During my research, I came a cross on these notions :

Definition 1: A structure $S$, is a pair $(X, \mathcal T)$ with $X$ a set and $\mathcal T$ a set of subsets of $X$, stable by arbitrary intersections, with $X \in \mathcal T$ and if $\mathcal U=\{ A \in \mathcal T\text{ | } A\neq X \text{ and } \forall C \in \mathcal T, A\subset C\text{ or } C \subset A \}$ then $O_{\mathcal T}= \bigcup \limits_{F \in \mathcal U} F$


Definition 2: Let $U \subset X$ we denote $\langle U\rangle_S=\bigcap \limits_{F \in \mathcal T, U \subset F} F$


Definition 3: We say that $S=(X,\mathcal T)$ a structure has a dimension if :

$ \forall U \subset X,U\neq \emptyset, U \cap O_{\mathcal T}= \emptyset$ and $A=\langle U\rangle_S$,

with $\forall V \subset A, V \cap O_{\mathcal T}=\emptyset$ , $\text{card}(V)>\text{card}(U)$, $\exists v_o \in V $

such that $v_o \in \langle v \text{ | } v \in V \text{ and } v \neq v_o\rangle_S $


Question: Is this generalization of the dimension already existing?

PS : in this case, a set $E$ has a dimension, with the structure $(E,\mathcal P(E) )$.

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    $\begingroup$ Perhaps read about matroids ... en.wikipedia.org/wiki/Matroid $\endgroup$ – Gerald Edgar Dec 14 '18 at 11:55
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    $\begingroup$ You didn't include it in your question, but I assume that for $C\in T$, you want to say that the dimension of $C$ is the size of any independent set with closure $C$ (and $\infty$ if there is no such set)? Consider $X = \{0,1\}$ and $T = \{\emptyset, \{0\}, \{0,1\}\}$. Now it's easy to check that $(X,T)$ is a "structure with a dimension". And we have $\text{dim}(\{0\}) = \text{dim}(\{0,1\}) = 1$. So you have have proper containments between sets of the same dimension. Is this a kind of behavior you want to allow? $\endgroup$ – Alex Kruckman Dec 14 '18 at 15:33
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    $\begingroup$ Oh, I liked the old definition much better... You can easily adjust my example to fit your new definition: $X = \{0,1,*\}$, $T = \{\{*\}, \{0,*\}, \{0,1,*\}\}$. Again $(X,T)$ is a "structure with dimension", with $O_T = \{*\}$, and $\dim(\{0,*\}) = \dim(\{0,1,*\}) = 1$, since the former is the closure of $\{0\}$, and the latter is the closure of $\{1\}$. $\endgroup$ – Alex Kruckman Dec 14 '18 at 16:25
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    $\begingroup$ To be clear, I'm not saying that proper containments between sets of the same dimension is necessarily a bad thing - I'm just curious if you intended that kind of behavior. I think it's the main thing making your definition different than the notion of a matroid (at least in the case when $X$ is finite). If you strengthen the definition of "has a dimension" to rule out this kind of behavior, I think you'll exactly end up with a definition that's equivalent to the definition of a matroid. $\endgroup$ – Alex Kruckman Dec 14 '18 at 16:28
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    $\begingroup$ Definition 3 is quite hard to read, the long sentence is cut somewhat arbitrarily at the middle, and you switch from symbolic notation to English words before the end. Also it's quite helpful, for readability, to denote sets of subsets with mathcal letters. $\endgroup$ – YCor Dec 15 '18 at 10:13

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