Noetherian Space is Compact

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Theorem

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


Then $\struct {X, \tau}$ is compact.


Proof

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

That is, $\bigcup_{i \mathop \in I} U_i = X$.

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.

By Set of Open Sets in Noetherian Space has Maximal Element, $W$ has a maximal element.

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

Aiming for a contradiction, suppose $U' \subsetneq X$.

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 $U'$ and contradicts maximality condition.

Hence $U'$ is a finite subcover.

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

$\blacksquare$