A sufficient condition for a collection of open sets of a manifold to contain all open sets

Question

Question

Let $k\geq 0$ be an integer and let $M$ be a topological $n$-manifold. Let $\mathcal{U}$ be a set of open sets of $M$ which satisfies the following closure properties:

(1). Let $U\subset M$ be an open set and let $A_0,\dots,A_k\subset U$ be pairwise disjoint closed subsets. Suppose that, for each nonempty finite subset $S\subset \{0,\dots,k\},$ the intersection $\bigcap_{i\in S}(U\setminus A_i)$ belongs to $\mathcal{U}$. Then $U$ belongs to $\mathcal{U}$.

(2). Let $U_0\subset U_1\subset \cdots $ be an increasing sequence of elements in $\mathcal{U}.$ Then $\bigcup_{i\geq0}U_i$ belongs to $\mathcal{U}$.

(3). Let $U\subset M$ be an open set homeomorphic to $\mathbb{R}^n\times S$, where $S$ is a finite set of cardinality $\leq k$. Then $U$ belongs to $\mathcal{U}$.

Is it true that $\mathcal{U}$ contains every open set of $M$?

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Remarks

• This is true if $M$ admits a smooth structure, as Weiss proved in Theorem 5.1 of [Wei99]. (This question was motivated by his proof.)

• Weiss's proof relies on the classical fact that every smooth manifold with boundary $N$ admits a very nice handle decomposition, in the sense that there is a filtration $$ \emptyset=N_{-1}\subset N_0 \subset \dots\subset N_n=N $$ such that $N_{i}$ is obtained from $N_{i-1}$ by attaching $i$-handles. It is well-known that in dimension $\neq 4$, every topological manifold admits a handle decomposition; but as far as I am aware, the ordering of the indices of the handles is completely random in this decomposition (i.e., we might attach a $5$-handle, and then a $3$-handle, say), so we do not seem to have a nice handle decomposition as above.

I have the impression that the proof should be pretty straightforward, but I cannot wrap my head around this problem. I appreciate any help. Thanks in advance.

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[Wei99] Michael Weiss "Embeddings from the point of view of immersion theory: Part I," Geom. Topol. 3(1), 67-101, (1999)

Answer 1

Yes, it is true that $\mathcal{U}$ contains all open sets of $M$. The proof is a minor modification of Weiss's argument, and proceeds in several steps.

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Step1

We show that every open set of $M$ homeomorphic to $\mathbb{R}^n\times S$ for some finite set $S$ belongs to $\mathcal{U}$, arguing by induction on the cardinality of $S$. This is easy and hence omitted.

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Step2

Let $U\subset M$ be an open set such that there is a compact manifold with boundary $N$ such that $U=\operatorname{Int}N.$ We will show that $U\in \mathcal{U}$ if $N$ admits a handle decomposition.

Let us say that a handle decomposition (starting from $\emptyset$) of $N$ has type $(a_0,\dots,a_n)$ if $d$-handles are used exactly $a_d$ times in the decomposition. We define the handle type of $N$ to be the minimal element $(a_0,\dots,a_n)\in \mathbb{Z}_{\geq0}^{n+1}$ for which $N$ admits a handle decomposition of type $(a_0,\dots,a_n)$; here we understand that $\mathbb{Z}_{\geq0}^{n+1}$ is equipped with the lexicographic ordering read from right to left. (Thus $(a_0,\dots,a_n)\leq(b_0,\dots,b_n)$ if and only if $a_i=b_i$ for all $i$ or $a_s<b_s$, where $s$ is the maximal integer such that $a_s\neq b_s$.) The proof is a transfinite induction on the handle type of $N$.

Let $(a_0,\dots,a_n)$ denote the handle type of $N$, and suppose that we have proved the claim for all indices smaller than $(a_0,\dots,a_n)$. We wish to show that $U=\operatorname{Int}N$ belongs to $\mathcal{U}$. The claim follows from Step1 if $a_1=\dots=a_n=0$, so assume that $a_i>0$ for some $i>0$. Let

$$ \emptyset=N_0\subset N_1\subset \cdots \subset N_a=N $$ be a handle decomposition of $N$ of type $(a_0,\dots,a_n)$, where we wrote $a=\sum_{i=0}^na_i$. By rechoosing a handle decomposition if necessary, we may assume that $N_{a}$ is obtained from $N_{a-1}$ by attaching a handle of positive index, say $\lambda>0$. Let $e:D^{\lambda}\times D^{n-\lambda}\to N_{a}$ denote the embedding of the $\lambda$-handle. Choose disjoint closed disks $C_0,\dots,C_k\subset \operatorname{Int}N$, and set $A_j=N_{a}\setminus e(C_j\times D^{n-j})$. For each nonempty finite subset $S\subset \{0,\dots,k\}$, the intersection $\bigcap_{j\in S}U\setminus {A_j}$ is the interior of a manifold with boundary obtained from $N_{a-1}$ by attaching $|S|-1$ $(\lambda-1)$-handles; hence it is an element of $\mathcal{U}$. Applying condition (1), we deduce that $U\in \mathcal{U},$ as required.

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Step3

Using condition (2) and Step2, we deduce that every open set $U\subset M$ which is the interior of a compact manifold with boundary admitting a handle decomposition belongs to $\mathcal{U}$. This includes the case where $n\neq 4$ or $U$ admits a smooth structure.

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Step4

Suppose $n=4$, and let $U\subset M$ be an open set. We claim that $U$ belongs to $\mathcal{U}$. For each path component $V\in \pi_0(U),$ pick $k+1$ distinct points $p_0^V,\dots,p_k^V$. We let $A_j$ denote the set of points $p_j^V,$ where $V$ ranges over the components of $U$. Since connected, noncompact 4-manifolds are smoothable, for each nonempty finite set $S\subset \{0,\dots,k\}$, the intersection $\bigcap_{j\in S}U\setminus A_j$ is smoothable. It follows from Step 3 and condition (1) that $U$ belongs to $\mathcal{U}$, and the proof is complete.