TITLE: A noetherian topological space is compact
QUESTION [10 upvotes]: Have to prove that every noetherian topological space $(X,\mathcal{T})$ is also compact.
I don't understand the proof I found:
Let $\{\mathcal{U}_\alpha\}_{\alpha\in\Lambda}$ be an open cover of $X$, and let,
$$A=\{\bigcup^n_{i=1}\mathcal{U}_{\alpha_i}:\alpha_i\in\Lambda,n\in\mathbb{N}\}$$
since $X$ is noether and every set in $A$ is open, it follows that $A$ has a maximal element $M$ (I'm not quite sure why should it belong to $A$, since the set who has maximal element is $\mathcal{T}$). Then if there is $x\in X\setminus M$ then $x\in\mathcal{U}_\alpha$ for some $\alpha\in \Lambda$, so $M\cup\mathcal{U}_\alpha\in A$ and $M$ wouldn't be maximal. So must be $X\subset M$ and since $M$ was a finite union of elements of the cover, we have the result.

REPLY [14 votes]: Perhaps written this way is a little bit clearer:
Recall that being Noetherian is equivalent to the property that every non-empty familly of open subsets has a maximal element.
Let $\mathcal{U} = \{U_\alpha\}_{\alpha\in \Lambda}$ be an open cover for $X$. Consider the collection $\mathcal{F}$ consisting of finite unions of elements from $\mathcal{U}$. Since $X$ is Noetherian, $\mathcal{F}$ must have a maximal element $U_{\alpha_1}\cup ... \cup U_{\alpha_n}$. Suppose that $U_{\alpha_1}\cup ... \cup U_{\alpha_n} \subsetneq X$. Then there exists some element $x\in X\setminus (U_{\alpha_1}\cup ... \cup U_{\alpha_n})$. Since $\mathcal{U}$ covers $X$, there exists some $\alpha\in \Lambda$ such that $x\in U_\alpha$. But then $U_{\alpha_1}\cup ... \cup U_{\alpha_n} \subseteq U_{\alpha_1}\cup ... \cup U_{\alpha_n} \cup U_\alpha$, which contradicts the maximality of $U_{\alpha_1}\cup ... \cup U_{\alpha_n}$.