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Open manifolds admit exhaustions with no caps

Statement

Assume the axiom of countable choice. Let $M$ be a nonempty connected open smooth $m$ -manifold without boundary: every connected component of a manifold is open and closed, so here no component is compact, and connectedness makes $M$ noncompact. Then there is a sequence $\emptyset \neq M_{0} \subseteq M_{1} \subseteq M_{2} \subseteq \dots$ of compact $m$ -submanifolds with boundary such that $M_{j} \subseteq int M_{j + 1}$ , $M = ⋃_{j} M_{j}$ , and for every $j$ the complement $M ∖ int M_{j}$ has no compact connected component. Call a compact connected component of $M ∖ int M_{j}$ a cap of $M_{j}$ ; the conclusion is that no $M_{j}$ has a cap. Consequently, for every $j$ and every connected component $C$ of the band $M_{j + 1} ∖ int M_{j}$ the outgoing boundary $C \cap \partial M_{j + 1}$ is nonempty; the incoming boundary $C \cap \partial M_{j}$ may be empty.

Facts & Assumptions

Given: ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )) and a nonempty connected open smooth $m$ -manifold $M$ without boundary and with no compact component.

[F1]

Under ${A C}_{ω}$ there is a smooth exhaustive function $h : M \to [0, \infty)$ with $h^{- 1} ([0, c])$ compact for every $c$ (Every smooth manifold admits a smooth proper exhaustion function).

[F2]

The regular values of a smooth function have null complement, hence are dense, so every nonempty open interval contains one (Regular values have null complement and are dense).

[F3]

For a regular value $c$ the sublevel $h^{- 1} ((- \infty, c])$ is a compact smooth manifold with boundary $h^{- 1} (c)$ and interior $h^{- 1} ((- \infty, c))$ (Regular sublevels are compact manifolds with boundary); interiors and boundaries are as in Interior and boundary of a manifold with boundary and Closed sublevel and level set of a smooth function.

[F4]

Components of a manifold are open and closed, manifolds are locally connected, and a compact locally connected space has finitely many components (Connected components, quasicomponents, and totally disconnected spaces, Locally connected and locally path-connected spaces: a neighbourhood base of open connected, respectively open path-connected, sets at every point, Smooth manifolds and their smooth charts, Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right); a clopen subset of a connected space is empty or the whole space (Separation of a topological space, connected and disconnected spaces, clopen sets, and connected subsets).

[F5]

The identity on a nonempty compact metric subset of $R$ attains a minimum (A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value).

Proof

technique · direct

1.1F1F2F5givenchoose

Choose $h$ as in [F1]. Choose any $x_{0} \in M$ . The nonempty compact sublevel ${h \leq h (x_{0})}$ has compact image in $R$ , by pulling any image cover back to a cover of the sublevel; the identity function on that image has a minimum by [F5], and all other points have larger values; thus $h$ attains a global minimum $m_{0}$ ; choose a regular value $b_{0} \geq m_{0}$ with $m_{0} < b_{0} < m_{0} + 1$ , and for every $j \geq 1$ choose a regular value $b_{j} \in (b_{0} + j, b_{0} + j + 1)$ . Each interval is nonempty and contains a regular value by [F2], and the countably many selections are licensed by [F1]'s ${A C}_{ω}$ . The sequence $(b_{j})$ is strictly increasing with $b_{j} \to \infty$ .

2.1F3step 1.1

For each $j$ put $K_{j} : = h^{- 1} ((- \infty, b_{j}])$ . By [F3] each $K_{j}$ is a nonempty compact smooth $m$ -manifold with boundary $h^{- 1} (b_{j})$ and interior $h^{- 1} ((- \infty, b_{j}))$ ; since $b_{j} < b_{j + 1}$ are regular, $K_{j} \subseteq int K_{j + 1}$ , and the $K_{j}$ exhaust $M$ because $h$ is exhaustive.

3.1F4givenstep 2.1

Fix $j$ and let $Z$ be a cap of $K_{j}$ , that is a compact connected component of $M ∖ int K_{j}$ . Its boundary in $M$ is $\partial Z = Z \cap h^{- 1} (b_{j})$ : a point of $Z$ with $h > b_{j}$ has a ball around it contained in the open set ${h > b_{j}} \subseteq M ∖ int K_{j}$ and, being connected, that ball lies in the component $Z$ , while a ball around a point of $Z \cap h^{- 1} (b_{j})$ meets ${h < b_{j}}$ because $b_{j}$ is a regular value. If $\partial Z = \emptyset$ , then every point of $Z$ is interior to $Z$ in $M$ , so $Z$ is open in $M$ , while $Z$ is closed in $M$ because it is a component of the closed set $M ∖ int K_{j}$ ; connectivity of $M$ then forces $Z = M$ ; but then $M$ would be compact, contradicting that $M$ has no compact component. Hence $\partial Z \neq \emptyset$ .

4.1F4step 3.1algebra

The cap $Z$ is a compact smooth $m$ -manifold with boundary $\partial Z$ : at a point with $h > b_{j}$ it is open in $M$ by the ball argument of step 3.1, and at a point of the level the local normal form of $h$ at the regular value $b_{j}$ (Local normal form for submersions) exhibits a neighbourhood of $Z$ as a half-space. Hence $\partial Z$ is a nonempty closed $(m - 1)$ -submanifold of the compact $(m - 1)$ -manifold $h^{- 1} (b_{j})$ (The boundary of a positive-dimensional manifold is a closed embedded smooth (n-1)-manifold, Embedded smooth submanifolds with boundary), hence a union of components of $h^{- 1} (b_{j})$ . Two distinct caps have disjoint boundaries: a point of $\partial Z_{1} \cap \partial Z_{2}$ has a neighbourhood in $M ∖ int K_{j}$ that is connected (a half-ball at the level set) and meets both caps, contradicting that they are distinct components. Since $h^{- 1} (b_{j})$ is compact and locally connected it has finitely many components by [F4], so sending a cap to the nonempty set of level components in its boundary injects the caps into the power set of a finite set: there are finitely many caps $Z_{1}, \dots, Z_{r}$ of $K_{j}$ .

5.1F3step 4.1construct

Put $K_{j}^{+} : = K_{j} \cup Z_{1} \cup \dots \cup Z_{r}$ . Give $K_{j}^{+}$ the smooth structure with boundary carried by the ambient charts of $M$ : a point of $int K_{j}$ or of $int Z_{i}$ has an open neighbourhood in $M$ contained in $K_{j}^{+}$ and serves as an interior chart; at a seam point in $\partial Z_{i} \subseteq \partial K_{j}$ the local regular-level chart has its lower half in $K_{j}$ and its upper half in $Z_{i}$ , so their union contains a full ambient neighbourhood and the seam point is interior too; a point of $\partial K_{j} ∖ (\partial Z_{1} \cup \dots \cup \partial Z_{r})$ has a half-space chart inherited from a boundary chart of $K_{j}$ , and a sufficiently small such chart avoids the caps because the caps meet $\partial K_{j}$ exactly in the closed sets $\partial Z_{i}$ ; transitions are restrictions of transition maps of $M$ . Hence $K_{j}^{+}$ is a compact smooth $m$ -manifold with boundary, with $\partial K_{j}^{+} = \partial K_{j} ∖ (\partial Z_{1} \cup \dots \cup \partial Z_{r})$ and $int K_{j}^{+} = int K_{j} \cup Z_{1} \cup \dots \cup Z_{r}$ .

6.1step 3.1step 5.1

The complement $M ∖ int K_{j}^{+}$ is obtained from $M ∖ int K_{j}$ by deleting the components $Z_{1}, \dots, Z_{r}$ , so every connected component of it is a connected component of $M ∖ int K_{j}$ other than the $Z_{i}$ , hence is noncompact by the definition of a cap. Therefore $K_{j}^{+}$ has no cap.

7.1F3step 2.1step 5.1step 6.1constructchoose

Define $M_{0} : = K_{0}^{+}$ , which is nonempty, compact, and cap-free by step 6.1. Given a cap-free compact $M_{j}$ , let $k_{j}$ be the least integer with $M_{j} \subseteq int K_{k_{j}}$ , which exists because the compact $M_{j}$ is contained in $M = ⋃_{k} int K_{k}$ ; set $M_{j + 1} : = K_{k_{j}}^{+}$ . Then $M_{j} \subseteq int K_{k_{j}} \subseteq int M_{j + 1}$ and $M_{j + 1}$ is compact and cap-free by step 6.1. The indices $k_{j}$ strictly increase, since $K_{k_{j}} \subseteq M_{j + 1}$ forces $k_{j + 1} > k_{j}$ ; hence $M = ⋃_{j} M_{j}$ and each $M_{j}$ is a nonempty compact $m$ -manifold with boundary.

8.1F4step 7.1algebra∎

Let $C$ be a connected component of the band $M_{j + 1} ∖ int M_{j}$ and suppose $C \cap \partial M_{j + 1} = \emptyset$ ; then $C \subseteq int M_{j + 1} ∖ int M_{j}$ . At a point $x \in \partial C$ we have $\partial C \subseteq \partial M_{j}$ , and since $\partial M_{j} \subseteq M_{j} \subseteq int M_{j + 1}$ there is a ball $B$ around $x$ contained in $int M_{j + 1}$ ; the set $B \cap (M ∖ int M_{j})$ is a half-ball, hence connected, and meets $C$ , so it lies in $C$ . Thus $C$ is open and closed in $M ∖ int M_{j}$ , and it is compact, so it is a compact component of $M ∖ int M_{j}$ , that is a cap of $M_{j}$ , contradicting cap-freeness. Hence $C \cap \partial M_{j + 1} \neq \emptyset$ . The band is compact and locally connected and therefore has finitely many components by [F4].

Depends on

Used by

Open manifolds admit handle filtrations without top-index handles Lemma

Dependency tree · two levels

72 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Liviu I. Nicolaescu, An Invitation to Morse Theory, 2nd ed., Ch. 2 §2.2 “The Topology of Sublevel Sets” (exhaustive Morse functions, compact sublevel sets, handle attachment across critical values), printed pp. 37–56 (standard reference, not scraped)
John Francis, The h-Principle, Lectures 5 & 6: The Hirsch–Smale theorem, Lemma 1.6 (a manifold has a handle decomposition without n-handles iff it is open) (standard reference, not scraped)