Subspaces of a Noetherian space and its compact open subsets

Let $W\subseteq X$ be an open subset and let $(P_a{)}_{a\in A}$ be an open cover of the space $W$, so that each $P_a$ is an open subset of $W$. We show that finitely many $P_a$ cover $W$. Suppose not. Recursively choose finite unions of members of the given cover: put $V_0:=\mathrm{\varnothing }$; since $P_{a_1}\cup \cdots \cup P_{a_n}=W$ fails for every finite list and $V_n\cap W=P_{a_1}\cup \cdots \cup P_{a_n}$ for a list of traces, the set $W\setminus V_n$ is nonempty, so we may choose $x_n\in W\setminus V_n$; because the family covers $W$ there is $a_{n+1}\in A$ with $x_n\in P_{a_{n+1}}$, and we put $V_{n+1}:=V_n\cup P_{a_{n+1}}$. Each $P_a$ is the trace of an open subset of $X$ by [F3], so each $V_n$ is an open subset of $X$, and $V_0\subseteq V_1\subseteq V_2\subseteq \cdots$ is an ascending chain with $V_{n+1}\mathrm{\ne }{V}_n$ for every $n$ because $x_n\in V_{n+1}\setminus V_n$. Such a chain does not stabilize, contradicting [F1]. Hence finitely many $P_a$ cover $W$, so $W$ is compact by [F2]; the recursion uses the Axiom of Dependent Choice, available by [F4].

Let $Y\subseteq X$ be a subspace and let $V_0\subseteq V_1\subseteq V_2\subseteq \cdots$ be an ascending chain of open subsets of $Y$. By [F3] each $V_n$ is a trace $V_n=U_n\cap Y$ of an open subset $U_n\subseteq X$; the assignment $n\mapsto U_n$ is a sequence of choices, so we record it as a construction using [F4]. Put $W_n:=U_1\cup \cdots \cup U_n$, an open subset of $X$ with $W_n\subseteq W_{n+1}$. By [F1] the chain $W_1\subseteq W_2\subseteq \cdots$ stabilizes, so there is $N$ with $W_n=W_N$ for every $n\ge N$. For such $n$ one has $V_n=W_n\cap Y$ and $V_{n+1}=W_{n+1}\cap Y$, because $W_n\cap Y=(U_1\cap Y)\cup \cdots \cup (U_n\cap Y)=V_1\cup \cdots \cup V_n=V_n$ by the ascending hypothesis; hence $V_n=V_{n+1}$. Therefore the given chain in $Y$ stabilizes and $Y$ is Noetherian.

Let $Y\subseteq X$ be a subspace. By [step 1.2] the space $Y$ is Noetherian, so [step 1.1] applies to it with $W=Y$: every open cover of $Y$ has a finite subcover, that is, $Y$ is compact by [F2]. Taking $Y:=U$ for an open $U\subseteq X$ in particular shows that every open subset of $X$ is compact.

Let $U_1,U_2\subseteq X$ be compact open subsets. The intersection $U_1\cap U_2$ is open in $X$, hence compact by [step 2.1] applied to $U:=U_1\cap U_2$. Thus the compact open subsets of $X$ are closed under finite intersections.

Assertion 1 is [step 1.2], assertion 2 is [step 2.1] and assertion 3 is [step 3.1]. The Axiom of Choice entered only through the Axiom of Dependent Choice of [F4], used for the recursive selection of the sequence $(V_n)$ in [step 1.1] and for the sequence of open sets representing the chain in [step 1.2]; no other selection is made. ∎