I worked this theory : 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 w.r.t. 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?