Proof of The Simplex, the Control Set and Their Product are Compact Separable Metric Spaces

Throughout, $x_{\gamma}$ denotes the $\gamma$-th component of a point $x\in\mathbb{R}^{l}$, the components determining the point by claim 1 of Euclidean Points as Tuples of Real Numbers, and $|x|=\bigl(\sum_{\gamma=1}^{l}x_{\gamma}^{2}\bigr)^{1/2}$ is the Euclidean norm. By Probability Simplex, a point $x\in\mathbb{R}^{l}$ lies in $\Delta^{l}$ exactly when $x_{\gamma}\ge0$ for every $\gamma$ and $\sum_{\gamma=1}^{l}x_{\gamma}=1$.

Claim 1. Let $e$ be the point with $e_{1}=1$ and $e_{\gamma}=0$ for $\gamma\ge2$, which exists by claim 2 of Euclidean Points as Tuples of Real Numbers. Then $e\in\Delta^{l}$, so $\Delta^{l}$ is nonempty.

Let $x\in\Delta^{l}$ and fix $\gamma$. Every other component is nonnegative, so $x_{\gamma}=1-\sum_{\delta\neq\gamma}x_{\delta}\le1$ by claim 3 of Elementary Order Arithmetic in an Ordered Field in the nonstrict form obtained by adjoining the case of equality; thus $0\le x_{\gamma}\le1$. Multiplying the inequality $x_{\gamma}\le1$ by the nonnegative number $x_{\gamma}$, using claim 10 of Elementary Order Arithmetic in an Ordered Field in the nonstrict form obtained by adjoining the case of equality, gives $x_{\gamma}^{2}\le x_{\gamma}$. Summing over $\gamma$ and using claim 3 of Elementary Order Arithmetic in an Ordered Field repeatedly, again in the nonstrict form obtained by adjoining the case of equality,

$$|x|^{2}=\sum_{\gamma=1}^{l}x_{\gamma}^{2}\le\sum_{\gamma=1}^{l}x_{\gamma}=1 ,$$

and therefore $|x|\le1$. Indeed, suppose instead that $1<|x|$. From $0\le1$ and $1<|x|$, claim 2 of Elementary Order Arithmetic in an Ordered Field gives $0<|x|$, so claim 10 of that lemma, applied to the inequality $1<|x|$ with the positive multiplier $|x|$, gives $|x|<|x|^{2}$; combining $1<|x|$ with $|x|<|x|^{2}$ by claim 2 again gives $1<|x|^{2}$, contradicting $|x|^{2}\le1$. Consequently every point of $\Delta^{l}$ lies within distance $1$ of the origin, so $\Delta^{l}$ is bounded in $(\mathbb{R}^{l},d_{\mathbb{R}^{l}})$.

For closedness we show that the complement of $\Delta^{l}$ is open in $(\mathbb{R}^{l},d_{\mathbb{R}^{l}})$; by Closed Subset of a Topological Space this is exactly what is required. Let $y\in\mathbb{R}^{l}\setminus\Delta^{l}$. Then either some component of $y$ is negative, or all are nonnegative and $s=\sum_{\gamma}y_{\gamma}\neq1$.

In the first case fix $\gamma$ with $y_{\gamma}<0$ and put $r=-y_{\gamma}>0$. If $|z-y|<r$ then $|z_{\gamma}-y_{\gamma}|\le|z-y|<r$ by claim 4 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$, hence $|z_{\gamma}-y_{\gamma}|<r$ by claim 2 of Elementary Order Arithmetic in an Ordered Field. The strict two-sided bound, claim 9 of Properties of the Absolute Value in an Ordered Field, then gives $z_{\gamma}-y_{\gamma}<r$, and claim 1 of Elementary Order Arithmetic in an Ordered Field gives $z_{\gamma}<y_{\gamma}+r=0$. Since every component of a point of $\Delta^{l}$ is nonnegative, the strict inequality $z_{\gamma}<0$ gives $z\notin\Delta^{l}$.

In the second case put $r=|s-1|/l>0$. Suppose $|z-y|<r$. By claim 4 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ each $|z_{\gamma}-y_{\gamma}|\le|z-y|$, so by the triangle inequality for the absolute value, claim 5 of Properties of the Absolute Value in an Ordered Field, applied $l-1$ times,

$$\Bigl|\sum_{\gamma}z_{\gamma}-s\Bigr|=\Bigl|\sum_{\gamma}(z_{\gamma}-y_{\gamma})\Bigr|\le\sum_{\gamma}|z_{\gamma}-y_{\gamma}|\le l\,|z-y|<l\,r=|s-1| .$$

If $z$ were in $\Delta^{l}$ we would have $\sum_{\gamma}z_{\gamma}=1$ and hence $\bigl|\sum_{\gamma}z_{\gamma}-s\bigr|=|1-s|=|s-1|$ by claim 4 of Properties of the Absolute Value in an Ordered Field, contradicting the strict inequality just obtained. So $z\notin\Delta^{l}$.

In either case the open ball of radius $r$ about $y$ misses $\Delta^{l}$, so the complement of $\Delta^{l}$ is open and $\Delta^{l}$ is closed.

Claim 2. By claim 1 and the implication from condition 2 to condition 1 in Heine-Borel Theorem in $\mathbb{R}^n$, $\Delta^{l}$ is a compact subset of $\mathbb{R}^{l}$ for the topology determined by the Euclidean distance. By claim 2 of The Restriction of a Metric to a Subset Induces the Subspace Topology the subsets of $\Delta^{l}$ open in $(\Delta^{l},d_{\Delta})$ are exactly the sets $\Delta^{l}\cap U$ with $U$ open in $(\mathbb{R}^{l},d_{\mathbb{R}^{l}})$, that is, the topology of $(\Delta^{l},d_{\Delta})$ is the subspace topology inherited from $\mathbb{R}^{l}$; so by Compact Topological Space and Compact Subset compactness of the subset $\Delta^{l}$ of $\mathbb{R}^{l}$ and compactness of the space $(\Delta^{l},d_{\Delta})$ are the same assertion. Hence $(\Delta^{l},d_{\Delta})$ is compact, and $\Delta^{l}$ is a sequentially compact subset of $(\Delta^{l},d_{\Delta})$ by Compactness and Sequential Compactness Agree for Subsets of a Metric Space. Separability of $(\Delta^{l},d_{\Delta})$ is claim 2 of A Totally Bounded Metric Space is Separable, applied to the compact metric space $(\Delta^{l},d_{\Delta})$.

Claim 3. By claim 3 of The Set of Controls with Values in a Compact Convex Set is Weakly Metrizable and Compact, $\mathcal{U}_{\mathcal{A}}$ is a sequentially compact and a compact subset of $(\mathcal{U}_{\mathcal{A}},\rho)$; since $\mathcal{U}_{\mathcal{A}}$ is the whole underlying set of that metric space, the space $(\mathcal{U}_{\mathcal{A}},\rho)$ is itself compact and sequentially compact. Separability is then claim 2 of A Totally Bounded Metric Space is Separable.

Claim 4. By claim 2 the set $\Delta^{l}$ is sequentially compact in $(\Delta^{l},d_{\Delta})$ and by claim 3 the set $\mathcal{U}_{\mathcal{A}}$ is sequentially compact in $(\mathcal{U}_{\mathcal{A}},\rho)$, so by claim 2 of Coordinatewise Convergence, Sequential Compactness and Density in a Product Metric Space the product $X=\Delta^{l}\times\mathcal{U}_{\mathcal{A}}$ is sequentially compact in $(X,d_{X})$; being the whole underlying set, it makes $(X,d_{X})$ a sequentially compact space, and a compact one by Compactness and Sequential Compactness Agree for Subsets of a Metric Space.

Let $D_{\Delta}$ and $D_{\mathcal{U}}$ be as in the statement. By claim 3 of Coordinatewise Convergence, Sequential Compactness and Density in a Product Metric Space the set $D_{\Delta}\times D_{\mathcal{U}}$ is dense in $X$ for the topology of subsets open in $(X,d_{X})$, and it is countable. Such $D_{\Delta}$ and $D_{\mathcal{U}}$ exist by the separability assertions of claims 2 and 3, so $(X,d_{X})$ is separable. $\blacksquare$