Let $E$ be a finite set.
$1)$ If $I$ is an independence system on $E$, then the family of maximal elements of $I$ is a basis system
$2)$ If $B$ is a basis system, then $I_{b \in B} 2^b$ is an independence system
And finally, I would like to understand why the above two constructions are mutual inverses
Thank you. I used to study mathematics when I was young, and now I'm just dabbling in random texts. This is a proposition on a text on algebraic combinatorics that I've been reading, and the proof is given as an exercise, and I have not been able to crack it. Thank you!
Suppose that $\mathscr{I}$ is an independence system on $E$, and let $\mathscr{M}$ be the family of maximal elements of $\mathscr{I}$. Plainly $\mathscr{M}\ne\varnothing$, so $\mathscr{M}$ will be a basis if and only if it has the following property: whenever $A,B\in\mathscr{M}$, and $a\in A\setminus B$, there is a $b\in B\setminus A$ such that $(A\setminus\{a\})\cup\{b\}\in\mathscr{M}$.
Suppose that $A,B\in\mathscr{M}$; I claim that $|A|=|B|$. If not, without loss of generality suppose that $|A|>|B|$; $\mathscr{I}$ has the augmentation property, so there is an $a\in A\setminus B$ such that $B\cup\{a\}\in\mathscr{I}$, contradicting the maximality of $B$, and the claim follows. Now suppose that $A,B\in\mathscr{M}$, and $a\in A\setminus B$. Let $A'=A\setminus\{a\}$; $A'\in\mathscr{I}$, and $|B|>|A'|$, so there is a $b\in B\setminus A'$ such that $(A\setminus\{a\})\cup\{b\}=A'\cup\{b\}\in\mathscr{I}$. Let $M\in\mathscr{I}$ be maximal such that $(A\setminus\{a\})\cup\{b\}\subseteq M$; then $M\in\mathscr{M}$, so $|M|=|A|=|(A\setminus\{a\})\cup\{b\}|$, so $(A\setminus\{a\})\cup\{b\}=M\in\mathscr{M}$, and $\mathscr{M}$ is a basis.
Now suppose that $\mathscr{B}$ is a matroid basis on $E$, and let $\mathscr{I}=\bigcup_{B\in\mathscr{B}}\wp(B)$; we want to show that $\mathscr{I}$ is an independence system on $E$. (Here I am guessing at the meaning of your $I_{b\in B}2^b$: the $I$ is rather obscure.) Clearly $\varnothing\in\mathscr{I}$, and $\mathscr{I}$ is downward-closed (meaning that if $J\subseteq I\in\mathscr{I}$, then $J\in\mathscr{I}$), so it only remains to show that $\mathscr{I}$ has the augmentation property.
It is very helpful to prove first that all members of $\mathscr{B}$ have the same cardinality; I’ll sketch the argument. Suppose that $M,N\in\mathscr{B}$, and $|M|<|N|$. Clearly there is an $a\in N\setminus M$, so by the exchange property there is a $b\in M\setminus N$ such that $(N\setminus\{a\})\cup\{b\}\in\mathscr{B}$. Let $N_1=(N\setminus\{a\})\cup\{b\}$; $|N_1|=|N|>|M|$, so you can invoke the exchange property again to get an $N_2\in\mathscr{B}$ such that $|N_2|=|N|$, and $N_2$ contains two elements of $M\setminus N$. Continue invoking the exchange property between $N_k$ and $M$; eventually you get $N_{|M\setminus N|}\in\mathscr{B}$, and at that point $M\subseteq N_{|M\setminus N|}$, so that it is impossible to invoke the exchange property one more time, even though $|N_{|M\setminus N|}|=|N|>|M|$. This is contrary to the definition of a basis, so we conclude that all members of $\mathscr{B}$ must have the same cardinality.
Now let $A,B\in\mathscr{I}$ be such that $|A|>|B|$; we want to show that there is an $a\in A\setminus B$ such that $B\cup\{a\}\in\mathscr{I}$. By definition there are $M,N\in\mathscr{M}$ such that $A\subseteq M$ and $B\subseteq N$. We can use the base exchange property repeatedly as in the last paragraph to replace all elements of $N\setminus B$ with elements of $M$, so we may as well assume that $N\setminus B\subseteq M$. $|N|=|M|$ by the preceding paragraph, and $|B|<|A|$, so $|N\setminus B|>|M\setminus A|$, and therefore $N\setminus B\nsubseteq M\setminus A$. Thus, $(N\setminus B)\cap A\ne\varnothing$, and we may choose $a\in(N\setminus B)\cap A$. But then $a\in A\setminus B$, and $B\cup\{a\}\subseteq N$, so $B\cup\{a\}\in\mathscr{I}$, as desired.