Let $(X,I)$ be a matroid ($X$ is a finite set).
By our defenition the pair $(X,I)$ is a matroid if
a)$\{\}\in I$
b)$ Z\in I$ and $Y \subset Z \implies Y\in I$
c)$Y,Z \in I $ and $ |Y|\lt |Z|$ then there exists an $z \in Z\setminus Y$ such that $(Y\cup z) \in I$
I need to prove that the condition c) is aquivalent to the condition
d)Every Base $B$ of $Y\subset X$ has the same cardinality
I have no problem showing the implication that $c) \implies d)$ but can anyone can give me a tip how to proof $d) \implies c)$
Would appreciate any help.
Let $E$ be a finite set with $\mathcal I \subseteq 2^E$ such that $(E,\mathcal I)$ is a matroid. For $X \subseteq E$ we define $\mathcal I_X := \{I \in \mathcal I:: I \subseteq X\}$.
(a) $\emptyset \in \mathcal I$
(b) For all $Z \in \mathcal I$ and $Y \subseteq Z$ it holds $Y \in \mathcal I$.
(c) For all $Y, Z \in \mathcal I$ with $|Y| < |Z|$ there exists $z \in Z\setminus Y$ such that $Y \cup \{z\} \in \mathcal I$.
(d) If $X \in 2^{E}$ and $I_1, I_2 \in \mathcal I_X$ are maximal, then $|I_1| = |I_2|$.
Then, $(E, \mathcal I)$ is a matroid iff (a), (b) and (d) hold for every $X \subseteq E$.
Case $\supset$: Let $M = (E, \mathcal I)$ be a matroid and $X \subseteq E$. Choose $I_1, I_2 \in \mathcal I_X$ as maximal and suppose $|I_1| < |I_2|$.
Then, $\exists e \in I_2\setminus I_1: I_1 \cup \{e\} \in \mathcal I(M)$ by (c). Moreover, $I_1 \cup \{e\} \subseteq I_1 \cup I_2 \subseteq X$, that is, $I_1 \subsetneq I_1 \cup \{e\} \in \mathcal I_X$ which is a contradiction to $I_1$ being maximal in $\mathcal I_X$. Thus, $|I_1| = |I_2|$.
Case $\subset$: Let $I_1, I_2 \in \mathcal I$ with $|I_1| < |I_2|$. Define $X := I_1 \cup I_2 \subseteq E$. Then, $I_1, I_2 \in \mathcal I_X$. Furthermore, there are maximal sets $I_1', I_2' \in \mathcal I_X$ with $I_\ell \subseteq I_\ell'$ for $\ell \in \{1,2\}$, that is, $|I_\ell| \leq |I_\ell'|$. Since $|I_1| < |I_2| \leq |I_2'| = |I_1'|$, it follows that $I_1$ cannot be maximal in $\mathcal I_X$, i.e., $\exists e \in X\setminus I_1: I_1 \cup \{e\} \subseteq I_1' \in \mathcal I$. Thus, $I_1 \cup \{e\} \in \mathcal I$ and $e \in (I_1 \cup I_2)\setminus I_1 = I_2\setminus I_1$.