I am trying to solve the following problem.(p133 of Taylor's ''introduction to functional analysis'')
Let $X$ be a real topological linear space. If $S \subset X,$ a support of $S$ is a hyperplane $M$ such that $S$ lies on one side of $M$ and $S \cap M \not = \emptyset$. A closed convex set with a nonempty interior is called a convex body. Show that a support of a convex body is closed.
I think the support does not intersect the interior of $S$. So the support is not dense in $X$, which implies it is closed.
But, I have some trouble with a proof that the support does not intersect the interior of $S$. Please give me some advice or reference.
Thank you.
Edit:
Thanks to the comments, I have proved it. I will sketch here.
Suppose $S$ be a neighborhood of zero and $S\subset \{x \in X| f(x) \geq c\}$ for some nonzero linear functional $f.$
For any $x \in \mathrm{int}S$, there exitsts a balanced neiborhoof $U$ of zero such that $x+U\subset S.$ Since $U$ is absorbing and $f$ is not identically zero, there exists $u \in U$ such that $f(u) \not = 0.$ We may assume $f(u) > 0.$ Since $x -u \in x+U \subset S,$ we have $f(x)-f(u)\geq c$, which implies $f(x)>c.$