A pair $M=(E,S)$ is a matroid iff
- $M$ is an independence system, meaning
1.1 $\forall A\in S, B\subseteq A: B\in S$
1.2 $S\subseteq 2^E$ - $\forall A,B\in S: |A|=|B|+1\implies \exists v\in A\setminus B: B\cup\{v\}\in S$
Let $E_0\subseteq E$ and define $S_0:=\{X\cap E_0 | X\in S\}$. How do I show that $M_0=(E_0,S_0)$ is a matroid?
Sofar I have managed to prove that $M_0$ is an independence system, but I am struggling with property 2.
This is how far I got:
Let $A,B\in S_0$ such that $|A|=|B|+1$.
By definition, $\exists X,Y\in S$ such that $A=X\cap E_0, B=Y\cap E_0$.
Since $M$ is a matroid and $A\subseteq X, B\subseteq Y$, it must hold that $A, B\in S$.
Furthermore, applying property 2, we can conclude that $\exists v\in A\setminus B: B\cup\{v\}\in S$.
However, the goal would be to find a $v\in A\setminus B$ such that $B\cup\{v\}\in S_0$.
Now, it's easy to see that $v\in (X\setminus Y)\cap E_0$.
And furthermore $B\cup\{v\}=(Y\cup\{v\})\cap E_0$.
If I were able to show that $Y\cup\{v\}\in S$ then I would be done. But I have no idea how to do that thus far.
My problem is that I don't really know much about how $X$ and $Y$ relate to each other (except that there for sure is one element in $X$ that isn't in $Y$), especially not their sizes.
Any hints or solutions are much appreciated, thank you.
I almost immediately realized the answer after posting the question.
Since $\forall X\in S: X\subseteq E_0\implies X\in S_0$ and we know that $B\cup\{v\}\in S$ and $B\cup\{v\}\subseteq E_0$, we can conclude $B\cup\{v\}\in S_0$. QED