Noetherian Topological Space is Compact

Theorem
Let $T = \struct {X, \tau}$ be a Noetherian topological space.

Then $T$ is compact.

Proof
Let $\family {U_i}_{i \mathop \in I}$ be a cover of $X$.

That is:
 * $\ds \bigcup_{i \mathop \in I} U_i = X$

Let $V$ be the collection of finite cover of $\family {U_i}_{i \mathop \in I}$.

Let $W = \set {\bigcup Y: Y \in V}$.

Then $W$ is a collection of open sets of $T$.

By, $W$ has a maximal element the subset relation.

Let $\ds U' = \bigcup_{j \mathop = 1}^n U_{i_j}$ be the maximal element.

$U' \subsetneq X$.

Let $x \in X \setminus U'$.

Let $U_{i_{n + 1} }$ be a neighborhood of $x$, where $i_{n + 1} \in I$.

Then $U' \cup U_{i_{n + 1} }$ is larger than $U'$.

This contradicts our hypothesis that $U'$ is maximal.

Hence $U'$ is a finite subcover.

This shows that $\struct {X, \tau}$ is compact.

Proof 2
Let $\CC \subseteq \tau$ be a cover of $X$.

Let:
 * $A := \set { \bigcup \eta : \eta \text{ finite subset of } \CC }$

Observe that $A \ne \O$ and $A \subseteq \tau$,

By, $A$ has a maximal element:
 * $U_1 \cup \cdots \cup U_n$

We shall show that $\set {U_1,\ldots,U_n} \subseteq \CC$ is a finite subcover, i.e.
 * $X = U_1 \cup \cdots \cup U_n$

Otherwise, there would exist:
 * $x \in X \setminus \paren {U_1 \cup \cdots \cup U_n}$

Then there must be some $V \in \CC$ such that:
 * $x \in V$

Therefore:
 * $U_1 \cup \cdots \cup U_n \subsetneq U_1 \cup \cdots \cup U_n \cup V$

Since $U_1 \cup \cdots \cup U_n \cup V \in A$, this would contradict the maximality.