Is the disjoint union of two Matroids a Matroid itself?

Question

Let $E_1$ and $E_2$ be two disjoint sets.

Moreover, assume that $(E_1, S_1)$ and $(E_2, S_2)$ are matroids.

Define $S:=\{X\cup Y | X\subseteq S_1 \text{ and} Y \subseteq S_2\}$.

S1,S2 and S are independent sets of Matroid.

Prove that $(E_1 \cup E_2, S)$ is a matroid.

Answer 1

You have three properties to check, using the definition here.

1. Clearly $\emptyset \in S$.
2. Suppose $A' \subseteq A \in S$. Then $A = X \cup Y$ for $X \in S_1$, $Y \in S_2$, so $$A' = A' \cap A = (A' \cap X) \cup (A' \cap Y).$$ But $A' \cap X \subseteq X$, so $A' \cap X \in S_1$; similarly $A' \cap Y \in S_2$. Thus $A' \in S$.
3. Let $A, B \in S$, $|A| > |B|$. Write $A = X \cup Y$, $B = X' \cup Y'$. Then we must have $|X| > |X'|$ or $|Y| > |Y'|$; suppose we have $|X| > |X'|$. Then there is $x \in X$ such that $X' \cup \{ x \} \in S_1$ by the augmentation property for the first matroid. In particular, $B \cup \{ x \} = (X' \cup \{ x \}) \cup Y' \in S$. So we have the augmentation property.