A noetherian topological space is compact 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.
 A: 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 $U_{\alpha_1}\cup ... \cup U_{\alpha_n}$.
A: You have the open cover $\mathscr{U}$ of $X$. Let
$$\mathscr{A}=\left\{\bigcup\mathscr{F}:\mathscr{F}\text{ is a finite subset of }\mathscr{U}\right\}\;;$$
we want to show that $X\in\mathscr{A}$, since that means that some finite $\mathscr{F}\subseteq\mathscr{U}$ covers $X$.
Let $\mathscr{C}$ be a chain in the partial order $\langle\mathscr{A},\subseteq\rangle$. Suppose that $\mathscr{C}$ does not have a maximal element; then for each $C\in\mathscr{C}$ there is a $C'\in\mathscr{C}$ such that $C\subsetneqq C'$, and we can recursively construct a family $\{C_n:n\in\Bbb N\}\subseteq\mathscr{C}$ such that $C_0\subsetneqq C_1\subsetneqq C_2\subsetneqq\ldots\;$, contradicting the hypothesis that $X$ is Noetherian. Thus, $\mathscr{C}$ has a maximal element. $\mathscr{C}$ was an arbitrary chain in $\mathscr{A}$, so Zorn’s lemma says that $\mathscr{A}$ has a maximal element, $M$. We’d like to show that $M=X$, since that means that $X\in\mathscr{A}$ and is therefore the union of some finite subfamily of $\mathscr{U}$.
Since $M\in\mathscr{A}$, there is a finite $\mathscr{F}\subseteq\mathscr{U}$ such that $M=\bigcup\mathscr{F}$. Suppose that $M\ne X$; then there is some $x\in X\setminus M$. $\mathscr{U}$ covers $X$, so there is some $U\in\mathscr{U}$ such that $x\in U$. Now let $\mathscr{G}=\mathscr{F}\cup\{U\}$, and let $G=\bigcup\mathscr{G}$. Clearly $\mathscr{G}$ is a finite subset of $\mathscr{U}$, so $G\in\mathscr{A}$. Moreover,
$$M=\bigcup\mathscr{F}\subseteq\bigcup\mathscr{G}=G\qquad\text{and}\qquad x\in G\setminus M\;,$$
so $M\subsetneqq G$. This contradicts the maximality of $M$ and shows that in fact we must have $M=X$, as desired: $\mathscr{F}$ is a finite subfamily of $\mathscr{U}$ that covers $X$, and since $\mathscr{U}$ was an arbitrary open cover of $X$, we’ve shown that $X$ is compact.
