# Tube lemma

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Lemma in topology

In [mathematics](/source/Mathematics), particularly [topology](/source/Topology), the **tube lemma**, also called Wallace's theorem, is a useful tool in order to prove that the [product](/source/Product_topology) of finitely many [compact spaces](/source/Compact_spaces) is compact.

## Statement

The lemma uses the following terminology:

- If X {\displaystyle X} and Y {\displaystyle Y} are [topological spaces](/source/Topological_space) and X × Y {\displaystyle X\times Y} is the product space, endowed with the [product topology](/source/Product_topology), a **slice** in X × Y {\displaystyle X\times Y} is a set of the form { x } × Y {\displaystyle \{x\}\times Y} for x ∈ X {\displaystyle x\in X} .

- A **tube** in X × Y {\displaystyle X\times Y} is a subset of the form U × Y {\displaystyle U\times Y} where U {\displaystyle U} is an open subset of X {\displaystyle X} . It contains all the slices { x } × Y {\displaystyle \{x\}\times Y} for x ∈ U {\displaystyle x\in U} .

**Tube Lemma**—Let X {\displaystyle X} and Y {\displaystyle Y} be topological spaces with Y {\displaystyle Y} compact, and consider the [product space](/source/Product_topology) X × Y . {\displaystyle X\times Y.} If N {\displaystyle N} is an open set containing a slice in X × Y , {\displaystyle X\times Y,} then there exists a tube in X × Y {\displaystyle X\times Y} containing this slice and contained in N . {\displaystyle N.}

Using the concept of [closed maps](/source/Closed_map), this can be rephrased concisely as follows: if X {\displaystyle X} is any topological space and Y {\displaystyle Y} a compact space, then the projection map X × Y → X {\displaystyle X\times Y\to X} is closed.

**Generalized Tube Lemma 1**—Let X {\displaystyle X} and Y {\displaystyle Y} be topological spaces and consider the product space X × Y . {\displaystyle X\times Y.} Let A {\displaystyle A} be a compact subset of X {\displaystyle X} and B {\displaystyle B} be a compact subset of Y . {\displaystyle Y.} If N {\displaystyle N} is an open set containing A × B , {\displaystyle A\times B,} then there exists U {\displaystyle U} open in X {\displaystyle X} and V {\displaystyle V} open in Y {\displaystyle Y} such that A × B ⊆ U × V ⊆ N . {\displaystyle A\times B\subseteq U\times V\subseteq N.}

**Generalized Tube Lemma 2**—Let X i , i ∈ I {\displaystyle X_{i},i\in I} be topological spaces and consider the product space ∏ i ∈ I X i . {\displaystyle \prod _{i\in I}X_{i}.} For each i ∈ I {\displaystyle i\in I} , let A i {\displaystyle A_{i}} be a compact subset of X i . {\displaystyle X_{i}.} If N {\displaystyle N} is an open set containing ∏ i ∈ I A i , {\displaystyle \prod _{i\in I}A_{i},} then there exists U i {\displaystyle U_{i}} open in X i {\displaystyle X_{i}} with U i = X i {\displaystyle U_{i}=X_{i}} for all but finite amount of i ∈ I {\displaystyle i\in I} , such that ∏ i ∈ I A i ⊆ ∏ i ∈ I U i ⊆ N . {\displaystyle \prod _{i\in I}A_{i}\subseteq \prod _{i\in I}U_{i}\subseteq N.}

## Examples and properties

1. Consider R × R {\displaystyle \mathbb {R} \times \mathbb {R} } in the product topology, that is the [Euclidean plane](/source/Euclidean_plane), and the open set N = { ( x , y ) ∈ R × R : | x y | < 1 } . {\displaystyle N=\{(x,y)\in \mathbb {R} \times \mathbb {R} ~:~|xy|<1\}.} The open set N {\displaystyle N} contains { 0 } × R , {\displaystyle \{0\}\times \mathbb {R} ,} but contains no tube, so in this case the tube lemma fails. Indeed, if W × R {\displaystyle W\times \mathbb {R} } is a tube containing { 0 } × R {\displaystyle \{0\}\times \mathbb {R} } and contained in N , {\displaystyle N,} W {\displaystyle W} must be a subset of ( − 1 / x , 1 / x ) {\displaystyle \left(-1/x,1/x\right)} for all x > 0 {\displaystyle x>0} which means W = { 0 } {\displaystyle W=\{0\}} contradicting the fact that W {\displaystyle W} is open in R {\displaystyle \mathbb {R} } (because W × R {\displaystyle W\times \mathbb {R} } is a tube). This shows that the compactness assumption is essential.

2. The tube lemma can be used to prove that if X {\displaystyle X} and Y {\displaystyle Y} are compact spaces, then X × Y {\displaystyle X\times Y} is compact as follows:

Let { G a } {\displaystyle \{G_{a}\}} be an [open cover](/source/Cover_(topology)) of X × Y {\displaystyle X\times Y} . For each x ∈ X {\displaystyle x\in X} , cover the slice { x } × Y {\displaystyle \{x\}\times Y} by finitely many elements of { G a } {\displaystyle \{G_{a}\}} (this is possible since { x } × Y {\displaystyle \{x\}\times Y} is compact, being [homeomorphic](/source/Homeomorphic) to Y {\displaystyle Y} ). Call the union of these finitely many elements N x . {\displaystyle N_{x}.} By the tube lemma, there is an [open set](/source/Open_set) of the form W x × Y {\displaystyle W_{x}\times Y} containing { x } × Y {\displaystyle \{x\}\times Y} and contained in N x . {\displaystyle N_{x}.} The collection of all W x {\displaystyle W_{x}} for x ∈ X {\displaystyle x\in X} is an open cover of X {\displaystyle X} and hence has a finite subcover { W x 1 , … , W x n } {\displaystyle \{W_{x_{1}},\dots ,W_{x_{n}}\}} . Thus the finite collection { W x 1 × Y , … , W x n × Y } {\displaystyle \{W_{x_{1}}\times Y,\dots ,W_{x_{n}}\times Y\}} covers X × Y {\displaystyle X\times Y} . Using the fact that each W x i × Y {\displaystyle W_{x_{i}}\times Y} is contained in N x i {\displaystyle N_{x_{i}}} and each N x i {\displaystyle N_{x_{i}}} is the finite union of elements of { G a } {\displaystyle \{G_{a}\}} , one gets a finite subcollection of { G a } {\displaystyle \{G_{a}\}} that covers X × Y {\displaystyle X\times Y} .

3. By part 2 and induction, one can show that the finite product of compact spaces is compact.

4. The tube lemma cannot be used to prove the [Tychonoff theorem](/source/Tychonoff_theorem), which generalizes the above to infinite products.

## Proof

The tube lemma follows from the generalized tube lemma by taking A = { x } {\displaystyle A=\{x\}} and B = Y . {\displaystyle B=Y.} It therefore suffices to prove the generalized tube lemma. By the definition of the product topology, for each ( a , b ) ∈ A × B {\displaystyle (a,b)\in A\times B} there are open sets U a , b ⊆ X {\displaystyle U_{a,b}\subseteq X} and V a , b ⊆ Y {\displaystyle V_{a,b}\subseteq Y} such that ( a , b ) ∈ U a , b × V a , b ⊆ N . {\displaystyle (a,b)\in U_{a,b}\times V_{a,b}\subseteq N.} For any a ∈ A , {\displaystyle a\in A,} { V a , b : b ∈ B } {\displaystyle \left\{V_{a,b}~:~b\in B\right\}} is an open cover of the compact set B {\displaystyle B} so this cover has a finite subcover; namely, there is a finite set B 0 ( a ) ⊆ B {\displaystyle B_{0}(a)\subseteq B} such that V a := ⋃ b ∈ B 0 ( a ) V a , b {\displaystyle V_{a}:=\bigcup _{b\in B_{0}(a)}V_{a,b}} contains B , {\displaystyle B,} where V a {\displaystyle V_{a}} is open in Y . {\displaystyle Y.} For every a ∈ A , {\displaystyle a\in A,} let U a := ⋂ b ∈ B 0 ( a ) U a , b , {\displaystyle U_{a}:=\bigcap _{b\in B_{0}(a)}U_{a,b},} which is an open set in X {\displaystyle X} since B 0 ( a ) {\displaystyle B_{0}(a)} is finite. Moreover, the construction of U a {\displaystyle U_{a}} and V a {\displaystyle V_{a}} implies that { a } × B ⊆ U a × V a ⊆ N . {\displaystyle \{a\}\times B\subseteq U_{a}\times V_{a}\subseteq N.} We now essentially repeat the argument to drop the dependence on a . {\displaystyle a.} Let A 0 ⊆ A {\displaystyle A_{0}\subseteq A} be a finite subset such that U := ⋃ a ∈ A 0 U a {\displaystyle U:=\bigcup _{a\in A_{0}}U_{a}} contains A {\displaystyle A} and set V := ⋂ a ∈ A 0 V a . {\displaystyle V:=\bigcap _{a\in A_{0}}V_{a}.} It then follows by the above reasoning that A × B ⊆ U × V ⊆ N {\displaystyle A\times B\subseteq U\times V\subseteq N} , and U ⊆ X {\displaystyle U\subseteq X} and V ⊆ Y {\displaystyle V\subseteq Y} are respectively open, which completes the proof.

## See also

- [Alexander's sub-base theorem](/source/Alexander's_sub-base_theorem) – Collection of subsets that generate a topologyPages displaying short descriptions of redirect targets

- [Tubular neighborhood](/source/Tubular_neighborhood) – Neighborhood of a submanifold

- [Tychonoff theorem](/source/Tychonoff_theorem) – Product of any collection of compact topological spaces is compactPages displaying short descriptions of redirect targets

## References

- [James Munkres](/source/James_Munkres) (1999). *Topology* (2nd ed.). [Prentice Hall](/source/Prentice_Hall). [ISBN](/source/ISBN_(identifier)) [0-13-181629-2](https://en.wikipedia.org/wiki/Special:BookSources/0-13-181629-2).

- [Joseph J. Rotman](/source/Joseph_J._Rotman) (1988). [*An Introduction to Algebraic Topology*](https://archive.org/details/introductiontoal0000rotm). Springer. [ISBN](/source/ISBN_(identifier)) [0-387-96678-1](https://en.wikipedia.org/wiki/Special:BookSources/0-387-96678-1). *(See Chapter 8, Lemma 8.9)*

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