Given a matroid $(N,\mathcal{M})$, for some $a \in N$, is this a matroid: $$ (N \setminus \{a\},\mathcal{M} \setminus \{S : S \subseteq N,\ a \in S\} )$$ Is the following reasoning correct:
- Since $\varnothing \in \mathcal{M}$, $\varnothing \in \mathcal{M} \setminus \{S : S \subseteq N,\ a \in S\}$
- Consider some $S_1 \in \mathcal{M} \setminus \{S : S \subseteq N,\ a \in S\}$. For any $S_2 \subset S_1$, $a \not\in S_2$. Since such an $S_2 \in \mathcal{M}$, it would also belong in $\mathcal{M} \setminus \{S : S \subseteq N,\ a \in S\}$.
- Consider any $S_1,S_2 \in \mathcal{M} \setminus \{S : S \subseteq N,\ a \in S\}$ such that $|S_1| > |S_2|$. Since they belong in $\mathcal{M}$, $\exists b \in S_1 \setminus S_2$ such that $S_2 \cup \{b\} \in \mathcal{M}$. Since $a \not\in S_1 \setminus S_2$, $b \ne a$ and hence, $S_1, S_2$ satisfy the augmentation property.
Your reasoning is exactly right! This matroid is known as the deletion. It is commonly notated as follows: if $M = (N, \mathcal{M})$, your deletion is typically denoted $M\setminus a$. It is one of two fundamental operations, along with contraction, that form minors of matroids. For more, see Chapter 3 of Oxley's book Matroid Theory.