Krull Dimension of Open Cover

Definition
Let $\struct {X, \tau}$ be a topological space.

Let $\CC \subseteq \tau$ be an open cover of $X$.

Then:
 * $\map \dim X = \set { \map \dim U : U \in \CC }$

where $\dim$ denotes the Krull dimension.

Proof
By Krull Dimension of Topological Subspace is Smaller:
 * $\forall U \in \CC : \map \dim X \ge \map \dim U$

Thus:
 * $\map \dim X \ge \set { \map \dim U : U \in \CC }$

Conversely, let:
 * $A_0 \subsetneq A_1 \subsetneq \cdots \subsetneq A_n$

be a chain of closed irreducible sets of $X$.

There exists an $U_0 \in \CC$ such that:
 * $U_0 \cap A_0 \ne \emptyset$

Let:
 * $\forall i = 0, \ldots, n : \tilde A_i := U_0 \cap A_i$

By Open Set of Irreducible Space is Irreducible, each $\tilde A_i$ is irreducible.

Moreover:
 * $\tilde A_0 \subsetneq \tilde A_1 \subsetneq \cdots \subsetneq \tilde A_n$

Indeed, $\tilde A_i = \tilde A_{i+1}$ would imply:
 * $A_{i+1} = A_i \cup \paren {A_{i+1} \setminus U_0}$

which contradicts the irreduciblity of $A_{i+1}$.

Thus:
 * $n \le \map \dim {U_0} \le \sup \set { \map \dim U : U \in \CC }$

As $\map \dim X$ is the spremum of such $n$, we have:
 * $\map \dim X \le \sup \set { \map \dim U : U \in \CC }$