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Uniform hitting for a compact closed family of ray segments

Statement

Assume Dependent Choice. Let $(Z, d)$ be a metric space with its metric topology, let $(B, T)$ be a topological space, and let $π : Z \to B$ be continuous. Write $C ([0, \infty), Z)$ for the set of continuous maps from the nonnegative real half-line, with the relative topology inherited from the metric $∣ s - t ∣$ on $R$ , to $Z$ . A complete forward path is an element of this set; its base path is $π \circ q$ . Let $F \subseteq C ([0, \infty), Z)$ and let $A \subseteq C ([0, \infty), Z)$ be a specified set of paths called admissible. Let $U \in T$ .

For maps on $[0, m]$ , uniform convergence means that, for every $ε > 0$ , eventually $d (q_{n} (t), v (t)) < ε$ for every $t \in [0, m]$ . Locally uniform convergence on $[0, \infty)$ means uniform convergence on each $[0, m]$ , for every positive integer $m$ . Assume the following two properties.

For every sequence $(q_{n})$ in $F$ and every positive integer $m$ , some subsequence of the restrictions $q_{n} ∣_{[0, m]}$ converges uniformly to a map $v : [0, m] \to Z$ . This quantifies over every sequence, including every subsequence of any previously selected sequence.
Every locally uniform limit $q : [0, \infty) \to Z$ of a sequence from $F$ belongs to $A$ .

If every such limiting path $q \in A$ satisfies $π (q (t)) \in U$ for some finite $t \geq 0$ , then there is $L < \infty$ such that every $q \in F$ satisfies $π (q (t)) \in U$ for some $t \in [0, L]$ . The formulation is abstract: a generalized-ray application supplies its own continuous phase paths and admissible class. Neither uniqueness through an initial point nor an admissible extension of an arbitrary finite segment is assumed.

Facts & Assumptions

[F1]

The phase metric and its topology are those of Metric space: d(x,y) = 0 iff x = y, symmetry, and the triangle inequality; pseudometric and ultrametric and The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement. The base topology and continuity of the projection are those of Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison and Continuity of a map of topological spaces at a point and globally. If $π (z) \in U$ , continuity provides an open neighbourhood $V$ of $z$ with $π (V) \subseteq U$ ; the metric topology then provides $δ > 0$ such that $d (w, z) < δ$ implies $w \in V$ . Time intervals carry their relative real topology as in Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace.

[F2]

Uniform convergence has the metric meaning of Uniform convergence, and the topology of uniform convergence: the metric topology of the uniform metric on Y^X and on C(X,Y). Claim 2 of A uniform limit of continuous functions is continuous, so C(X,Y) is closed in Y^X under the uniform metric makes every uniform limit of continuous maps on the nonempty interval $[0, m]$ continuous. Completeness of $Z$ is unnecessary because the finite-interval hypothesis supplies limits in $Z$ .

[F3]

Assume The axiom of dependent choice: a relation in which every element is related to something admits an ℕ-indexed chain. It selects the avoiding sequence by extension of finite lists, and is also the stated choice hypothesis of the diagonal supplier.

[F4]

Use the abstract part of Diagonal gluing without a single-valued reflected flow with $A_{0} (I) = C (I, Z)$ for every nonempty time interval $I \subset [0, \infty)$ . These are auxiliary interval classes, distinct from the specified complete-path class $A$ . Restrictions are continuous by the subspace-topology supplier. The local admissibility contract follows from the definition of continuity: if near each $t \in I$ a map is continuous on a relative interval neighbourhood, every open neighbourhood of its value has a preimage containing a neighbourhood of $t$ in $I$ . Thus the map is continuous at every $t$ . The hereditary extraction hypothesis will be checked for the actual avoiding sequence; no geometric ray construction is used in this abstract application.

Proof

Given: The metric phase space, topological base, continuous projection, path classes, two compactness/closedness assumptions, limiting-path hitting assumption, and Dependent Choice. Goal: A finite common hitting bound for $F$ .

1.1givenF3

Suppose no common bound exists. For every positive integer $n$ there is a path in $F$ whose base avoids $U$ throughout $[0, n]$ . Consider finite lists of paths whose $n$ th entry has this property, including the empty list. Extending a list by one suitable entry is an entire relation on these lists, since failure of a common bound supplies that entry. Dependent Choice starting at the empty list yields a sequence $(q_{j})_{j \geq 1}$ with $π (q_{j} (t)) \notin U$ for $0 \leq t \leq j$ .

2.1givenF2F4step 1.1

Set $a_{j} = q_{j} ∣_{[0, j]}$ . These curves belong to $A_{0} ([0, j])$ by restriction continuity. Fix a positive integer $m$ and any strictly increasing indices $j_{k} \geq m$ . The sequence $(q_{j_{k}})$ consists of members of $F$ , so the finite-interval compactness hypothesis gives a further subsequence converging uniformly on $[0, m]$ to a map into $Z$ . That limit is continuous by the uniform limit theorem and therefore belongs to $A_{0} ([0, m])$ . This verifies the hereditary extraction property of the diagonal supplier, including extraction from every previously selected subsequence.

3.1givenF3F4step 2.1

Apply the diagonal supplier to $(a_{j})$ and the interval classes $A_{0} (I)$ . It gives strictly increasing indices $n_{k}$ and one continuous path $q : [0, \infty) \to Z$ such that $q_{n_{k}} \to q$ uniformly on every bounded time interval, since $a_{n_{k}}$ and $q_{n_{k}}$ agree there whenever $n_{k}$ exceeds the endpoint. The closedness assumption in the Statement puts $q$ in $A$ , and the limiting-path hitting assumption supplies a finite $t \geq 0$ with $π (q (t)) \in U$ .

4.1F1F2step 1.1step 3.1∎

Put $z = q (t)$ and choose $δ > 0$ as in F1 so that $d (w, z) < δ$ implies $π (w) \in U$ . Choose an integer $m > t$ . Uniform convergence on $[0, m]$ gives $d (q_{n_{k}} (t), z) < δ$ for all sufficiently large $k$ , hence $π (q_{n_{k}} (t)) \in U$ . But $n_{k} \to \infty$ , so also $n_{k} \geq t$ for large $k$ , contradicting the avoidance property in step 1.1. The contradiction proves that a finite common bound exists.

Depends on

Used by · 1

Non-trapping and geometric control from infinity

Dependency tree · 2 levels

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