Convex closures and hulls of finitely many compact convex sets

Let $x,y\in \bar{C}$ for convex $C$, and $0<t<1$. The affine map $(a,b)\mapsto (1-t)a+tb$ is continuous by the vector operations. For an open neighborhood $O$ of $(1-t)x+ty$, its preimage contains a product neighborhood $P\times Q$ of $(x,y)$. There exist $a\in P\cap C$ and $b\in Q\cap C$ by the closure test. Their convex combination is in $O\cap C$. Thus $(1-t)x+ty\in \bar{C}$. At $t=0,1$ the assertion follows from membership of $x,y$; for $C=\mathrm{\varnothing }$ its closure is empty.

For $x,y\in \mathrm{int}C$ and $0<t<1$, the open set $(1-t)\mathrm{int}C+t\mathrm{int}C$ contains their combination and lies in $C$. It is open as a union of translates of a nonzero dilate of an open set. The endpoints $t=0,1$ are immediate, and an empty interior is convex vacuously. If $y\in \mathrm{int}C$ and $x\in C$, then for $0<s\le 1$ the open set $(1-s)x+s\mathrm{int}C$ lies in $C$. Hence $(1-s)x+sy$ belongs to its interior. Continuity of the orbit at $s=0$ shows every neighborhood of $x$ contains such a point, so $x\in \overline{\mathrm{int}C}$.

Let $C$ be balanced. For $0<\mathrm{\mid }\lambda \mathrm{\mid }\le 1$, the homeomorphism $D_{\lambda }:x\mapsto \lambda x$ carries $\bar{C}$ onto $\overline{\lambda C}$. Indeed, pull an open neighborhood back by $D_{\lambda }$ for one inclusion and use its inverse for the other. Since $\lambda C\subseteq C$, the closure test gives $\overline{\lambda C}\subseteq \bar{C}$. For $\lambda =0$ and $C\mathrm{\ne }\mathrm{\varnothing }$, balance gives $0\in C$, so $0\bar{C}=\{0\}\subseteq \bar{C}$; for empty $C$ the dilation has empty image. Thus the closure is balanced.

Suppose $n\ge 1$ and each $K_j$ is nonempty. The simplex $\mathrm{\Delta }=\{t\in {\mathbb{R}}^n:t_j\ge 0,{\sum }_j{t}_j=1\}$ is closed: coordinate maps and their finite sum are continuous, and the conditions are inverse images of closed real rays and $\{1\}$. It is bounded since $0\le t_j\le 1$ and ${\sum }_j{t}_j^2\le n$. It is compact by Heine–Borel. Finite product compactness makes $\mathrm{\Delta }\times {\prod }_j{K}_j$ compact. The map $\mathrm{\Phi }(t,x)={\sum }_j{t}_j{x}_j$ is continuous: coordinate projections and inclusions are continuous by preimages of basic opens, scalar multiplication is jointly continuous, and iterating addition preserves continuity. Therefore its image is compact.

Every value of $\mathrm{\Phi }$ is a convex combination from the union. Conversely, write a hull point as ${\sum }_{\mathrm{\ell }=1}^m{a}_{\mathrm{\ell }}{y}_{\mathrm{\ell }}$. Assign each $y_{\mathrm{\ell }}$ the least index $j$ for which $y_{\mathrm{\ell }}\in K_j$, and let $t_j$ be the sum of the weights with that label. For $t_j>0$, their normalized combination $x_j=t_j^{-1}{\sum }_{\mathrm{\ell }\text{ labelled }j}{a}_{\mathrm{\ell }}{y}_{\mathrm{\ell }}$ belongs to $K_j$ by finite convexity. For the finitely many zero $t_j$, choose any $x_j\in K_j$ using finite choice. Then $t\in \mathrm{\Delta }$ and ${\sum }_j{t}_j{x}_j={\sum }_{\mathrm{\ell }}{a}_{\mathrm{\ell }}{y}_{\mathrm{\ell }}$. This proves equality of the two sets.

The hull is compact by step 1.4 and step 2.1, and Hausdorffness gives closedness. Each singleton is compact since any cover has one member covering its sole point, and convex since its every combination is that point; hence finite point hulls are covered. Delete empty members of a finite family in their original order. If none remain, its union and hull are empty, and the empty subcover proves compactness. For one nonempty convex member the hull is that member. All choices made above are finite.