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Source. Theorems 1 and 2, printed p. 148 (published PDF), specialized from the complete different-matroid arguments on pp. 150–152.

Statement. Let $M=(E,\mathcal F)$ be finite and $k\ge1$.

1. $E$ is a union of $k$ independent sets, equivalently an indexed partition into $k$ independent sets with empty parts allowed, if and only if $|A|\le k r(A)$ for every $A\subseteq E$.
2. There are $k$ pairwise disjoint bases if and only if

$$|E\setminus A|\ge k\bigl(r(E)-r(A)\bigr) \qquad(A\subseteq E). \tag{1}$$

Equivalently, $E$ has an indexed partition into $k$ spanning sets, with the same empty-part convention.

Proof. Apply Theorem 1c with every $M_i=M$ to obtain part 1 for partitions. An independent cover can be made disjoint by assigning each element to one of its containing members; heredity preserves independence. A partition is already a cover.

For part 2, Theorem 2c with all matroids equal gives $|B|\ge k(r(E)-r(E\setminus B))$ for every $B$. Setting $B=E\setminus A$ gives (1).

Disjoint bases extend to a partition into spanning sets by assigning all unused elements to, for example, the first part. Conversely, choosing a base within each part of a spanning partition gives disjoint bases. All choices are finite. $\square$

The source describes part 2 using at least $k$ spanning parts and also discusses the maximum number of disjoint bases. For positive rank, combining spanning parts reduces an at-least-$k$ partition to $k$ parts. In rank zero the exact fixed-$k$ indexed formulation above is preferable: empty bases exist for every $k$, so there is no finite maximum packing number under this convention. If a partition is defined to require nonempty parts, its rank-zero formulation needs the additional cardinality restriction and is not used here.