In
mathematics, particularly
topology
In mathematics, topology (from the Greek words , and ) is concerned with the properties of a geometric object that are preserved under continuous deformations, such as stretching, twisting, crumpling, and bending; that is, without closing ...
, the tube lemma is a useful tool in order to prove that the finite
product
Product may refer to:
Business
* Product (business), an item that serves as a solution to a specific consumer problem.
* Product (project management), a deliverable or set of deliverables that contribute to a business solution
Mathematics
* Produ ...
of
compact space
In mathematics, specifically general topology, compactness is a property that seeks to generalize the notion of a closed and bounded subset of Euclidean space by making precise the idea of a space having no "punctures" or "missing endpoints", ...
s is compact.
Statement
The lemma uses the following terminology:
* If
and
are
topological space
In mathematics, a topological space is, roughly speaking, a geometrical space in which closeness is defined but cannot necessarily be measured by a numeric distance. More specifically, a topological space is a set whose elements are called po ...
s and
is the product space, endowed with the
product topology
In topology and related areas of mathematics, a product space is the Cartesian product of a family of topological spaces equipped with a natural topology called the product topology. This topology differs from another, perhaps more natural-seemi ...
, a slice in
is a set of the form
for
.
* A tube in
is a subset of the form
where
is an open subset of
. It contains all the slices
for
.
Using the concept of
closed map
In mathematics, more specifically in topology, an open map is a function between two topological spaces that maps open sets to open sets.
That is, a function f : X \to Y is open if for any open set U in X, the image f(U) is open in Y.
Likewise, a ...
s, this can be rephrased concisely as follows: if
is any topological space and
a compact space, then the projection map
is closed.
Examples and properties
1. Consider
in the product topology, that is the
Euclidean plane
In mathematics, the Euclidean plane is a Euclidean space of dimension two. That is, a geometric setting in which two real quantities are required to determine the position of each point ( element of the plane), which includes affine notions of ...
, and the open set
The open set
contains
but contains no tube, so in this case the tube lemma fails. Indeed, if
is a tube containing
and contained in
must be a subset of
for all
which means
contradicting the fact that
is open in
(because
is a tube). This shows that the compactness assumption is essential.
2. The tube lemma can be used to prove that if
and
are compact spaces, then
is compact as follows:
Let
be an open cover of
. For each
, cover the slice
by finitely many elements of
(this is possible since
is compact, being
homeomorphic
In the mathematical field of topology, a homeomorphism, topological isomorphism, or bicontinuous function is a bijective and continuous function between topological spaces that has a continuous inverse function. Homeomorphisms are the isomorp ...
to
).
Call the union of these finitely many elements
By the tube lemma, there is an open set of the form
containing
and contained in
The collection of all
for
is an open cover of
and hence has a finite subcover
. Thus the finite collection
covers
.
Using the fact that each
is contained in
and each
is the finite union of elements of
, one gets a finite subcollection of
that covers
.
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
In mathematics, Tychonoff's theorem states that the product of any collection of compact topological spaces is compact with respect to the product topology. The theorem is named after Andrey Nikolayevich Tikhonov (whose surname sometimes is tra ...
, which generalizes the above to infinite products.
Proof
The tube lemma follows from the generalized tube lemma by taking
and
It therefore suffices to prove the generalized tube lemma.
By the definition of the product topology, for each
there are open sets
and
such that
For any
is an open cover of the compact set
so this cover has a finite subcover; namely, there is a finite set
such that
contains
where observe that
is open in
For every
let
which is an open in
set since
is finite.
Moreover, the construction of
and
implies that
We now essentially repeat the argument to drop the dependence on
Let
be a finite subset such that
contains
and set
It then follows by the above reasoning that
and
and
are open, which completes the proof.
See also
*
*
*
References
*
* {{cite book
, author = Joseph J. Rotman
, author-link = Joseph J. Rotman
, year = 1988
, title = An Introduction to Algebraic Topology
, url = https://archive.org/details/introductiontoal0000rotm
, url-access = registration
, publisher = Springer
, isbn = 0-387-96678-1
''(See Chapter 8, Lemma 8.9)''
Topology
Lemmas
Articles containing proofs