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 ${U_i}$. $\blacksquare$