Inductive dimension

The small and large inductive dimension are two dimension terms considered in the mathematical sub-area of topology . These terms do not use any algebraic constructions to define a dimension, as is known from the theory of vector spaces , but only the topological space considered itself. It is an alternative to Lebesgue's coverage dimension , which is denoted by and used here for comparison purposes becomes. ${\displaystyle \mathrm {dim} }$

motivation

The idea of ​​the inductive dimension is based on the observation that the edge of a -dimensional sphere is -dimensional, whereby -dimensional is to be understood here in the sense of differential geometry (see manifold ) or simply purely graphically. This suggests the idea of ​​tracing the term dimension of  a set back to the term dimension of  the edge of this set and thus striving for an inductive definition. ${\displaystyle n}$ ${\displaystyle (n-1)}$ ${\displaystyle n}$ ${\displaystyle n}$ ${\displaystyle n-1}$

Since a one-point space, which should definitely have the dimension 0, has an empty border, the dimension of the empty set must be defined as −1. An implementation of the idea of ​​inductive definition then leads to the following two variants:

definition

The small inductive dimension

The small inductive dimension of a topological space is defined as follows: ${\displaystyle \mathrm {ind} (X)\,}$ ${\displaystyle X}$

• ${\displaystyle \mathrm {ind} (\emptyset ):=-1}$
• ${\displaystyle \mathrm {ind} (X)\leq n}$if for every point and every open environment of there is an open environment of with and . ${\displaystyle x\in X}$ ${\displaystyle U}$ ${\displaystyle x}$ ${\displaystyle V}$ ${\displaystyle x}$ ${\displaystyle {\overline {V}}\subset U}$ ${\displaystyle \mathrm {ind} (\partial V)\leq n-1}$

That explains what . One further defines: ${\displaystyle \mathrm {ind} (X)\leq n}$

• ${\displaystyle \mathrm {ind} (X)\,=\,n}$if and not ${\displaystyle \mathrm {ind} (X)\leq n}$ ${\displaystyle \mathrm {ind} (X)\leq n-1}$
• ${\displaystyle \mathrm {ind} (X)\,=\,\infty }$if the inequality holds for none . ${\displaystyle n\in \mathbb {N} }$ ${\displaystyle \mathrm {ind} (X)\leq n}$

The great inductive dimension

If one replaces the point from the definition of the small inductive dimension by any closed set , one obtains the concept of the large inductive dimension. More precisely: The large inductive dimension of a topological space is defined as follows: ${\displaystyle x\in X}$ ${\displaystyle \mathrm {Ind} (X)\,}$ ${\displaystyle X}$

• ${\displaystyle \mathrm {Ind} (\emptyset ):=-1}$
• ${\displaystyle \mathrm {Ind} (X)\leq n}$if there is an open environment of with and for every closed set and every open environment of . ${\displaystyle A\subset X}$ ${\displaystyle U}$ ${\displaystyle A}$ ${\displaystyle V}$ ${\displaystyle A}$ ${\displaystyle {\overline {V}}\subset U}$ ${\displaystyle \mathrm {Ind} (\partial V)\leq n-1}$

That explains what . One further defines: ${\displaystyle \mathrm {Ind} (X)\leq n}$

• ${\displaystyle \mathrm {Ind} (X)\,=\,n}$if and not ${\displaystyle \mathrm {Ind} (X)\leq n}$ ${\displaystyle \mathrm {Ind} (X)\leq n-1}$
• ${\displaystyle \mathrm {Ind} (X)\,=\,\infty }$if the inequality holds for none . ${\displaystyle n\in \mathbb {N} }$ ${\displaystyle \mathrm {Ind} (X)\leq n}$

Remarks

• Since the single-point subsets are closed in -rooms, it follows immediately for such spaces . ${\displaystyle T_{1}}$ ${\displaystyle \mathrm {ind} (X)\leq \mathrm {Ind} (X)}$
• Is a discrete space is so . ${\displaystyle X}$ ${\displaystyle \mathrm {dim} (X)=\mathrm {ind} (X)=\mathrm {Ind} (X)\,=\,0}$
• The statement can be reformulated as follows: Every point has a surrounding basis made up of closed sets with edges of the small inductive dimension . In particular, in this case each point has an environment base made up of closed sets, so that this term only makes sense in regular spaces . ${\displaystyle \mathrm {ind} (X)\leq n}$ ${\displaystyle x\in X}$ ${\displaystyle \leq n-1}$
• The statement can be reformulated as follows: For every two disjoint , closed subsets there are open environments and with , and . In particular, in this case two disjoint, closed sets can be separated by open sets , so that this term only makes sense in normal spaces . ${\displaystyle \mathrm {Ind} (X)\leq n}$ ${\displaystyle A,B\subset X}$ ${\displaystyle U\supset A}$ ${\displaystyle V\supset B}$ ${\displaystyle U\cap V=\emptyset }$ ${\displaystyle \partial U\leq n-1}$ ${\displaystyle \partial V\leq n-1}$
• While with the small inductive dimension each point of the room can be assigned a dimension in an obvious way, this is not possible with the large inductive dimension, this relates to the entire room.

Theorems about the inductive dimension

Comparisons

If there is a metric space , then according to a theorem of M. Katětov ${\displaystyle X}$

${\displaystyle \mathrm {ind} (X)\leq \mathrm {Ind} (X)\leq \mathrm {dim} (X)}$.

A sentence by PS Alexandrow says for compact Hausdorff spaces :

${\displaystyle \mathrm {dim} (X)\leq \mathrm {ind} (X)\leq \mathrm {Ind} (X)}$.

One has equality for separable metrizable spaces:

${\displaystyle \mathrm {ind} (X)\,=\,\mathrm {Ind} (X)\,=\,\mathrm {dim} (X)}$.

K. Nagami constructed a normal space for which , and is true. ${\displaystyle X}$ ${\displaystyle \mathrm {ind} (X)=0}$ ${\displaystyle \mathrm {dim} (X)=1}$ ${\displaystyle \mathrm {Ind} (X)=2}$

Compactification

It denotes the Stone-Čech compactification of . Then applies ${\displaystyle \beta X}$ ${\displaystyle X}$

• N. Wendenisow: If it is normal, then the following applies . ${\displaystyle X}$ ${\displaystyle \mathrm {Ind} (X)=\mathrm {Ind} (\beta X)}$
• JR Isbell : Is normal, so true . ${\displaystyle X}$ ${\displaystyle \mathrm {dim} (X)=\mathrm {dim} (\beta X)}$
• An analogous statement for the small inductive dimension is wrong.

Partial set

${\displaystyle \mathrm {Ind} }$and satisfy the partial set theorem for totally normal spaces , that is ${\displaystyle \mathrm {dim} }$

• Is totally normal and , then (or ) applies . ${\displaystyle X}$ ${\displaystyle Y\subset X}$ ${\displaystyle \mathrm {Ind} (Y)\leq \mathrm {Ind} (X)}$ ${\displaystyle \mathrm {dim} (Y)\leq \mathrm {dim} (X)}$

Sum rate

The large inductive dimension is sufficient for the sum theorem for completely normal rooms , that is

• CH Dowker : Is completely normal and a sequence of closed sets with , then applies . ${\displaystyle X}$ ${\displaystyle (F_{n})_{n\in \mathbb {N} }}$ ${\displaystyle X=\bigcup _{n\in \mathbb {N} }F_{n}}$ ${\displaystyle \mathrm {Ind} (X)\leq \sup _{n\in \mathbb {N} }\mathrm {Ind} (F_{n})}$

• For general normal rooms, the sums rate applies neither for nor for , not even if one restricts oneself to compact Hausdorff rooms . ${\displaystyle \mathrm {ind} }$ ${\displaystyle \mathrm {Ind} }$

Product set

It is said that a dimension concept fulfills a product proposition if the dimension of the product space of two spaces can be estimated against the sum of the dimensions of these two spaces. Notice . ${\displaystyle \mathbb {R} ^{n}\times \mathbb {R} ^{m}\cong \mathbb {R} ^{n+m}}$

• If and are not empty regular Hausdorff rooms, then applies . ${\displaystyle X}$ ${\displaystyle Y}$ ${\displaystyle \mathrm {ind} (X\times Y)\leq \mathrm {ind} (X)+\mathrm {ind} (Y)}$
• If perfectly normal and metrizable and both are non-empty, then the following applies . ${\displaystyle X}$ ${\displaystyle Y}$ ${\displaystyle \mathrm {Ind} (X\times Y)\leq \mathrm {Ind} (X)+\mathrm {Ind} (Y)}$
• An analogous statement applies to the coverage dimension if and both are metrizable or if they are paracompact and compact. ${\displaystyle \mathrm {dim} }$ ${\displaystyle X}$ ${\displaystyle Y}$ ${\displaystyle X}$ ${\displaystyle Y}$

literature

• Keiô Nagami: Dimension Theory (= Pure and Applied Mathematics. Vol. 37). Academic Press, New York NY et al. 1970, ISBN 0-12-513650-1 .

Individual evidence

1. ↑ Keiō Nagami: A normal space Z with ind Z = 0, dim Z = 1, Ind Z = 2 . (PDF, English)