[proofplan] We construct the box $L = \prod_{i=1}^n (0, \ell_i)$ by a constrained minimisation argument over the space of "projection arrays." Consider arrays $\{x_A : A \subseteq [n]\}$ of non-negative reals satisfying three families of constraints: (i) $x_A \leq |K_A|$, (ii) $|K|^k \leq \prod_{A \in \mathcal{A}} x_A$ for every irreducible uniform $k$-cover $\mathcal{A}$ of $[n]$, and (iii) $x_A \leq \prod_{i \in A} x_{\{i\}}$ for all $A$. The array $x_A = |K_A|$ is feasible (by the [Uniform Cover Inequality](/theorems/2609)). By the [Finiteness of Irreducible Covers](/theorems/2610), the constraint system is finite, so a componentwise-minimal feasible array exists. We show this minimal array equals $\prod_{i \in A} x_{\{i\}}$ for all $A$, so the box $L = \prod_{i=1}^n (0, x_{\{i\}})$ has $|L| = |K|$ and $|L_A| \leq |K_A|$ for all $A$. [/proofplan] [step:Define the feasible set of projection arrays] For a bounded open set $K \subseteq \mathbb{R}^n$ with $K \neq \varnothing$, consider the set $\mathcal{F}$ of all arrays $x = (x_A)_{A \subseteq [n]}$ of non-negative reals satisfying: **(C1)** For all $A \subseteq [n]$: $x_A \leq |K_A|$. **(C2)** For every irreducible uniform $k$-cover $\mathcal{A}$ of $[n]$: $|K|^k \leq \prod_{A \in \mathcal{A}} x_A$. **(C3)** For all $A \subseteq [n]$: $x_A \leq \prod_{i \in A} x_{\{i\}}$. The array $x_A := |K_A|$ satisfies (C1) with equality, satisfies (C2) by the [Uniform Cover Inequality](/theorems/2609) (which gives $|K|^k \leq \prod_{A \in \mathcal{A}} |K_A|$ for every uniform $k$-cover, and in particular for every irreducible one), and satisfies (C3) because a partition into singletons $\{i\}$ for $i \in A$ gives $|K_A| \leq \prod_{i \in A} |K_{\{i\}}|$ (the elementary projection bound from partitions). Therefore $\mathcal{F} \neq \varnothing$. [guided] Why these three families of constraints? Constraint (C1) ensures the array is bounded above by the actual projections of $K$, so the eventual box will have projections no larger than $K$'s. Constraint (C2) ensures the array cannot be made too small -- the volume $|K|$ is protected by the cover inequalities. Constraint (C3) encodes the structure of a box: for a box $L = \prod (0, \ell_i)$, the projection $|L_A| = \prod_{i \in A} \ell_i$, so the singletons determine all projections. Constraint (C3) forces the array to be "box-like." The feasibility of $x_A = |K_A|$ is the starting point: the actual projections of $K$ satisfy all three constraints. The idea is to decrease the $x_A$ values as much as possible while maintaining feasibility, until we reach a minimal array that is exactly the projection array of some box. [/guided] [/step] [step:Show that a componentwise-minimal feasible array exists] The feasible set $\mathcal{F}$ is a subset of $[0, M]^{2^n}$ where $M = \max_{A \subseteq [n]} |K_A|$ (by constraint (C1)). The constraints (C1) and (C3) define closed conditions (non-strict inequalities), and the constraints (C2) also define a closed set (since the map $x \mapsto \prod_{A \in \mathcal{A}} x_A$ is continuous and the condition $|K|^k \leq \prod x_A$ is a closed inequality). By the [Finiteness of Irreducible Covers](/theorems/2610), there are only finitely many irreducible uniform covers of $[n]$, so constraint family (C2) consists of finitely many inequality constraints. Therefore $\mathcal{F}$ is the intersection of finitely many closed sets in $[0, M]^{2^n}$, hence closed. Since $[0, M]^{2^n}$ is compact, $\mathcal{F}$ is compact and non-empty. We seek a componentwise-minimal element of $\mathcal{F}$. Define the objective function $\Phi(x) := \sum_{A \subseteq [n]} x_A$, which is continuous on the compact set $\mathcal{F}$. By the extreme value theorem, $\Phi$ attains its minimum on $\mathcal{F}$. Let $x^*$ be a minimiser. We verify that $x^*$ is componentwise-minimal: if some $x_A^*$ could be strictly decreased while maintaining all constraints, then $\Phi$ would decrease, contradicting minimality. (More precisely, if $y \in \mathcal{F}$ with $y \leq x^*$ componentwise, then $\Phi(y) \leq \Phi(x^*)$, so $\Phi(y) = \Phi(x^*)$, and since all components are non-negative, $y = x^*$.) [guided] The finiteness of irreducible covers is essential here. With finitely many constraints, the feasible set is a closed subset of a compact cube, hence compact. If there were infinitely many irreducible covers, the intersection of infinitely many closed sets need not be well-behaved -- more seriously, the minimisation would not be over a standard finite-dimensional optimisation problem. The sum $\Phi(x) = \sum x_A$ serves as a proxy for "total size" of the array. Minimising $\Phi$ ensures we push every component as low as possible. Any array that is componentwise below $x^*$ and still feasible would have a smaller sum, contradicting the minimality of $\Phi(x^*)$. So $x^*$ is componentwise-minimal. [/guided] [/step] [step:Show that the minimal array satisfies $x_A^* = \prod_{i \in A} x_{\{i\}}^*$ for all $A$] At the minimal array $x^*$, constraint (C3) states $x_A^* \leq \prod_{i \in A} x_{\{i\}}^*$ for all $A$. We claim equality holds. Suppose for contradiction that $x_A^* < \prod_{i \in A} x_{\{i\}}^*$ for some $A$ with $|A| \geq 2$. We show that $x_A^*$ could be decreased further, contradicting minimality. Consider the role of $x_A^*$ in the constraints. In (C1), the constraint $x_A^* \leq |K_A|$ is unaffected by decreasing $x_A^*$. In (C3), decreasing $x_A^*$ preserves the inequality $x_A^* \leq \prod_{i \in A} x_{\{i\}}^*$. The only constraints that could be violated by decreasing $x_A^*$ are those in (C2) where $A$ appears in the cover. But $x_A^*$ appears in (C2) constraints only in products $\prod_{B \in \mathcal{A}} x_B^*$. Since $x_A^* > 0$ (otherwise the product in some (C2) constraint would be zero while $|K|^k > 0$), decreasing $x_A^*$ by a sufficiently small amount preserves all (C2) constraints by continuity. This contradicts the componentwise minimality of $x^*$, unless $x_A^*$ is already at the lowest value consistent with all constraints. But the analysis shows that the binding constraint on $x_A^*$ must come from (C2) -- and for (C2) constraints to be tight, the product structure forces $x_A^* = \prod_{i \in A} x_{\{i\}}^*$. More precisely: at minimality, for each $A$, at least one constraint involving $x_A$ on the right-hand side of an inequality must hold with equality. The constraints in (C2) give $|K|^k = \prod_{B \in \mathcal{A}} x_B^*$ for some irreducible cover $\mathcal{A}$ containing $A$. Combined with (C3) applied to each $B$: $\prod_{B \in \mathcal{A}} x_B^* \leq \prod_{B \in \mathcal{A}} \prod_{i \in B} x_{\{i\}}^* = \prod_{i=1}^n (x_{\{i\}}^*)^k$ (since $\mathcal{A}$ is a uniform $k$-cover). With equality forced in (C2), we need $x_B^* = \prod_{i \in B} x_{\{i\}}^*$ for every $B \in \mathcal{A}$, and in particular for $B = A$. [guided] Why must the minimal array be "multiplicative" (i.e., $x_A = \prod_{i \in A} x_{\{i\}}$)? This is the box structure: a box's projection onto any coordinate set $A$ factors as a product of its one-dimensional projections. The constraints force this factorisation at the minimum. The argument is by contradiction: if $x_A^* < \prod_{i \in A} x_{\{i\}}^*$, then there is "slack" in constraint (C3) for this $A$. The question is whether $x_A^*$ participates as a binding factor in any (C2) constraint. If it does, then decreasing $x_A^*$ would violate that (C2) constraint. But then the (C2) constraint is tight: $|K|^k = \prod_{B \in \mathcal{A}} x_B^*$. Substituting the (C3) upper bounds: $\prod_{B \in \mathcal{A}} x_B^* \leq \prod_{B \in \mathcal{A}} \prod_{i \in B} x_{\{i\}}^*$. Since $\mathcal{A}$ is a uniform $k$-cover, each $i$ appears exactly $k$ times, so $\prod_{B \in \mathcal{A}} \prod_{i \in B} x_{\{i\}}^* = \prod_{i=1}^n (x_{\{i\}}^*)^k$. For the (C2) equality to hold, we need equality throughout, i.e., $x_B^* = \prod_{i \in B} x_{\{i\}}^*$ for all $B \in \mathcal{A}$. If $x_A^*$ does not participate bindingly in any (C2) constraint, then it can be decreased -- contradicting minimality. Either way, we reach $x_A^* = \prod_{i \in A} x_{\{i\}}^*$. [/guided] [/step] [step:Construct the box and verify the claimed properties] Set $\ell_i := x_{\{i\}}^*$ for each $i \in [n]$, and define the box \begin{align*} L := \prod_{i=1}^n (0, \ell_i). \end{align*} **Volume:** By the previous step, $x_{[n]}^* = \prod_{i=1}^n x_{\{i\}}^* = \prod_{i=1}^n \ell_i = |L|$. From (C2) applied to the elementary uniform $1$-cover $\{[n]\}$ (the single set $[n]$ is an irreducible $1$-cover of $[n]$), we get $|K| \leq x_{[n]}^* = |L|$. From (C1), $x_{[n]}^* \leq |K_{[n]}| = |K|$. Therefore $|L| = |K|$. **Projections:** For any $A \subseteq [n]$, the projection of the box is $L_A = \prod_{i \in A} (0, \ell_i)$, which has measure $|L_A| = \prod_{i \in A} \ell_i = \prod_{i \in A} x_{\{i\}}^* = x_A^*$. By (C1), $x_A^* \leq |K_A|$, so $|L_A| \leq |K_A|$. This establishes that $L$ is a box with $|L| = |K|$ and $|L_A| \leq |K_A|$ for all $A \subseteq [n]$. [guided] The box $L$ is read off directly from the minimal array: the side lengths are the singleton values $x_{\{i\}}^*$, and the multiplicativity $x_A^* = \prod_{i \in A} x_{\{i\}}^*$ ensures that $x_A^*$ is exactly the $A$-projection of this box. Two properties must be checked: 1. **Volume equals $|K|$:** The (C2) constraint for the cover $\{[n]\}$ gives $|K| \leq x_{[n]}^* = |L|$, and the (C1) constraint gives $|L| = x_{[n]}^* \leq |K_{[n]}| = |K|$. So $|L| = |K|$. 2. **Projections bounded by $K$'s:** $|L_A| = x_A^* \leq |K_A|$ by (C1). The Box Theorem is thus a consequence of the existence of a minimal feasible array, which itself relies on the compactness of the feasible set -- guaranteed by the finiteness of irreducible covers. [/guided] [/step]