Lebesgue covering dimension

The Lebesgue'sche covering dimension ( after Henri Léon Lebesgue ) is a geometrically very graphic, topological characterization of the dimension.

Definition

A topological space has the dimension if the smallest natural number such that there are any open covering a finer open cover, so that each point lies in more than the quantities. Is there no such so called of infinite dimension.

The dimension is denoted by. Since there are a number of other dimension terms, we speak more precisely of the Lebesgue covering dimension.

Explanation

It is an open cover of if every open and the union is. The overlap is called finer than when each is included in any.

Clearly, the overlap in the above definition, a size limit on coverage amounts dar. In this sense, there is so to any size restriction always overlaps, at which maximum overlap each volume. In fact, the overlap dimension in compact metric spaces can be rephrased as follows. A compact metric space has the dimension if the smallest natural number such that, for every an open cover such that for all and each point lies in more than the quantities. It refers to the diameter of.

The above definition is purely topological, ie it is only by open sets of the question. Therefore, the Lebesgue'sche coverage dimension is a topological invariant, that is homeomorphic spaces have the same dimension.

Examples

Simple examples

• Every finite space is 0 -dimensional, because each point is in a minimal open set. Are the minimal open sets, so is finer than any covering and each point lies in exactly one.

• The Cantor discontinuum is a 0 -dimensional compact Hausdorff space with uncountably many points.

• A route around the unit interval is one-dimensional. How does the upper part of the adjacent drawing plausible, one can always find arbitrarily fine open coverings, which intersect more than two sets. Therefore, the dimension. These overlaps are unavoidable; easily considered one that otherwise might not be contiguous. Therefore, the dimension is even.

• The drawing also shows that there are always so to plane figures such as circles or rectangles surfaces arbitrarily fine coverings, in which each point is contained in at most three quantities, the dimension is so. Easily generalized to the higher dimensions, so has about a bullet in the dimension. That in fact here there is equality, is a more difficult proposition, whose proof combinatorial arguments are used.

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

Set ( spheres, boxes, simplices )

This set is historically significant. It was far from clear whether one can distinguish the unit cube in and for topologically, that is, whether they can prove to be non- homeomorphic. It had the mathematician surprised when Georg Cantor had given bijective maps between different dimensional spaces, however these were discontinuous. Giuseppe Peano had constructed continuous and surjective mappings of after, they were not bijective, see Peano curve. So it was not excluded that a clever combination of these structures could lead to a homeomorphism between cubes of different dimensions. That this is not actually possible to show the above theorem, which was first proved by Luitzen Egbertus January Brouwer.

Embedding theorem of Menger - Nöbeling

It turns into a can embed the question of whether homeomorphic finite topological spaces, ie if it is homeomorphic to a subset of the are. Such as the circle shown may be necessary to embed one-dimensional space the plane. An example of a node indicates that for the embedding of a one-dimensional space may even be necessary. Is there an upper limit? This question is answered the following set of Menger - Nöbeling.

Inheritance of dimension

Is a compact metric space and a subspace, then.

For quotient spaces, that is, with surjective continuous mappings, has a surprising behavior. Every compact metric space is the continuous image of the 0 -dimensional Cantor'schen discontinuum.

Are metrizable and so true. Equality does not apply in general, a counter- example.

Compared with other dimension terms

Is a normal room, the Lebesgue'sche dimension is always less than or equal to the large dimension of the inductive. For metrizable spaces equality holds.