Here is the question I am trying to understand its solution:
Let $C_1$ and $C_2$ be circuits of a matroid $M$ such that $C_1 \cup C_2 = E(M)$ and $C_1 - C_2 = \{e\}.$ Prove that if $C_3$ is a circuit of $M,$ then either $C_3 = C_1$ or $C_2 - C_1 \subseteq C_3.$
Here is a link to the solution given here in this site:
Property of circuits of a matroid made up of two circuits that differ by one element
The OP gave many steps of the solution as follows:
Either $C_3 = C_1$ or not. If $C_3 = C_1$ we're done, so suppose $C_3 \ne C_1$.
We have $C_3 \subseteq E(M) = C_1 \cup C_2$.
If $C_3 \subseteq C_2$, then by (C2)—the second axiom of circuits—$C_3 = C_2$, which satisfies $C_3 \supseteq C_2 \setminus C_1$, and we're done.
Now suppose $C_3 \nsubseteq C_2$. This means there is an element $x$ in $C_3$ that is not in $C_2$, which must be in $E(M) \setminus C_2 = (C_1 \cup C_2) \setminus C_2 = C_1$. Hence $x \in C_1 \setminus C_2$, so by hypothesis $x = e$. Thus we have $e \in C_1 \cap C_3$. By (C3)—the third axiom of circuits—there is a circuit: $$ C_4 \subseteq (C_1 \cup C_3) - e = (C_1 - e) \cup (C_3 - e) \subseteq C_2 \cup C_3 $$ (...?)
But then there was an answer posted to complete the steps given by the OP, which is below:
Great effort so far. So $C_4\subseteq (C_1\cup C_3)\setminus e\subseteq C_2$ because we are taking the only element out of $C_2$ and so $C_4=C_2.$ Also, $C_1\setminus \{e\}=C_1\cap C_2$ and so $C_2=(C_1\cap C_2)\cup (C_2\setminus C_1)=C_4\subseteq (C_1\cap C_2)\cup (C_3\setminus \{e\})$ and so $C_2\setminus C_1\subseteq C_3\setminus \{e\}\subseteq C_3.$
My questions about this answer is as follows:
1- Why $(C_1\cup C_3)\setminus e\subseteq C_2$? I draw a venn diagram with $C_3$ taking e from $C_1$ and passing by the intersection between $C_1 $ and $C_3$ and having some part in $C_2,$ does this venn diagram violates anything from the assumptions prior to it?
2- Also, why $C_1\setminus \{e\}=C_1\cap C_2$?
Could someone clarify this questions for me please?
