Though the following was found in Hartshorne’s Algebraic Geometry (read, ‘terror maths’), this is really point-set topology.

Definition: A topological space {X} is called Noetherian if it satisfies the descending chain condition for closed subsets: for any sequence {Y_1 \supseteq Y_2 \supseteq \ldots} of closed subsets of {X}, there is an integer {r} such that {Y_r=Y_{r+1}=\ldots}.

We have the following properties of Noetherian spaces:

Proposition: A Noetherian space is compact.

Proof: It is easy to show that the Noetherian property is equivalent to every nonempty family of open subsets having a maximal element. (The proof resembles a similar one in ring theory).

Let {X} be the Noetherian space and {U_\alpha} be an open cover of {X}; {\alpha \in J}. Define {\mathcal S} to be the family of finite union of {U_\alpha}‘s. Then {\mathcal S} is a family of open subsets of {X} and hence must have a maximal element, {U}. I claim, {U=X} because if there is an element {x\in X\backslash U}, then since {U_\alpha} is an open cover of {x}, I can find an open set {U_x} containing {x}. Now, {U \cup U_x \in \mathcal S} since it is the finite union of elements of the cover of X, and contains {U} properly. This is a contradiction to the maximality of {U}. Hence {X=U}, a finite union of elements of the open cover.\blacksquare

Lemma: (Proof easy) If {Y} is any subset of a topological space {X}, then dim {Y \leq } dim {X}.

Theorem: If {X} is a topological space covered by a family of open subsets {\{U_i\}}, then dim {X} = {\sup} dim {U_i}.

Proof:

We only prove the case that dim {X<\infty}.

Using the above lemma, dim {U_i \leq } dim {X} is clear, and so taking supremum gives one side of the equality, {\sup} dim {U_i\leq} dim {X}. Consider the chain

\displaystyle X_1 \subsetneq X_2 \subsetneq \ldots \subsetneq X_n

of closed subsets of {X}, where {X_1} is a singleton. We show that this chain is preserved when interescted with {U=U_i}, for some {i}. This would mean that dim {X \leq} dim {U_i}. Consider
\displaystyle (X_1 \cap U) \subseteq (X_2 \cap U) \subseteq \ldots \subseteq (X_n \cap U)

and assume that {X_j\cap U = X_{j+1}\cap U} for some {j}. Now,

\displaystyle X_{j+1} = (X_{j+1} \cap U) \cup (X_{j+1} \cap U^c)

\displaystyle = (X_j \cap U) \cup (X_{j+1} \cap U^c)

\displaystyle \subseteq X_j \cup (X_{j+1} \cap U^c) \subseteq X_{j+1}

Hence, {X_{j+1} \subseteq U^c} or, {X_{j+1} \cap U = \phi} and thus, {X_1 \cap U = \ldots = X_{j+1} \cap U = \phi}. In particular, {X_1 \cap U = \phi} for every {U\in \{U_i\}}. But since {\{U_i\}} is a cover of {X}, some {U_i} must contain the singleton {X_1}, thus giving us a contradiction. Hence there exists a {U_i} such that
\displaystyle (X_1 \cap U_i) \subsetneq (X_2 \cap U_i) \subsetneq \ldots \subsetneq (X_n \cap U_i)

thus proving that dim {X\leq } dim {U_i}.   \blacksquare

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