Let $X$ be real vector space and let $F$ be a linear subspace of the space of all linear functional $X \to \mathbb{R}$. And let $U_F$ be the weakest topology on $X$ such that all elements of $F$ are continuous.
Then I want to prove that if $E$ is a linear subspace, its closure $\overline{E}$ is given by
$\cap_{\Lambda \in F, \Lambda E \subseteq Ker(\Lambda)} Ker(\Lambda)$
The part in the proof I don't understand is proving if $x \notin \overline{E}$, then $x \notin \cap_{\Lambda \in F, \Lambda E \subseteq Ker(\Lambda)} Ker(\Lambda)$
The proof say if $x \notin \overline{E}$, there exists open and convex $U$ that contains $x$ and $U \cap E = \emptyset$. As $U$ and $E$ are convex, we can find linear functional $\lambda \colon X \to \mathbb{R}$ with $\Lambda(x) > \sup_{y \in E} \Lambda(y)$.
Then the proof says as $E$ is a linear subspace, $E \subseteq Ker(\Lambda)$, which I don't understand why.
If someone could explain that part to me, I would be grateful.
Thanks!
Best Answer
Ok, here’s the point:
Suppose that there exists a $\phi \in X^*$ such that $\phi$ separates $x$ and $E$. This is probably just simple Hahn Banach theorem.
The definition of $\phi$ separating $x$ and $E$ is that $$ \phi(x) > \sup_{y \in E} \phi(y).$$
Suppose $\phi(y) \neq 0$ for some $y \in E$. Then, since $E$ is a linear subspace, the element $ty \in E$ for all $t \in \mathbb{R}$. However, since $\phi$ is linear this means that $$ \phi(ty) = t\phi(y) $$ which goes to plus $\infty$ as $t$ gets very large. I.e. there is some $t’$ such that $$ t’\phi(y) > \phi(x)$$ which is a contradiction since $t’y \in E$.
Thus, $\phi(y) = 0$ for all $y \in E$ which means that $E \subset \ker \phi$. This is what we wanted to show.