Matroid: adding an arbitrary element?

Question

Let $M$ be a matroid and $I\in M$. For any $x\not\in I$, it is true that there exists $y\in I$ such that $(I\setminus\{y\})\cup \{x\}\in M$?

If $I\cup\{x\}\in M$, then we know any subset of it is also in $M$. What if $I\cup\{x\}$ is not an independent set?

Answer 1

For any $x\notin I$, it is true that there exists $y\in I$ such that $(I\setminus\{y\})\cup\{x\}$ is independent?

As I stated in the comments, the answer is no. If $\{x\}$ is a circuit then $(I\setminus\{y\})\cup\{x\}$ contains a dependent set and hence is dependent itself.

What if $I\cup\{x\}$ is not an independent set?

In that case there exists a unique circuit $C\subseteq I\cup\{x\}$ such that $x\in C$. This is known as the Unique Circuit Property.

Proof:

Since $I\cup\{x\}$ is not independent, then either it is a circuit or it contains a circuit $C$. Clearly $x\in C$. For uniqueness, if $C'\subseteq I\cup\{x\}$ is also a circuit such that $C\neq C'$, then by the circuit elimination axiom there exists a circuit $C''\subseteq (C\cup C')\setminus\{x\}$. But if $C''$ doesn't contain $x$ then it must be contained in $I$, contradicting that it is a circuit.