Tube lemma

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This article is about the statement of a simple but indispensable lemma in topology


Let X be a compact space and A any topological space. Consider X \times A endowed with the product topology. Suppose a \in A and U is an open subset of X \times A containing the entire slice X \times \{ a \}. Then, we can find an open subset V of A such that:

a \in V, and X \times V \subseteq U

In other words, any open subset containing a slice contains an open cylinder that contains the slice.


Given: A compact space X, a topological space A. a \in A, and U is an open subset of X \times A containing the slice X \times \{ a \}.

To prove: There exists an open subset V of A such that a \in V X \times V is contained in U.


  1. A collection of open subsets inside U whose union contains X \times \{ a \}: For each x \in X, we have (x,a) \in U, so by the definition of openness in the product topology, there exists a basis open subset M_x \times N_x \subseteq U containing (x,a). In particular, we get a collection M_x \times N_x, x \in X of open subsets contained in U, whose union contains X \times \{ a \}.
  2. This collection yields a point-indexed open cover for X: Note that since M_x \times N_x is a basis open set containing (x,a), M_x is an open subset of X containing X, so the M_x, x \in X, form an open cover of X.
  3. (Given data used: X is compact): This cover has a finite subcover: Indeed, since X is compact, we can choose a finite collection of points \{ x_1, x_2, \dots, x_n \} \subseteq X such that X is the union of the M_{x_i}s.
  4. If V is the intersection of the corresponding N_{x_i}s, then V is open in A, a \in V, and X \times V \subseteq U: First, V is open since it is an intersection of finitely many open subsets of Y. Second, each N_{x_i} contains a, so a \in V. Third, if (x,v) \in X \times V, then there exists x_i such that x \in M_{x_i}. By definition, v \in N_{x_i}, so (x,v) \in M_{x_i} \times N_{x_i} \subseteq V. Thus, (x,v) \in U.


Textbook references

  • Topology (2nd edition) by James R. Munkres, More info, Page 168, Lemma 26.8, Chapter 3, Section 26 (the proof is given before the theorem, as Step 1 of the proof of Theorem 26.7 on page 167)