Let $A$ be a convex and closed subset of $\mathbb R²$ which not contains $0$. Show that there is a vectorial line separated from A.

Question

I have not studied Hahn-Banach theorem yet, so I have to do without. I made a drawing, it's quite clear on it :)

Thank you for your help.

Answer 1

Let $r = \inf \{d(0,a) \mid a \in A\}$. By assumption $r > 0$. By definition $A \subset B(0,r)^c$.

Moreover there is a unique point $p \in A \cap \partial B(0,r)$. For the unicity, if there is two points on the boundary, then by convexity the segment between them is inside the circle, contradicting definition of $r$. The existence is because $A$ is closed.

As you said, take $L$ perpendicular to the segment $\overline{ Op}$. Assume there is a point $q \in L \cap A$. Then, by convexity the segment $\overline{qp} \subset A$. Now the segment $\overline{qp}$ intersect $\partial B(0,r)$ in some point $p' \neq p$ which is a contradiction by unicity of the point of the boundary.

Remark : if you are not convinced that $\overline{qp}$ intersect $\partial B(0,r)$ in another point that $p$, here is an algebraic proof. By rotating we can assume $p = (1,0)$ and $q = (0,y)$. The line between $p$ and $q$ is parametrized by $(1+t, - yt)$ (for example). Now we want to solve the equation $f(t) = 0$ where $f(t) = (1+t)^2 + (yt)^2 - 1$. A little calculation gives $t = \frac{-2}{y^2 + 1}$ is the desired point.