Why every component of a loopless matroid is flat?
I know that the rank of a loop in a matroid is zero and I read that: A loop is an element of a matroid that is not contained in any independent set (is this because it has rank zero so it may be considered as dependent set?), or equivalently, an element that is contained in all flats (why is this correct?).
Also, can the previous paragraph help me by any means in understanding the first question I proposed?
Edit:
I also know that an element $e$ is a loop of $M$ if and only if $e$ is not in any of the bases of $M.$
Yes, $X$ is independent iff $r(X)=|X|$.
Well, a loop $e$ has rank zero, so if $e\not \in X$, submodularity tells you that $r(X\cup e)\leq r(X)+r(e)=r(X)$ and so $r(X)=r(X\cup e)$ so $e$ is in every clausure, so its in every flat. If $e$ is not in some flat, then you can use submodularity too to check that its rank is bigger than zero.
Let $A$ be a component and suppose that $A\subset cl_M(A)$. Now, consider $x\in cl_M(A)\setminus A$, so there exists $A'\subseteq A$ such that $A'\cup \{x\}$ is a circuit, and notice that $A'$ is non-empty because we have no loops, then $x$ is in the component $A$ (recall that the definition is that they are equivalence classes under the relation of being in a circuit) which is a contradiction, then $A=cl_M(A)$ and so $A$ is a flat.