[example: Three Sources of Independence]
Let the ground set be the column labels of $A$, the edge set of $G$, or the finite set $E$ in the partitioned example. These give three independence rules:
\begin{align*}
I \text{ is independent exactly when the columns } \{A_i:i\in I\}\text{ are linearly independent over }k.
\end{align*}
\begin{align*}
F \text{ is independent exactly when } F\text{ contains no cycle in }G.
\end{align*}
\begin{align*}
I \text{ is independent exactly when } |I\cap E_j|\le b_j\text{ for every }j.
\end{align*}
Each rule is hereditary. For column sets, if $J\subset I$ and
\begin{align*}
\sum_{j\in J} c_jA_j=0,
\end{align*}
then extending the coefficients by $c_i=0$ for every $i\in I\setminus J$ gives
\begin{align*}
\sum_{i\in I} c_iA_i=0.
\end{align*}
Since the columns indexed by $I$ are linearly independent, every coefficient is zero, so every $c_j=0$. For forests, a cycle contained in $J\subset F$ would also be a cycle contained in $F$. For the partition rule, if $J\subset I$, then for each block $E_\ell$,
\begin{align*}
|J\cap E_\ell|\le |I\cap E_\ell|\le b_\ell.
\end{align*}
The same three examples also illustrate exchange. For vectors, if $I$ and $J$ are linearly independent and $|I|<|J|$, then
\begin{align*}
\dim \operatorname{span}\{A_i:i\in I\}=|I|.
\end{align*}
Also,
\begin{align*}
\dim \operatorname{span}\{A_j:j\in J\}=|J|.
\end{align*}
Since $|I|<|J|$, the span of the columns indexed by $J$ cannot be contained in the span of the columns indexed by $I$. Choose $e\in J$ with $A_e\notin \operatorname{span}\{A_i:i\in I\}$. If a relation
\begin{align*}
c_eA_e+\sum_{i\in I} c_iA_i=0
\end{align*}
had $c_e\ne 0$, then
\begin{align*}
A_e=-c_e^{-1}\sum_{i\in I} c_iA_i,
\end{align*}
contradicting $A_e\notin \operatorname{span}\{A_i:i\in I\}$. Hence $c_e=0$, and then [linear independence](/page/Linear%20Independence) of $I$ forces every $c_i=0$, so $I\cup\{e\}$ is independent.
For forests, let $F$ and $H$ be forests on the same vertex set $V$ with $|F|<|H|$. A forest on $V$ with edge set $S$ has $|V|-|S|$ connected components, so
\begin{align*}
|V|-|F|>|V|-|H|.
\end{align*}
Thus $F$ has more connected components than $H$. If every edge of $H\setminus F$ joined two vertices already connected in $F$, then adding all edges of $H$ could not reduce the number of connected components below that of $F$, contradicting the displayed inequality. Therefore some $e\in H\setminus F$ joins two different components of $F$, and adding such an edge cannot create a cycle.
For the partition rule, suppose $I$ and $J$ satisfy all capacity inequalities and $|I|<|J|$. If every block had $|I\cap E_\ell|\ge |J\cap E_\ell|$, then summing over the partition would give $|I|\ge |J|$, a contradiction. Hence some block $E_\ell$ satisfies
\begin{align*}
|I\cap E_\ell|<|J\cap E_\ell|\le b_\ell.
\end{align*}
Choose $e\in (J\cap E_\ell)\setminus I$. Then
\begin{align*}
|(I\cup\{e\})\cap E_\ell|=|I\cap E_\ell|+1\le |J\cap E_\ell|\le b_\ell,
\end{align*}
and all other block counts are unchanged. Thus in all three settings, removing elements preserves independence, and a larger independent set can donate an element to a smaller one without creating redundancy.
[/example]