User:RaisinBread/Sandbox

Work in progress
Stone Theorem

Theorem
Let $K$ be a compact metric space and $\left(\mathcal{C}(K,\R),\|\cdot\|_{\infty}\right)$ be the Banach space of continuous functions from $K$ to $\R$.

Let $\mathcal{F}$ be a vector subspace of $\mathcal{C}(K,\R)$.

If $\mathcal{F}$ is such that:
 * $(1): \forall f,g\in \mathcal{F}$, the functions $(f\lor g)(x)=\max(f(x),g(x))$ and $(f\land g)(x)=\min(f(x),g(x))$ are also in $\mathcal{F}$.
 * $(2): \forall x,y\in K:x\neq y$, $\exists f\in\mathcal{F}:f(x)\neq f(y)$.
 * $(3):$ The unit function $f_1(x)=1$ is an element of $\mathcal{F}$.

Then $\mathcal{F}$ is everywhere dense in $\left(\mathcal{C}(K,\R),\|\cdot\|_{\infty}\right)$.

Lemma
Let $x,y\in K:x\neq y$.

$\forall a,b\in\R,\exists g\in\mathcal{F}:g(x)=a$ and $g(y)=b$.

Suppose $a=b$.

Let $g=a\cdot f_1$, where $f_1$ is the unit function.

$g\in\mathcal{F}$, since $\mathcal{F}$ is closed under linear combinations.

$\forall z\in K$, $g(z)=a=b$, so $g(x)=a$ and $g(y)=b$.

Suppose $a\neq b$.

$\exists f\in\mathcal{F}:f(x)\neq f(y)$.

Let $g=\lambda\cdot f +\mu\cdot f_1$,where $f_1$ is the unit function and $\lambda,\mu\in\R:$

$g\in\mathcal{F}$, since $\mathcal{F}$ is closed under linear combinations.

Furthermore,

Hence, in all cases, there exists a satisfying function $g\in\mathcal{F}$.

Let $\epsilon >0,f\in\mathcal{C}(K,\R)$ and $x,y\in K:x\neq y$.

By the lemma, $\exists g_{x,y}\in\mathcal{F}:g_{x,y}(x)=f(x)$ and $g_{x,y}(y)=f(y)$.

$\mathcal{C}(K,\R)$ is closed under linear combinations, so $g_{x,y}-f$ is continuous.

Furthermore, $(g_{x,y}-f)(x)=0=(g_{x,y}-f)(y)$

Thus, by the definition of continuity on metric spaces, There exists an open neighborhood of $y$, $N_{\delta}(y)$ such that: