Noetherian Topological Space is Compact/Proof 1

Theorem

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

Then $T$ is compact.

Proof

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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 $\family {U_i}_{i \mathop \in I}$.

This article, or a section of it, needs explaining. In particular: Exactly what is meant by the above? Does it mean "the set / collection of all finite covers of $X$"? If not, can it be explained? If so, is there a reason why we cannot use the word "Set" here? If not, we need to change that word "collection" to a specific link to set, and make "finite cover" plural. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code.

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

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

This article, or a section of it, needs explaining. In particular: Again, is there any reason to use "collection" and not "set"? Is there a possibility that $W$ is itself not a set? You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code.

By Definition 4 of Noetherian Topological Space, $W$ has a maximal element with respect to the subset relation.

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 $U'$.

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

Hence $U'$ is a finite subcover.

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

$\blacksquare$