LemmaStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-07-29 (claude-sonnet-5)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Tube lemma: if $K$ is a compact subset of a metric space $X$ , $Z$ is a topological space and $N$ is open in $X \times Z$ with $K \times {z_{0}} \subseteq N$ , then $K \times W \subseteq N$ for some open $W ∋ z_{0}$

Statement

Let $(X, d)$ be a metric space carrying its metric topology (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, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not), let $(Z, T_{Z})$ be a topological space (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison), and give $X \times Z$ the product topology (The product set $\prod_{i \in I} X_{i}$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space). Let $K \subseteq X$ be a compact subset (Open cover, subcover, compact metric space, and compact subset of a metric space), let $z_{0} \in Z$ , and let $N \subseteq X \times Z$ be open with

$K \times {z_{0}} \subseteq N .$

Then there is an open $W \subseteq Z$ with $z_{0} \in W$ and

$K \times W \subseteq N .$

The set $K \times W$ is the tube of the name. The case $K = \emptyset$ is included and is settled by $W = Z$ . No choice principle is used at all: the cover produced below is indexed by pairs of open sets, so the ambient form of compactness returns the second entries together with the indices and nothing has to be selected afterwards.

Facts & Assumptions

Given: A metric space $(X, d)$ with its metric topology, a topological space $(Z, T_{Z})$ , the product $X \times Z$ with the product topology, a compact $K \subseteq X$ , a point $z_{0} \in Z$ and an open $N \subseteq X \times Z$ with $K \times {z_{0}} \subseteq N$ .

[A1]

$K \times {z_{0}} \subseteq N$ , that is $(a, z_{0}) \in N$ for every $a \in K$ .

[L1]

For a two-element index set the basic product-open sets are exactly the boxes: the sets $U \times W$ with $U$ open in $X$ and $W$ open in $Z$ form a basis for the product topology on $X \times Z$ (The product set $\prod_{i \in I} X_{i}$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space, A family is a basis for a unique topology iff it covers the set and every point of an intersection of two members lies in a member inside that intersection; finite intersections of any subbasis form a basis, Basis and subbasis for a topology, and the topology generated by a family of sets).

[L2]

$K$ is a compact subset of $X$ exactly when for every set $I$ and every family $(U_{i})_{i \in I}$ of open subsets of $X$ with $K \subseteq ⋃_{i \in I} U_{i}$ there are $n \in N$ and $i_{0}, \dots, i_{n} \in I$ with $K \subseteq U_{i_{0}} \cup \dots \cup U_{i_{n}}$ , or else $K = \emptyset$ (A subset of a metric space is open in the subspace metric exactly when it is the trace of an open set of the ambient space, and it is compact as a metric space in its own right exactly when every family of open subsets of the ambient space covering it, indexed or not, has finitely many members covering it, claim 3).

[L3]

$Z$ is open in $Z$ , and an intersection of finitely many open subsets of $Z$ is open (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison, axioms (T1) and (T3) iterated).

Proof

technique · direct

1.1

If $K = \emptyset$ then $K \times Z = \emptyset \subseteq N$ and $Z$ is an open set containing $z_{0}$ , so $W : = Z$ settles the claim; assume from here on that $K \neq \emptyset$ .

L3given

1.2

Let $P$ be the set of all pairs $(U, W)$ with $U$ open in $X$ , $W$ open in $Z$ , $z_{0} \in W$ and $U \times W \subseteq N$ ; this is a set cut out by a property of the pair, and nothing is selected in forming it.

constructL1

2.1

The family $(U)_{(U, W) \in P}$ , indexed by $P$ and assigning to each pair its first entry, is a family of open subsets of $X$ and it covers $K$ : for $a \in K$ we have $(a, z_{0}) \in N$ by [A1], so by [L1] there are $U$ open in $X$ and $W$ open in $Z$ with $(a, z_{0}) \in U \times W \subseteq N$ , and then $(U, W) \in P$ with $a \in U$ .

A1L1step 1.2

3.1

Since $K \neq \emptyset$ is compact, there are $n \in N$ and pairs $(U_{0}, W_{0}), \dots, (U_{n}, W_{n}) \in P$ with $K \subseteq U_{0} \cup \dots \cup U_{n}$ .

step 1.1step 2.1L2

4.1

Each index returned by step 3.1 is itself a pair, so its second entry $W_{j}$ is given with it and nothing is chosen; put $W : = W_{0} \cap \dots \cap W_{n}$ , which contains $z_{0}$ because every $W_{j}$ does, and is open in $Z$ as an intersection of $n + 1 \geq 1$ open sets.

step 3.1L3

5.1

$K \times W \subseteq N$ : given $a \in K$ and $z \in W$ , step 3.1 gives $j \leq n$ with $a \in U_{j}$ , and $z \in W \subseteq W_{j}$ , so $(a, z) \in U_{j} \times W_{j} \subseteq N$ by the defining property of $P$ .

step 1.2step 3.1step 4.1

6.1

Steps 4.1 and 5.1 exhibit an open $W ∋ z_{0}$ with $K \times W \subseteq N$ , which with step 1.1 proves the lemma in both cases.

step 1.1step 4.1step 5.1∎

Remarks

Why the pairs and not the open sets. A single open $U \subseteq X$ may be the first entry of many admissible pairs, and recovering a suitable $W$ from $U$ alone would be a selection over an infinite family. Indexing the cover by the pairs rather than by the sets is what makes the second entries come back with the indices, and it is the same device the ambient form of compactness uses (A subset of a metric space is open in the subspace metric exactly when it is the trace of an open set of the ambient space, and it is compact as a metric space in its own right exactly when every family of open subsets of the ambient space covering it, indexed or not, has finitely many members covering it).
Compactness of the metric factor is what the lemma is about. No hypothesis whatever is placed on $Z$ , and none is needed: the finite intersection of the $W_{j}$ is taken in $Z$ and uses only axiom (T3). What compactness of $K$ buys is that finitely many boxes already cover the slice $K \times {z_{0}}$ , so that finitely many second entries have to be intersected.
The hypothesis cannot be moved to the other factor. With $K$ replaced by a non-compact set the conclusion fails: the region under the graph of a positive function tending to $0$ contains a whole slice and no tube around it. Nothing on this page needs that witness, and it is not constructed here.
The general tube lemma, for a compact factor in an arbitrary topological product, is now available in this library, on an earlier page (Tube lemma: if $K$ is compact and an open $N \subseteq X \times Z$ contains $K \times {z_{0}}$ , then $N$ contains $K \times W$ for some open $W ∋ z_{0}$ ). The proof above is the metric special case of that general lemma, narrowed to a metric factor $X$ and written independently of it: nothing above cites the general statement, and nothing needs to, since compactness of a metric-space subset is the same notion under either reading (A subset of a metric space is open in the subspace metric exactly when it is the trace of an open set of the ambient space, and it is compact as a metric space in its own right exactly when every family of open subsets of the ambient space covering it, indexed or not, has finitely many members covering it).

Depends on

Used by

If f : X × Z → Y is continuous then its transpose F : Z → C(X,Y), F(z)(x) = f(x,z), is continuous for the compact-open topology, with no hypothesis on X beyond being metric Theorem

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45 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

Tube lemma (Wikipedia) (standard reference, not scraped)
J. Munkres, Topology, 2nd ed., §26 (standard reference, not scraped)

Tube lemma: if KK is a compact subset of a metric space XX, ZZ is a topological space and NN is open in X×ZX×Z with K×{z0}⊆NK×{z0​}⊆N, then K×W⊆NK×W⊆N for some open W∋z0W∋z0​