A smooth Urysohn lemma for a closed set in an open set

If $M=\mathrm{\varnothing }$ or $A=\mathrm{\varnothing }$, take $f=0$, using the empty open neighbourhood of $A$. Otherwise put $O_0=U$ and $O_1=M\setminus A$. Form the set of all tuples consisting of a smooth coordinate ball $B$, its chart data, and a label $a\in \{0,1\}$ such that $\bar{B}$ is compact and lies in $O_a$. These balls cover $M$ by [F1]. Fix a countable basis. For each nonempty basis member contained in one eligible ball, [A1] chooses one such tuple. The selected balls cover $M$: every point lies in a basis member inside an eligible ball. Enumerate the selected tuples as $(B_n,a_n)$, repeating one if the nonempty family is finite. This avoids choosing a ball at every point.

Put $H_r={\bigcup }_{n\le r}\overline{B_n}$. These sets are compact and closed by [L1], and their interiors cover $M$. Put $r_1=1$ and let $r_{m+1}$ be the least integer greater than $r_m$ for which $H_{r_m}\subseteq \mathrm{int}{H}_{r_{m+1}}$. Such an integer exists by compactness and the increasing open cover $(\mathrm{int}{H}_r)$. Set $K_m=H_{r_m}$ and $K_0=K_{-1}=\mathrm{\varnothing }$. Then $K_m\subseteq \mathrm{int}{K}_{m+1}$ and their interiors cover $M$. This recursion uses least integers, not dependent choice.

The sets $C_m=K_m\setminus \mathrm{int}{K}_{m-1}$ are compact and cover $M$. For each $m$, form all tuples $(W,V,n)$ of nested smooth coordinate balls with compact closures satisfying $\bar{W}\subseteq V,\bar{V}\subseteq B_n\cap \mathrm{int}{K}_{m+1}\setminus K_{m-2}\mathrm{.}$ Their $W$-sets cover $C_m$: for $x\in C_m$, choose one $B_n$ containing $x$ and apply the small-ball construction twice inside the displayed open neighbourhood of $x$. This is an existence argument at one point, not a simultaneous point-indexed selection. Compactness provides a finite ordered list of such tuples covering $C_m$.

Apply [A1] to the countable family of nonempty sets of finite ordered covering lists, taking the empty list when $C_m$ is empty. Label a tuple from annulus $m$ by $(m,j)$. The resulting family $(W_{m,j})$ covers $M$, and $(V_{m,j})$ is locally finite: if $x\in \mathrm{int}{K}_t$, no tuple from $m\ge t+2$ meets $\mathrm{int}{K}_t$, since its $V$ misses $K_{m-2}\supseteq K_t$. Only finitely many tuples occur in the remaining finitely many annuli. The pairs $(m,j)$ form a subset of ${\mathbb{N}}^2$, hence an at most countable index set $I$.

For each $i=(m,j)\in I$, [L2] supplies a nonempty set of smooth bumps $g_i:M\to [0,1]$ equal to one near $\overline{W_i}$ and supported in $V_i$. Use [A1] to choose this countable family. Its supports are locally finite, and $G={\sum }_{i\in I}{g}_i$ is smooth and strictly positive by [L3], because the $W_i$ cover $M$. Attach to $i$ the original label $a_{n(i)}$ from its tuple.

For $a=0,1$, put $S_a={\bigcup }_{i:a_{n(i)}=a}\mathrm{supp}(g_i),h_a={\sum }_{i:a_{n(i)}=a}{g}_i\mathrm{.}$ The sums are smooth by [L3]. Each $S_a$ is closed: around a point outside $S_a$ choose a neighbourhood meeting only finitely many constituent supports, and remove those finitely many closed sets. The resulting neighbourhood misses $S_a$. Moreover $S_a\subseteq O_a$, and the nonzero locus of $h_a$ is contained in $S_a$, so $\mathrm{supp}(h_a)\subseteq S_a$. Thus taking a closure after regrouping does not enlarge the support outside its assigned open set.

Since $h_0+h_1=G>0$, define $f=h_0\mathrm{/}G$. It is smooth by [L4] and lies in $[0,1]$. Its nonzero locus lies in the closed set $S_0\subseteq U$, so $\mathrm{supp}(f)\subseteq U$. On the open set $M\setminus S_1$, which contains $A$, one has $h_1=0$ and therefore $f=h_0\mathrm{/}G=1$. This proves the required neighbourhood and support assertions, including the empty cases in step 1.1.