Krull Dimension of Open Cover
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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:
- $\map \dim X \le \sup \set { \map \dim U : U \in \CC }$
$\blacksquare$
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