In this tutorial the I-matroid is defined using three constraints, but, I think, the third one is confusing:
If $U,V \in I$ with $|U| > |V|$, then there is an $x \in U − V$ such that $V \cup \{x\} ∈ I$.
I mean if $U,V\in I $ then for any $x \in I$ we can get $V\cup\{x\}\in I$. What is the point I misunderstand?
I mean if $U,V\in I$ then for any $x\in I$ we can get $V\cup\{x\}\in I$.
As it stands this doesn’t make sense: $x$ should be an element of the underlying set $S$, so that $V\cup\{x\}$ is at least a subset of $S$ and hence conceivably an element of $I$. However, it simply isn’t necessarily true that if $V\in I$ and $x\in S$, then $V\cup\{x\}\in I$. If that were the case, $I$ would simply be $\wp(S)$: $\varnothing\in I$ by the first condition, and we can get every subset of $S$ by repeatedly adding single elements.
What is true is that if $x\in U\in I$, then $U\setminus\{x\}\in I$; this is an immediate consequence of the second condition. We can remove elements from an independent set and still have an independent set, but we cannot necessarily add elements to an independent set and still have an independent set.
The third condition, sometimes called the exchange condition, is actually very important. In general if $U$ is an independent set, and $x\in S\setminus U$, there is no guarantee that $U\cup\{x\}$ is independent: adding an element to an independent set can destroy independence. The exchange condition, however, says that under circumstances adding an element does not destroy independence. Specifically, it says that if $U$ and $V$ are independent sets, and $|U|>|V|$ (i.e., $U$ has more elements than $V$), then there is at least one element $x$ of $U$ that (a) is not already in $V$ (i.e., $x\in U\setminus V$), and (b) can be added to $V$ without destroying independence (i.e., $V\cup\{x\}\in I$). Thus, an independent set $V$ that is not of maximal size amongst independent sets can always be expanded to a larger independent set. Moreover, the added element can be chosen from any independent set $U$ that is larger than $V$.