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:
 * $\forall f_1,f_2\in \mathcal{F}$, the functions $(f_1\lor f_2)(x)=\max(f_1(x),f_2(x))$ and $(f_1\land f_2)(x)=\min(f_1(x),f_2(x))$ are also in $\mathcal{F}$
 * $\forall x,y\in K:x\neq y$, $\exists f\in\mathcal{F}:f(x)\neq f(y)$
 * 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.

$\forall z\in K$, $g(z)=a\cdot f_1=a$

Hence, $g(x)=a$ and $g(y)=a=b$.

Suppose $a\neq b$.

Let $f\in\mathcal{F}$ be such that $f(x)\neq f(y)$.

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,

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

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, $g_x^y(x)-f(x)=0=g_x^y(y)-f(y)$

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

By Metric Space Compact iff Complete and Totally Bounded, $K$ is totally bounded.

Thus, $\forall \mu>0,\exists k_1,...,k_n\in K$ and open $\mu$-neighborhoods of these points such that:

By combining $(1)$ and $(2)$, we can define $y_1,...,y_n\in K\setminus\{x\}$ and $g_x^{y_1},...,g_x^{y_n}\in\mathcal{F}$ having the following properties:
 * $\forall i\leq n,$ $g_x^{y_i}(x)=f(x)$ and $g_x^{y_i}(y_i)=f(y_i)$
 * $\forall i\leq n,$ $\exists \delta_i>0: z\in N_{\delta_i}(y_i)\implies g_x^{y_i}(z)-f(z)<\epsilon$
 * $\displaystyle\bigcup_{i=1}^n N_{\mu}(y_i)=K$, where $\mu=\min(\delta_1,...,\delta_n)$

Let $g_x=g_x^{y_1}\land ...\land g_x^{y_n}\in\mathcal{F}$, where, $\forall z\in K$, $(g_x^{y_1}\land ... \land g_x^{y_n})(z)=\min(g_x^{y_1}(z),...,g_x^{y_n}(z))$.

$\forall i\leq n$, $z\in N_{\delta_i}(y_i)\implies g_x(z)-f(z)<g_x^{y_i}(z)-f(z)<\epsilon$.

Since the $\delta_i$-neighborhoods of the points $y_i$ cover $K$, it follows that $\forall z\in K$, $g_x(z)-f(z)<\epsilon$.