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:

• 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 $S\subset \{0,\dots,n\},$ the intersection $\bigcap_{i\in S}(U\setminus A_i)$ belongs to $\mathcal{U}$. Then $U$ belongs to $\mathcal{U}$.

• 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 $\mathcal{U}$.

• 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 $M$?

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.

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