Proof of The Join of Two Compact Subsets of $\mathbb{R}^n$ is Compact

We realise $J(A,B)$ as the image of a compact parameter set in $\mathbb{R}^{2n+1}$ under a continuous function. Throughout, a subset of $\mathbb{R}^{q}$ is compact if and only if it is closed and bounded, by Convex Sets in $\mathbb{R}^n$: Standing Notation (v).

Coordinates. Let $\iota$ denote the concatenation functions $\mathbb{R}^{n}\times\mathbb{R}^{n}\to\mathbb{R}^{2n}$ and $\mathbb{R}^{2n}\times\mathbb{R}^{1}\to\mathbb{R}^{2n+1}$ of Euclidean Space and Real Matrices: Standing Notation (viii). They are bijections by Concatenation Identifies a Product of Euclidean Spaces with a Euclidean Space (iii), so every $w\in\mathbb{R}^{2n+1}$ can be written uniquely as

$$w=\iota\bigl(\iota(\alpha(w),\beta(w)),\hat{s}\bigr)\qquad\text{with }\alpha(w),\beta(w)\in\mathbb{R}^{n},\ \hat{s}\in\mathbb{R}^{1};$$

let $\sigma(w)\in\mathbb{R}$ be the coordinate of $\hat{s}$. For $w,w'\in\mathbb{R}^{2n+1}$, Concatenation Identifies a Product of Euclidean Spaces with a Euclidean Space (vi) and Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (iv) give

$$d_{E}(\alpha(w),\alpha(w'))\le d_{E}(w,w'),\qquad d_{E}(\beta(w),\beta(w'))\le d_{E}(w,w'),\qquad|\sigma(w)-\sigma(w')|\le d_{E}(w,w').$$

Hence, given $\varepsilon>0$, the choice $\delta=\varepsilon$ meets the requirement of Continuous Map Between Metric Spaces (i) at every point, and the functions $\alpha$, $\beta$ and $\sigma$ are continuous on every subset of $\mathbb{R}^{2n+1}$.

The parameter set. Let

$$P=\{w\in\mathbb{R}^{2n+1}:\ \alpha(w)\in A,\ \beta(w)\in B,\ 0\le\sigma(w)\le1\}.$$

If $A$ or $B$ is empty, then $P=\emptyset$ and $J(A,B)=\emptyset$, which is compact. Suppose from now on that $A$ and $B$ are both nonempty. Since $A$ and $B$ are compact, they are closed and bounded. Fix $a_{0}\in A$ and $b_{0}\in B$, take bounds $M_{A}$ and $M_{B}$ for the distances within $A$ and within $B$ from Bounded Subset of a Metric Space (i), and let $R=\max(\lVert a_{0}\rVert+M_{A},\lVert b_{0}\rVert+M_{B})\ge0$; then $\lVert a\rVert\le R$ for all $a\in A$ and $\lVert b\rVert\le R$ for all $b\in B$ by Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (vi). For $w\in P$, Concatenation Identifies a Product of Euclidean Spaces with a Euclidean Space (v) gives $\lVert w\rVert^{2}=\lVert\alpha(w)\rVert^{2}+\lVert\beta(w)\rVert^{2}+\sigma(w)^{2}\le2R^{2}+1\le(2R+1)^{2}$, so for $w,w'\in P$, by Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (ii), Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (vi) and Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (v),

$$d_{E}(w,w')=\lVert w-w'\rVert\le\lVert w\rVert+\lVert w'\rVert\le2(2R+1),$$

and $P$ is bounded. To see that $P$ is closed, let $w\notin P$; we find $r>0$ with $B(w,r)\cap P=\emptyset$. If $\alpha(w)\notin A$, then, since $\mathbb{R}^{n}\setminus A$ is open (Closed Subset of a Topological Space (i)), there is $r>0$ with $B(\alpha(w),r)\cap A=\emptyset$ (Open Subset of a Metric Space (i)), and the first inequality above shows $\alpha(w')\in B(\alpha(w),r)$, so $w'\notin P$, for every $w'\in B(w,r)$. The case $\beta(w)\notin B$ is the same. If $\sigma(w)<0$, take $r=-\sigma(w)$; if $\sigma(w)>1$, take $r=\sigma(w)-1$; in both cases the third inequality keeps $\sigma(w')$ outside $[0,1]$ for $w'\in B(w,r)$. So $\mathbb{R}^{2n+1}\setminus P$ is open, $P$ is closed, and $P$ is compact.

The parametrization. Define $\Phi:P\to\mathbb{R}^{n}$ by

$$\Phi(w)=(1-\sigma(w))\,\alpha(w)+\sigma(w)\,\beta(w).$$

The function $w\mapsto1-\sigma(w)$ is continuous on $P$ by the continuity of $\sigma$ and the continuity of constants and of constant multiples and sums of functions into $\mathbb{R}$, part of the background of Metric Spaces: Balls, the Metric Topology, and the Real and Complex Numbers as Metric Spaces (vi); so $\Phi$ is continuous on $P$ by Continuity of Vector- and Matrix-Valued Maps: the Componentwise Criterion, Norms and Algebraic Operations (iv) and Continuity of Vector- and Matrix-Valued Maps: the Componentwise Criterion, Norms and Algebraic Operations (v). For $a\in A$, $b\in B$ and $s\in[0,1]$, let $\hat{s}\in\mathbb{R}^{1}$ be the point with coordinate $s$; then $w=\iota(\iota(a,b),\hat{s})$ lies in $P$ and $\Phi(w)=(1-s)a+sb$, so $J(A,B)\subseteq\Phi(P)$. Conversely, for $w\in P$ we have $\alpha(w)\in A$, $\beta(w)\in B$ and $\sigma(w)\in[0,1]$, so $\Phi(w)\in J(A,B)$. Hence $\Phi(P)=J(A,B)$.

Finally, The Continuous Image of a Compact Subset of a Metric Space is Compact (i), applied to the metric spaces $(\mathbb{R}^{2n+1},d_{E})$ and $(\mathbb{R}^{n},d_{E})$, the compact set $P$ and the continuous function $\Phi$, shows that $J(A,B)=\Phi(P)$ is compact. This proves (i).