Hahn Banach separation theorem

Question

Let $M$ be a closed subspace of $X$ and $x \in X/M$, then show that there exists $F \in X^*$ such that $F(m)=0$ for all $m \in M$ but $F(x) \neq 0$.

I want to prove this result by using the Hahn Banach separation theorem.

I already proved that Let $X$ is a Hausdorff locally convex topological vector space over $\mathbb{R}$. Let $A, B \subset X$ be non empty convex subsets and $A$ is open, and $A,B$ are disjoint. Then for any $a \in A$ and $b \in B$ there a linear function $F$ such that $F(a) < \gamma \leq F(b)$.

Can anyone suggest the option for the set $A$ and $B$ or the proof of this part?

Answer 1

According to the Hahn Banach theorem about disjoint convex closed and convex compact sets,we can separate $M$ and $x$ by a hyperplane [$f=\alpha$], which means $\forall m\in M$, we have $$f(m)\le \alpha \le f(x).$$ So $\forall \lambda >0$ we have $f(\lambda m)\le\alpha$, which implies $f(m)\le \frac{\alpha}{\lambda}$ . So we obtain $f(m)\le 0$. Similarly we can obtain $f(m)\ge 0$.

Answer 2

Since $M$ is closed, its complement is open, and $x$ is an interior point of $X\setminus M$. Since $X$ is locally convex, there is an open and convex set $A$ such that $x\in A \subset X\setminus M$. Now apply the separation theorem with $B:=M$. Conclude that $F$ is zero on $M$ by using that $M$ is a subspace (as in the answer by @YZY).