the rank of a matroid $\mid X \mid + r(E\backslash X)-r(E) \geq 0 $

Question

If I have the matroid $M=(E,I)$ and $\mid X \mid + r(E\backslash X)-r(E) \geq 0 $ . Can someone give me any idea about how I can prove it ?I was thinking to: I know that $\mid X \mid \geq r(X)$ then I replace : $$r(X)+ r(E \backslash X)- r(E) \geq 0$$ $$r(X \cup (E \backslash X))+ r(X \cap (E \backslash X))- r(E) \geq 0$$ $$r(E) -r (E)\geq 0$$ and $ 0 \geq 0$ .\ I try to prove the inequality

Answer 1

One of the basic properties of the rank function is that

$$r(A\cup B)+r(A\cap B)\le r(A)+r(B)\tag{1}$$

for any $A,B\subseteq E$. Take $A=X$ and $B=E\setminus X$; then $A\cup B=E$ and $A\cap B=\varnothing$, so $(1)$ becomes

$$r(E)+r(\varnothing)\le r(X)+r(E\setminus X)\;.$$

Now use the fact that $r(A)\le|A|$ for all $A\subseteq E$ to deduce the desired inequality.