore $(\dfrac{\sqrt{b_n}}{k})_{n\in\mathbb{N}}-(b_n)_{n\in\mathbb{N}}=(c_n)_{n\in\mathbb{N}}+(d_n)_{n\in\mathbb{N}}$.

Now, $f((\dfrac{\sqrt{b_n}}{k}))-f((b_n))=f((c_n))+f((d_n))=f((c_n))\geq 0$ then $$\frac{1}{k}f((\sqrt{b_n}))\geq f((b_n))\geq 0.$$

Since $k$ is arbitrary. This inequality implies that $f((b_n)_{n\in\mathbb{N}})=0$. Now, $(b_n)_{n\in\mathbb{N}}$ is any element of $B$. Thus, $f(b)=0$ for every $b\in B$.

Since $f\in V^*$ and $V=\text{span }A\cup B$ then $f=0$.