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As I understand it, the small and the large inductive dimension differ only in the second property (taken from here). The second property of the small inductive dimension is as follows:

Definition:

  1. $\text{ind}(X) \le n$ if for every point $p \in X$, $p$ has "arbitrarily small" neighborhoods $U$ with $\text{ind}(\partial U) \le n-1$, where $\partial U$ denotes the boundary of $U$

The second property of the large inductive dimension is as follows:

Definition:

  1. $\text{Ind}(X) \le n$ if for every closed set $A \subseteq X$ and each open set $V \subseteq X$ which contains $A$ there exists an open set $U \subseteq X$ such that $$ A\subseteq \overline{U} \subseteq V \qquad\text{and}\qquad \text{Ind}(\partial U) \le n-1. $$

What is the difference between the two definitions taking into account property $2.$? What is the rationale behind the property $2.$ for each dimension?

Carmen González
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    I see that @XanderHenderson has discussed this exact issue extensively in the comments following his answer, including some references and suggestions for you to follow up on. Is there anything in those references which you followed up on and did not understand regarding this issue? – Lee Mosher Mar 02 '21 at 17:28
  • @LeeMosher I read the literature recommendations. The definitions he put in the post were taken from pages 17 and 18 of the book "Dimensions, embeddings, and attractors" by Robinson, James C., but they in the literature do not explain where they got these two properties from (which is what distinguishes the large inductive dimension). – Carmen González Mar 02 '21 at 19:09
  • @LeeMosher [cont.] In Engelking's “Dimension theory. A revised and enlarged translation of "Teoria wymiaru"", in proposition 1.6.2 of this page, the author says: "1.6.2. Proposition. A normal space $X$ satisfies the inequality $\text{Ind}X 0$ if and only if for every pair $A, B$ of disjoint closed subsets of $X$ there exists a partition $L$ between $A$ and $B$ such that $\text{Ind} L < n-1$. Applying induction with respect to $\text{Ind}X$, one can easily prove the following theorem, which justifies the names of the small and the large inductive dimensions." – Carmen González Mar 02 '21 at 19:11
  • @LeeMosher [cont.2] I didn't understand what he meant by that and I still don't understand the differences between the properties 2 because I can't find the demonstration or any intuitive example in the literature. – Carmen González Mar 02 '21 at 19:11
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    "small" and "large" are because for $T_4$ spaces $\textrm{ind}(X) \le \textrm{Ind}(X)$. They're called inductive dimensions because you essentially define them by recursion, starting with the base case $\textrm{ind}(\emptyset) = \textrm{Ind}(\emptyset) = -1$.Examples where there is a gap between these inductive dimensiosn are findable but all very nasty. – Henno Brandsma Mar 02 '21 at 23:01

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