A noetherian topological space is compact

Solution 1:

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.

Solution 2:

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 $U_{\alpha_1}\cup ... \cup U_{\alpha_n}$.