I worked this theory : https://mathoverflow.net/questions/317668/a-new-generalization-of-the-dimension

I have a theorem about dimensions which is more general and simple  than for matroids.

> **Definition 1:** A *structure* $S$, is a pair $(X, \mathcal T)$ with $X$ a set and $\mathcal T$ a set of subsets of $X$ which is stable with respect to arbitrary intersections, with $X \in  \mathcal T$.


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


>**Definition 3:** For a structure $S=(X,\mathcal T)$, we say the set $U\neq \emptyset$ is *free* if 
$$
\forall u \in U,\ u \notin  \langle v \mid v \in U,v\neq u \rangle_S
$$

> **Definition 4:** For a structure $S=(X,\mathcal T)$, we say this structure *has a dimension* if $\forall U \subset X$ free and $v \notin \langle U\rangle_S$, the set $U \cup \{v\}$ is free.


> **Definition 5:** For a structure $S=(X,\mathcal T)$ and $F \in \mathcal T$, we denote $\dim(F)=n$ if the largest free set of $F$ has a cardinality of $n$.


> **Theorem:** For a structure $S=(X,\mathcal T)$ with a dimension and $E,F \in \mathcal T$, if $\dim(E)=\dim(F)<\infty$ and $E \subset F$ then $E=F$


**Example:** $S=(\mathbb R,\mathcal F)$, the closed sets of reals, is a structure with a dimension, and the $\dim(\mathbb R)=\text{card}(\mathbb N)$, because if $A \subset \mathbb R$ with $\text{card}(A)>\text{card}(\mathbb N)$ then it exists $(a_n) \in A^{\mathbb N}$ injective with $\lim a_n=c$ and $c \in A$.

*Question: Is this generalization of the dimension already known?*