The closure/interior of a open/closed half space Let $X$ be a real topological vector space.
A subset $S$ of $X$ is called an open half-space if there is a continuous linear  functional $f:X\longrightarrow \mathbb{R}$ such that $S=\{x\in X: f(x)>a\}$ for some $a$. S is a closed half-space if there is a continuous linear functiona $f:X\longrightarrow \mathbb{R}$ such that $S=\{x\in X: f(x)\geq a\}$ for some $a$.
I want to show that: The closure of an open half-space is a closed half-space and the interior of a closed half-space is an open half-space.
I am trying to use the following result:  Let $X$ be a TVS and let $A$ be a convex subset of $X$. Then, if $a\in int A$ and $b\in cl A$, then $[a,b)=\{tb+(1-t)a, 0\leq t<1\}\subset int A$. But am having problems with inequalities.
 A: I am assuming that $f \neq 0$.
Let $C = \{ x | f(x) \ge a \}$ and $U = \{ x | f(x) > a \}$.
We have $U \subset C$, $C$ is closed and $U$ is open.
Let $x_0$ be such that $f(x_0) > a$.
Suppose $F$ is closed and $U \subset F$.
Suppose $f(x_1) = a$ and let $y_n = {1 \over n} x_0 + (1-{1 \over n} ) x_1$, note that $y_n \in U$ for all $n$ and $y_n \to x_1$. Since $F$ is closed we see that $x_1 \in F$ and hence $C \subset F$ and so $C = \overline{U}$.
Now suppose $V$ is open, $V \subset C$ and pick $v \in V$. Since $V$ is open, we must have $v-t x_0 \in V$ for some $t>0$ and so $f(v-tx_0) \ge a$ from which we get $f(v) > (1+t) a \ge a$. In particular, $v \in U$ and so $C^\circ = U$.
A: Let me try to answer one part of your question, with the hope that the other is similar. Try to prove that the interior of a closed half-space is an open half-space.
So let $S=\{x\in X: f(x)\ge a\}$ be a closed half space and $V$ be its interior.
If $f$ is a constant function then $S=X=V=\{x\in X: f(x)>a-1\}$, we are done (and $a=0$ but nevermind).
Assume that $f$ is not constant. Then $T=\{x\in X: f(x)>a\}$ is a non-empty open half-space. Clearly
$T\subseteq V$. Suppose $x\in V\setminus T$. Then $f(x)=a$. Take a symmetric
neighborhood $U$ of $0$ such that $x+U\subseteq V$. Take $z\in U$ such that $f(z)>0$. Then $x+z\in V$ but $f(x+z)=a+f(z)>a$, a contradiction. Hence $T=V$.
