I am trying to understand, why the Separation Theorem from Hahn-Banach needs one of the sets to be open.
The theorem states:
$$\text{Re}\langle x',x_1 \rangle < \alpha \leq\text{Re}\langle x', x_{2} \rangle$$
for $x_1 \in C_1, x_2 \in C_2$ and $C_1,C_2$ convex, disjoint sets with $C_1$ open.
Now I am looking at the example in $c_{00}$ referenced here:
https://mathoverflow.net/questions/37551/a-counter-example-to-hahn-banach-separation-theorem-of-convex-sets
My trouble lies in understanding why $\mathcal{l}(x) = 0$ follows from $\pm \mathcal{l}(x) + \delta \mathcal{l}(y) \geq 0$.
Best Answer
Since the inequality $$ \pm l(x)\leq \delta\,l(y) $$ holdsl for all $\delta>0$, you get that $\pm l(x)\leq0$. This forces $l(x)=0$.
The above is argued in the real case. In the complex case you would get $\operatorname{Re} l(x)=0$. As this can be done for the element $ix$, we also get $$0=\operatorname{Re}l(ix)=-\operatorname{Im} l(x).$$