Lebesgue coverage dimension

The Lebesgue cover dimension (after Henri Léon Lebesgue ) is a geometrically very clear, topological characterization of the dimension .

definition

A topological space has the dimension if is the smallest natural number such that for every open cover there is a finer open cover , so that every point from lies in at most the sets . If there is no such thing , it is called of infinite dimension. ${\displaystyle X}$ ${\displaystyle n}$ ${\displaystyle n}$ ${\displaystyle (U_{i})_{i}}$ ${\displaystyle (V_{j})_{j}}$ ${\displaystyle X}$ ${\displaystyle n+1}$ ${\displaystyle V_{j}}$ ${\displaystyle n}$ ${\displaystyle X}$

The dimension of is denoted by. Since there are a number of other dimension terms, one speaks more precisely of the Lebesgue cover dimension. ${\displaystyle X}$ ${\displaystyle \dim(X)}$

Explanation

There is an open coverage of when each is open and the union is. The overlap is called finer than when each is contained in any one . ${\displaystyle (U_{i})_{i}}$ ${\displaystyle X}$ ${\displaystyle U_{i}}$ ${\displaystyle X}$ ${\displaystyle U_{i}}$ ${\displaystyle (V_{j})_{j}}$ ${\displaystyle (U_{i})_{i}}$ ${\displaystyle V_{j}}$ ${\displaystyle U_{i}}$

The coverage in the above definition clearly represents a size restriction for coverage quantities . In this sense, there is always coverage for any size restriction in which at most quantities overlap. In fact, the coverage dimension for compact metric spaces can be reformulated as follows. A compact metric space has the dimension , if is the smallest natural number, such that there is an open cover for each , so that for all and each point out lies in at most the sets . Here referred to the diameter of . ${\displaystyle (U_{i})_{i}}$ ${\displaystyle n+1}$ ${\displaystyle n}$ ${\displaystyle n}$ ${\displaystyle \varepsilon >0}$ ${\displaystyle (V_{j})_{j}}$ ${\displaystyle \operatorname {diam} (V_{j})<\varepsilon }$ ${\displaystyle j}$ ${\displaystyle X}$ ${\displaystyle n+1}$ ${\displaystyle V_{j}}$ ${\displaystyle \operatorname {diam} (V_{j})}$ ${\displaystyle V_{j}}$

The above definition is purely topological, that is, we are only talking about open sets. The Lebesgue coverage dimension is therefore a topological invariant, so homeomorphic spaces have the same dimension.

Examples

Example for the Lebesgue coverage dimension

Simple examples

• Every finite Hausdorff space is 0-dimensional, because every point lies in a minimal open set. If the minimal open sets are, then is finer than any cover and every point lies in exactly one . ${\displaystyle x}$ ${\displaystyle V_{1},\dotsc ,V_{k}}$ ${\displaystyle (V_{j})_{j}}$ ${\displaystyle V_{j}}$

• Every discrete space (e.g. the set of whole numbers ) is 0-dimensional, because every point lies in a minimal open set. ${\displaystyle x}$

• The Cantor'sche Diskontinuum is a 0-dimensional compact Hausdorff space with uncountable many points.

• A segment, for example the unit interval , is one-dimensional. As the upper part of the adjacent drawing makes plausible, one can always find any fine open overlaps in which no more than two sets intersect. Hence the dimension . These overlaps are inevitable; it is easy to think that otherwise it could not be coherent . Hence the dimension is even . ${\displaystyle [0,1]}$ ${\displaystyle \leq 1}$ ${\displaystyle [0,1]}$ ${\displaystyle =1}$

• The adjacent drawing also shows that flat figures such as circular areas or rectangles etc. always have any fine overlaps where each point is contained in a maximum of 3 quantities. So the dimension is . It is easy to generalize this to higher dimensions, for example a sphere has the dimension . The fact that equality does indeed exist here is a more difficult proposition that combinatorial arguments are used to prove . ${\displaystyle \leq 2}$ ${\displaystyle {\mathbb {R} }^{n}}$ ${\displaystyle \leq n}$

• The Hilbert cube is an example of an infinitely dimensional, compact, metric space.

Theorem (spheres, cuboids, simplizes)

Spheres , non-degenerate cuboids or non-degenerate simplices in have the Lebesgue cover dimension . ${\displaystyle {\mathbb {R} }^{n}}$ ${\displaystyle n}$

This sentence is historically significant: It was not clear for a long time whether the unit cubes in and , which are each provided with the product topology , can be differentiated for topological, i.e. whether they can be proven to be non- homeomorphic . The mathematicians were surprised when Georg Cantor had given bijective mappings between different-dimensional spaces, which, however, were discontinuous. Giuseppe Peano had constructed continuous and surjective maps from to , these were not bijective, see Peano curve . So it could not be ruled out that a clever combination of these constructions could lead to a homeomorphism between cubes of different dimensions. That this is actually not possible is shown by the above sentence, which was first proven by Luitzen Egbertus Jan Brouwer . ${\displaystyle {\mathbb {R} }^{n}}$ ${\displaystyle {\mathbb {R} }^{m}}$ ${\displaystyle n\not =m}$ ${\displaystyle [0,1]}$ ${\displaystyle [0,1]^{2}}$

Embedding set from Menger-Nöbeling

The question arises whether finite-dimensional topological spaces can be embedded homeomorphically in one , i.e. H. whether they are homeomorphic to a subset of the . As the circular line shows, the plane may be required to embed a one-dimensional space . The following sentence by Menger - Nöbeling answers the question about an upper limit for this dimension . ${\displaystyle {\mathbb {R} }^{n}}$ ${\displaystyle {\mathbb {R} }^{n}}$ ${\displaystyle {\mathbb {R} }^{2}}$

One- dimensional compact metric space allows homeomorphic embeddings in the . ${\displaystyle n}$ ${\displaystyle {\mathbb {R} }^{2n+1}}$

Dimension inheritance

Is a compact metric space and a subspace so is . ${\displaystyle X}$ ${\displaystyle Y\subset X}$ ${\displaystyle \dim(Y)\leq \dim(X)}$

With quotient spaces , i.e. H. with surjective continuous mappings, a surprising behavior results: every compact metric space is a continuous picture of the 0-dimensional Cantor's discontinuum .

If and are metrizable, then applies . Equality generally does not apply, which is a counterexample . ${\displaystyle X}$ ${\displaystyle Y}$ ${\displaystyle \dim(X\times Y)\leq \dim(X)+\dim(Y)}$ ${\displaystyle X=Y=l^{2}(\mathbb {Q} )}$

The following estimate, known as the Hurewicz formula, applies : If normal, metrizable and a continuous , closed and surjective mapping, then applies ${\displaystyle X}$ ${\displaystyle Y}$ ${\displaystyle f\colon X\to Y}$

${\displaystyle \mathrm {dim} (X)\leq \mathrm {dim} (Y)+\sup _{y\in Y}\mathrm {dim} (f^{-1}(y))}$.

Note that the above estimate for the dimension of the Cartesian product of metric spaces easily follows from this.

Comparison with other dimension terms

If a room is normal , the Lebesgue dimension is always smaller than or equal to the large inductive dimension . Equality applies to metrizable rooms. ${\displaystyle X}$

See also

• Asymptotic dimension

literature

• Wolfgang Franz : Topology. Volume 1: General Topology ( Göschen Collection. 1181, ZDB -ID 842269-2 ). de Gruyter, Berlin 1960.

swell

1. ^ Paul Erdös : The Dimension of the Rational Points in Hilbert Space. In: Annals of Mathematics . 2nd Series, Vol. 41, No. 4, 1940, pp. 734-736, doi : 10.2307 / 1968851 .
2. ^ AR Pears: Dimension Theory of General Spaces , Cambridge University Press (1975), ISBN 0-521-20515-8 , Chapter 9, Theorem 2.6