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 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 $n$.
Definition 6: For a structure $S=(X,\mathcal T)$ has a good dimension
(i) if $S$ has a dimension
(ii) for all $F,E \in \mathcal T^2$ if $F \subset E$ then $\dim(F) \leq \dim(E)$???
Theorem 1: 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 $E=F$
Theorem 2: For a structure $S=(X,\mathcal T)$ has a good dimension, with $V \subset\langle U\rangle_S$, if $\text{card}(U)<\text{card}(V)$ then $V$ is not free.
Example 1: $S=(\mathbb R,\mathcal F)$, the closed sets of reals, is a structure with a good 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 $c \in A$.
Example 2: $S=(X=C([0,1],\mathbb R),\mathcal F)$, the closed sets for the uniform norm $||.||_{\infty}$, it's a structure with a good dimension, we know $\langle \mathbb Q[x] \rangle_S=X$ and $\text{card}(\mathbb Q[x])=\text{card}(\mathbb N)$, so by theorem2, if $V \subset X$ with $\text{card}(V)>\text{card}(\mathbb N)$ then $V$ has an accumulation point, for the uniform norm.
Question: Is this generalization of the dimension already known?
