A vector space $V$ with the closure operation defined by taking a subset $A \subseteq V$ to the linear span of $A$ is a quintessential example of a pregeometry. To any vector space $V$ we may define the dimension of $V$ to be cardinality of a basis, a maximal linearly independent set. It follows from Zorn’s Lemma that bases exist and some simple manipulations permit one to see that all bases have the same cardinality so that $\dim V$ is a well-defined cardinal.
We may transpose the definition of a basis to any pregeometry and then the proofs of the existence of bases and that they all have the same cadinality lift from linear algebra to general pregeometries.
Definition 1: If $(X,\text{cl})$ is a pregeometry and $A \subseteq X$, then we say that $A$ is independent if for every $a \in A$ we have that $a \notin \text{cl}(A \smallsetminus \{ a \})$.
Definition 2: If $(X,\text{cl})$ is a pregeometry and $B \subseteq X$ we say that $B$ is a basis of $X$ if $B$ is a maximal (with respect to inclusion) independent subset of $X$.
Let us show that all pregeometries have bases.
Proposition 3: If $(X,\text{cl})$ is a pregeometry then it has a basis.
Proof: Let $\mathcal{B}$ be the set of independent subsets of $X$. Let us check that the hypotheses of Zorn’s Lemma hold for $\mathcal{B}$. That is, we need to show that if $\mathcal{C} \subseteq \mathcal{B}$ is a chain in $\mathcal{B}$, then $\bigcup \mathcal{C} \in \mathcal{B}$. So, consider such a chain $\mathcal{C}$ and any element $b \in \bigcup \mathcal{C}$. Suppose that $b \in \text{cl} (\bigcup \mathcal{C} \smallsetminus \{ b \})$. By the finite character of the closure, there is some finite $C_0 \subseteq \bigcup \mathcal{C} \smallsetminus \{ b \}$ so that $b \in \text{cl}(C_0)$. As $\mathcal{C}$ is a chain and the finite set $C_0 \cup \{b \}$ is a subset of $\bigcup \mathcal{C}$, there is some element $C \in \mathcal{C}$ with $C_0 \cup \{ b \} \subseteq C$. But we have $b \in C$ and $b \in \text{cl}(C_0) \subseteq \text{cl}(C \smallsetminus \{ b \})$ which would mean that $C$ is not independent, contrary to the hypothesis that $\mathcal{C} \subseteq \mathcal{B}$. Therefore, $\bigcup C$ is independent. That is, $\bigcup \mathcal{C} \in \mathcal{B}$ showing that $\mathcal{B}$ satisfies the hypotheses of Zorn’s Lemma from which we deduce that there are maximal elements of $\mathcal{B}$, i.e. bases. $\Box$.
In vector spaces we know that bases are not only independent, but they span the full space. Properly interpreted, the same is true for general pregeometries.
Proposition 4: If $B$ is a basis of the pregeometry $(X,\text{cl})$, then $X = \text{cl}(B)$.
Proof: Let $a \in X$ and suppose that $a \notin \text{cl}(B)$. The set $Ba$ strictly contains $B$. As $B$ is a maximal independent set, $Ba$ is not independent. Thus, for some $b \in Ba$ we must have $b \in \text{cl}(Ba \smallsetminus \{ b \})$. We already know that $a \notin \text{cl}(B) = \text{cl}(Ba \smallsetminus \{ a \})$. (Note that here we are using the fact that $B \subseteq \text{cl}(B)$ so that we know that $a \notin B$.) Hence, $b \neq a$. As $B$ is independent, $b \notin \text{cl}(B \smallsetminus \{ b \})$. So we have $b \in \text{cl}((B \smallsetminus \{ b \} a) \smallsetminus \text{cl}(B \smallsetminus \{ b \})$. Therefore, by exchange we have $a \in \text{cl}((B \smallsetminus \{ b \})b) = \text{cl}(B)$, contrary to our hypothesis about $a$. $\Box$
Let us show now our main theorem that all bases have the same cardinality.
Theorem 5: If $(X,\text{cl})$ is a pregeometry and $B_1$ and $B_2$ are two bases, then $|B_1| = |B_2|$.
Proof: Without loss of generality $|B_1| \leq |B_2|$.
We start by showing that if $\aleph_0 \leq |B_1|$, then $|B_1| = |B_2|$. By Proposition 4, we have $B_1 \subseteq X = \text{cl}(B_2)$. Thus, for every $a \in B_1$ we have $a \in \text{cl}(B_2)$. By the finite character of the closure, there is some finite set $C(a) \subseteq B_2$ so that $a \in \text{cl}(C(a))$. (We are once again using the Axiom of Choice to find a function $C:B_1 \to \mathcal{P}^{<\omega}(B_2)$ so that for every $a \in B_1$ we have $a \in \text{cl}(C(a))$.) Let $B’_2 := \bigcup_{a \in B_1} C(a)$. Then $|B’_2| \leq |B_1| \cdot \aleph_0 = |B_1|$. On the other hand, $B_2 \subseteq X = \text{cl}(B_1) \subseteq \text{cl}(\text{cl}(B’_2))$. Thus, if $b \in B_2 \smallsetminus B’_2$ we would have $b \in \text{cl}(B’_2) \subseteq \text{cl}(B_2 \smallsetminus \{ b \})$, violating the independence of $B_2$. Therefore $B’_2 = B_2$ and we have established that $|B_1| \leq |B_2| = |B’_2| \leq |B_1|$.
Let us consider now the case that $B_1$ is finite, say of size $m$. We will construct by recursion on $n \leq m$ a sequence of sets $C_n$ having the properties that
- $|C_n| = m$ for all $n$
- $|C_n \cap B_2| \geq n$
- $C_n$ is a basis
We start with $C_0 := B_1$ for which it is clear that Conditions 1, 2, and 3 all hold. We are now given $C_n$ with $n < m$ and must produce $C_{n+1}$ satisfying Conditions 1, 2, and 3. If $|C_n \cap B_2| \geq n+1$, then we set $C_{n+1} := C_n$ which already satisfies Conditions 1 and 3. If $|C_n \cap B_2| < n+1 \leq m = |B_1| \leq |B_2|$, then take some $b \in B_2 \notin C_n$. Since $B_2$ is independent, $b \notin \text{cl}(B_2 \cap C_n)$. Let $D \subseteq C_n$ be a maximal subset of $C_n$ containing $C_n \cap B_2$ with $b \notin \text{cl}(D)$. Since $C_n$ is a basis, $b \in \text{cl}(C_n)$. Thus, $C_n \smallsetminus D \neq \varnothing$. Let $a \in C_n \smallsetminus D$. By maximality of $D$, $b \in \text{cl}(Da) \smallsetminus \text{cl}(D)$. By the exchange property, we have $a \in \text{cl}(Db)$. Let $C_{n+1} := (C_n \smallsetminus \{a\})b$. By construction, as $a \in C_n$ and $b \notin C_n$, we have $|C_{n+1}| = (|C_n| – 1) + 1 = (m-1) + 1 = m$, so that Conditon 1 holds. Because $b \notin C_n$ and $a \notin B_2$ ($D$ contains $C_n \cap B_2$ and $a \in C_n \smallsetminus D$), we see that $|C_{n+1} \cap B_2| = |(C_n \cap B) \cup \{ b \}| = |(C_n \cap B)| + 1 = n + 1$. That is, Conditon 2 holds. Finally, if $d \in C_{n+1}$ then either $d \in C_n \smallsetminus \{a \}$ or $d = b$. In the first case, if $d \in \text{cl}(C_{n+1} \smallsetminus \{ d \})$, then $d \in \text{cl}((C_n \smallsetminus \{a, d\})b)$. As $C_n$ is a basis, $d \notin \text{cl}(C_n \smallsetminus \{ d \})$, and, hence, also by monotonicity, $d \notin \text{cl}(C_n \smallsetminus \{a,d\})$. Then by exchange we have $b \in \text{cl}(C_n \smallsetminus \{a \})$. However, $D \subseteq C_n \smallsetminus \{ a \}$ and we know that $a \in \text{cl}(Db)$. That is, $a \in \text{cl}(Db) \subseteq \text{cl}((C_n \smallsetminus \{a \})b) \subseteq \text{cl}(\text{cl}(C_n \smallsetminus \{a \})) = \text{cl}(C_n \smallsetminus \{a \})$, contrary to the inductive hypothesis that $C_n$ is a basis, and in particular independent. Now in the second case that $d = b$, we would have $b \in \text{cl}(C_{n+1} \smallsetminus \{ b \}) = \text{cl}(C_n \smallsetminus \{a\})$ and we would reach a contradiction exactly as in the previous case.
Once we reach $n=m$ we have that $C_m \subseteq B_2$ is a basis with $|C_m| = m$. As $B_2$ is also an independent set, by maximality the only way this can happen is for $C_m = B_2$. That is $|B_2| = m = |B_1|$. $\Box$
Definition 6: If $(X,\text{cl})$ is a pregeometry then we define $\dim X := |B|$ for any basis $B$ of $X$.
We extend the notion of dimension to subsets of $X$ by noting that subset of pregeometries are naturally pregeometries.
Proposition 7: If $(X,\text{cl})$ is a pregeometry and $Y \subseteq X$ is a subset, then the function $\text{cl} \upharpoonright Y:\mathcal{P}(Y) \to \mathcal{P}(Y)$ defined by $A \mapsto \text{cl}(A) \cap Y$ defines a pregeometry $(Y,\text{cl}\upharpoonright Y)$.
Proof: The verification is straightforward and left to the reader. $\Box$
Definition 8: If $(X,\text{cl})$ is a pregeometry and $A \subseteq X$, then $\dim A$ is the dimension of $(A,\text{cl} \upharpoonright A)$.
There is also a relative notion of dimension.
Proposition 9: If $(X,\text{cl})$ is a pregeometry and $Y \subseteq X$ is a subset, then the function $\text{cl}_Y :\mathcal{P}(X) \to \mathcal{P}(X)$ defined by $A \mapsto \text{cl}(A \cup Y)$ defines a pregeometry $(X,\text{cl}_Y)$.
Proof: Once again the verification is straightforward and left to the reader. $\Box$