In the mathematical field of topology, the inductive dimension of a topological spaceX is either of two values, the small inductive dimension ind or the large inductive dimension Ind. These are based on the observation that, in n-dimensional Euclidean spaceRn, -dimensional spheres have dimension n − 1. Therefore it should be possible to define the dimension of a spaceinductively in terms of the dimensions of the boundaries of suitable open sets. The small and large inductive dimensions are two of the three most usual ways of capturing the notion of "dimension" for a topological space, in a way that depends only on the topology. The other is the Lebesgue covering dimension. The term "topological dimension" is ordinarily understood to refer to Lebesgue covering dimension. For "sufficiently nice" spaces, the three measures of dimension are equal.
Formal definition
We want the dimension of a point to be 0, and a point has empty boundary, so we start with Then inductively, ind is the smallest nsuch that, for every and every open setU containing x, there is an open set V containing x, such that the closure of V is a subset of U, and the boundary of V has small inductive dimension less than or equal ton − 1. For the large inductive dimension, we restrict the choice of V still further; Ind is the smallest n such that, for every closed subsetF of every open subsetU of X, there is an open Vin between, such that the boundary of V has large inductive dimension less than or equal to n − 1.
Relationship between dimensions
Let be the Lebesgue covering dimension. For any topological space X, we have Urysohn's theorem states that when X is a normal space with a countable base, then Such spaces are exactly the separable and metrizableX. The Nöbeling–Pontryagin theorem then states that such spaces with finite dimension are characterised up to homeomorphism as the subspaces of the Euclidean spaces, with their usual topology. The Menger–Nöbeling theorem states that if is compact metric separable and of dimension, then it embeds as a subspace of Euclidean space of dimension. Assuming only X metrizable we have or assuming X compact and Hausdorff Either inequality here may be strict; an example of Vladimir V. Filippov shows that the two inductive dimensions may differ. A separable metric spaceX satisfies the inequality if and only if for every closed sub-space of the space and each continuous mappingthere exists a continuous extension.
