Property iii \(\Rightarrow\) Property ii: Let
\(C_1,C_2\in \mathcal{C}\) such that
\(C_1\neq C_2\) and let
\(z\in C_1\cap C_2\text{.}\) By definition of
\(\mathcal{C}\) and the fact that
\(C_1,C_2\in \mathcal{C}\text{,}\) we cannot have
\(C_2\subseteq C_1\text{.}\) So, we can choose
\(x\in C_2\setminus C_1\text{.}\) Now, we are done by applying
Property iii.
Property ii \(\Rightarrow\) Property i: Suppose that
Property i and let
\(Y,Z\in\mathcal{I}\) with
\(|Y|< |Z|\) such that there does not exist
\(z\in Z\setminus Y\) with
\(Y\cup\{z\}\in\mathcal{I}\) and let
\(|Y\cap Z|\) be maximum with respect to this. By our choice of
\(Y\) and
\(Z\text{,}\) we cannot have
\(Y\subseteq Z\) and so we can choose
\(y\in Y\setminus Z\text{.}\) Since
\(|Y\cap Z|\) is maximal, we know that
\(Z\cup\{y\}\notin\mathcal{I}\text{.}\) So, there is a set
\(C_1\in\mathcal{C}\) contained in
\(Z\cup\{y\}\text{.}\) We claim that there cannot be a set
\(C_2\in\mathcal{C}\) distinct from
\(C_1\) such that
\(C_2\subseteq Z\cup\{y\}\text{.}\) Indeed, if there was, then, since
\(Z\in\mathcal{I}\) and
\(\mathcal{I}\) is a downset, we know that
\(y\in C_1\cap C_2\text{.}\) So, by property
Property ii, there is a set
\(C_3\in\mathcal{C}\) contained in
\((C_1\cup C_2)\setminus\{y\}\subseteq Z\) which is a contradiction.
Now, since \(C_1\nsubseteq Y\text{,}\) we can choose \(x\in C_1\cap (Z\setminus Y)\text{.}\) Now, define \(Z':=(Z\setminus\{x\})\cup\{y\}\text{.}\) Then \(Z'\) cannot contain any set of \(\mathcal{C}\) since \(x\in C_1\) and \(C_1\) is the unique such set contained in \(Z\cup\{y\}\text{.}\) So, \(Z'\) is in \(\mathcal{I}\) and satisfies \(|Z'|=|Z|\) and \(|Y\cap Z'|>|Y\cap Z|\text{.}\) So, by our choice of \(Y\) and \(Z\) there is an element of \(Z'\setminus Y\) that can be added to \(Y\) while maintaining containment in \(\mathcal{I}\text{;}\) this is a contradiction since \(Z'\setminus Y\subseteq Z\setminus Y\text{.}\)
Property i \(\Rightarrow\) Property iii: Let
\(C_1,C_2\in\mathcal{C}\text{,}\) \(z\in C_1\cap C_2\) and
\(x\in C_2\setminus C_1\text{.}\) We work in the restriction
\(M':=M\mid (C_1\cup C_2)\text{.}\) Let
\(\mathcal{B}'\) be the set of bases of
\(M'\text{.}\) By minimality of
\(C_1\) and
\(C_2\text{,}\) we can let
\(B_1,B_2\in\mathcal{B}'\) with
\(B_1\supseteq C_1\setminus\{z\}\) and
\(B_2\supseteq C_2\setminus\{x\}\text{.}\) Since
\(\mathcal{I}\) is a downset that does not contain
\(C_1\) nor
\(C_2\text{,}\) we have
\(z\notin B_1\) and
\(x\notin B_2\text{.}\)
By
Theorem 8.2.1 and since we are assuming
Property i, we have that
Property b holds. Suppose that
\(x\in B_1\text{.}\) Then
\(x\in B_1\setminus B_2\) which, by
Property b, implies that there exists
\(y\in B_2\setminus B_1\) such that
\((B_1\setminus \{x\})\cup\{y\}\in\mathcal{B}'\text{.}\) Define
\(B_1':=(B_1\setminus \{x\})\cup\{y\}\text{.}\) Then, since
\(B_1\) contains
\(C_1\setminus\{z\}\) and
\(x\notin C_1\text{,}\) we have that
\(B_1'\) is an element of
\(\mathcal{B}'\) containing
\(C_1\setminus\{z\}\text{.}\) So, by replacing
\(B_1\) with
\(B_1'\) if necessary, we can assume that
\(x\notin B_1\text{.}\)
Since \(x\notin B_1\text{,}\) by maximality of \(B_1\text{,}\) we have \(B_1\cup \{x\}\notin\mathcal{I}\text{.}\) So, we can let \(C_3\in\mathcal{C}\) such that \(C_3\subseteq B_1\cup\{x\}\text{.}\) Then \(x\in C_3\) because \(C_3\nsubsetq B_1\text{.}\) Also, \(z\notin C_3\) as \(z\notin B_1\text{.}\) Finally, \(C_3\subseteq C_1\cup C_2\text{,}\) as this is the ground set of \(M'\) and so \(B_1\cup\{x\}\subseteq C_1\cup C_2\text{.}\)
