Increasing-cover and decreasing-closed-set criteria

Assume (i), and let $(U_n)$ be an increasing open cover. Take a locally finite open refining cover $\mathcal{R}$. For each $R\in \mathcal{R}$ let $a(R)$ be the least $n$ with $R\subseteq U_n$, and set $C_n=X\setminus \bigcup \{R:a(R)>n\}$. This is closed. If $x\in C_n$, some $R$ contains $x$ because $\mathcal{R}$ covers; necessarily $a(R)\le n$, so $x\in U_n$. If $x\in X$, choose an open neighborhood meeting only finitely many members of $\mathcal{R}$. At least one meets it because $x$ is covered. The maximum $N$ of their assigned indices exists, and the neighborhood is contained in $C_N$. Thus the interiors cover and (ii) holds. All index assignments are least natural numbers, without choice.

Assume (ii) and let $(O_n)$ be any countable open cover; finite nonempty covers can be extended by empty entries. Put $U_n={\bigcup }_{i\le n}{O}_i$, obtain closed $C_n\subseteq U_n$ as in (ii), and replace $C_n$ by $D_n={\bigcup }_{i\le n}{C}_i$. Then $D_n$ is closed and increasing, is contained in $U_n$, and its interiors cover. With $D_{-1}=\mathrm{\varnothing }$ put $V_n=U_n\setminus D_{n-1}$. For each $x$, the least $n$ with $x\in D_n$ gives $x\in V_n$, so these open sets cover. If a neighborhood $N_x$ lies in $D_N$, it misses every $V_n$ with $n\ge N+1$. The family $\{V_n\cap O_i:i\le n<\omega \}$ covers $X$, refines $(O_n)$, and near $x$ has at most ${\sum }_{n\le N}(n+1)$ possibly meeting indexed members. It is the locally finite open refining cover required for (i). If the original cover is empty, $X$ is empty and its empty refining cover suffices.

Complementation proves (ii) implies (iii): given decreasing closed $F_n$ with empty intersection, use $U_n=X\setminus F_n$, and put $G_n=X\setminus C_n$. Then $F_n\subseteq G_n$ and ${\bigcap }_n\overline{G_n}=X\setminus {\bigcup }_n\mathrm{int}{C}_n=\mathrm{\varnothing }$. Conversely, given increasing $U_n$ covering $X$, put $F_n=X\setminus U_n$, take the expansions of (iii), and set $C_n=X\setminus G_n$. These are closed subsets of $U_n$, and the same identity says their interiors cover. The identity uses $\overline{X\setminus C}=X\setminus \mathrm{int}C$, which follows because a point has a neighborhood contained in $C$ exactly when it is outside that closure.

Condition (iii) implies (iv), since $G_n\subseteq \overline{G_n}$. Suppose now that $X$ is normal and (iv) holds. For a given decreasing closed sequence with empty intersection, take its open expansions $G_n$ with empty intersection. For each $n$, the closed set $F_n$ lies in the open $G_n$, so normal shrinking provides an open $H_n$ with $F_n\subseteq H_n\subseteq \overline{H_n}\subseteq G_n$. AC selects these witnesses for all $n$ simultaneously. Hence ${\bigcap }_n\overline{H_n}\subseteq {\bigcap }_n{G}_n=\mathrm{\varnothing }$, proving (iii). Steps 1.1–1.3 prove the other equivalences; empty sets cause no exception to these inclusions. QED.