I'm currently working through Conway's Functional Analysis, and have stumbled across a result that I can't seem to find the justification for myself:
Let $\mathcal{X}$ be a Real Locally Convex Space and $A$ a subet of $\mathcal{X}$ . The closed linear span of $A$ is the intersection of all closed hyperplanes containing $A$.
It seems to be a corollary to the result of the closed convex hull of such $A$ being the intersection of all of all the closed half-spaces containing $A$, but I can't seem to work it out myself. Any help would be appreciated.
Solved myself - First note that all closed hyperplanes in $\mathcal{X}$ are kernels of elements of $\mathcal{X}^*$, and vice versa. The closed linear span of $A$ clearly lies in this intersection. To show opposite containment, we note that since the closed linear span of $A$ is a closed convex subset of $\mathcal{X}$, it is strictly separate from any point $x_0\notin A$. ie, there exists $f\in\mathcal{X}^*$ and $\alpha\in\mathbb{R}$ such that $f(x_0)<\alpha$ and $f(a)>\alpha$ for all $a\in A$. Then the fact that $x_0$ is not in the intersection easily follows.