Kuratowski-Wojdyslawski embedding theorem

The target $C_b(M)$ is Banach for the supremum norm. Indeed, let $(f_n)$ be a Cauchy sequence in the supremum norm. Then each scalar sequence $(f_n(z){)}_n$ is Cauchy, so define $f(z):={\lim }_n{f}_n(z)$. Fix $N$ with $\mathrm{\parallel }{f}_n-f_m{\mathrm{\parallel }}_{\mathrm{\infty }}<1$ for $n,m\ge N$; letting $m\to \mathrm{\infty }$ gives $\mathrm{\mid }f(z)\mathrm{\mid }\le \mathrm{\parallel }{f}_N{\mathrm{\parallel }}_{\mathrm{\infty }}+1$ for all $z$, so $f$ is bounded. Given $\epsilon >0$, fix $N$ with $\mathrm{\parallel }{f}_n-f_m{\mathrm{\parallel }}_{\mathrm{\infty }}<\epsilon$ for $n,m\ge N$; letting $m\to \mathrm{\infty }$ yields $\mathrm{\parallel }{f}_n-f{\mathrm{\parallel }}_{\mathrm{\infty }}\le \epsilon$ for $n\ge N$, so $f_n\to f$ uniformly. By [L2], $f$ is continuous. Hence every Cauchy sequence converges in $C_b(M)$.

By [L1], $K_o$ is an isometric embedding of $M$ into the Banach space $C_b(M)$.

Let $f={\sum }_{j=1}^m{\lambda }_j{K}_o(x_j)$ be a point of the algebraic convex hull of $K_o[M]$, so each ${\lambda }_j\ge 0$ and ${\sum }_j{\lambda }_j=1$. Put $\mathrm{\Delta }:={\sum }_{i=1}^m{\sum }_{j=1}^m{\lambda }_i{\lambda }_{j}d(x_i,x_j)\mathrm{.}$ If $\mathrm{\Delta }=0$, then every pair with ${\lambda }_i{\lambda }_j>0$ satisfies $d(x_i,x_j)=0$, hence all support points coincide and $f=K_o(x_j)$ lies in the image.

Assume $\mathrm{\Delta }>0$, and let $y\in M$. Evaluating at $y$ gives $f(y)-K_o(y)(y)={\sum }_{j=1}^m{\lambda }_{j}d(x_j,y)\mathrm{.}$ For each $i,j$, the triangle inequality yields $d(x_i,x_j)\le d(x_i,y)+d(y,x_j)$. Multiplying by ${\lambda }_i{\lambda }_j$ and summing over $i,j$ gives $\mathrm{\Delta }\le 2{\sum }_{j=1}^m{\lambda }_{j}d(x_j,y)=2(f(y)-K_o(y)(y))\mathrm{.}$ Therefore $\mathrm{\parallel }f-K_o(y){\mathrm{\parallel }}_{\mathrm{\infty }}\ge f(y)-K_o(y)(y)\ge \frac{\mathrm{\Delta }}{2}>0.$ Since this holds for every $y\in M$, the distance from $f$ to $K_o[M]$ is at least $\mathrm{\Delta }\mathrm{/}2$. So no such $f$ lies in the closure of $K_o[M]$.

Steps 3.1 and 4.1 show that every point of the algebraic convex hull that lies in the closure of $K_o[M]$ is already in $K_o[M]$. Thus $K_o[M]$ is closed in its algebraic convex hull.

For the final claim, let $(X,\mathcal{T})$ be metrizable. If $X=\mathrm{\varnothing }$, the unique map $\mathrm{\varnothing }\to \{0\}$ into any Banach space is a homeomorphic embedding, and its image is closed in its algebraic convex hull. Assume now that $X\mathrm{\ne }\mathrm{\varnothing }$. By [L4], choose a metric $d$ inducing $\mathcal{T}$. By [L3], the bounded metric $d_b=d\mathrm{/}(1+d)$ induces the same topology. Applying steps 1.1 to 5.1 to $(X,d_b)$ gives an isometric embedding into a Banach space whose image is closed in its algebraic convex hull. Because $d$ and $d_b$ are topologically equivalent, that same map is a homeomorphic embedding for the original topology $\mathcal{T}$.