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Source interface. Rado (1949), printed pp. 340–342 (canonical PDF), uses elementary finite-rank and finite-independence facts attributed to Whitney (1935), especially in the proof of Lemma 2 and equation (11). The deductions needed here are expanded directly from axioms (R1)–(R3). This is not a separately numbered result of Rado's paper and does not claim to reconstruct all of Whitney's equivalence theorem.

Statement. For finite subsets of $M$:

1. $0\le r(A)\le |A|$, and $A\subseteq B$ implies $r(A)\le r(B)\le r(A)+|B\setminus A|$.
2. A finite $A$ is independent if and only if $r(A)=|A|$.
3. If $r(A\cup\{x\})=r(A)$ and $A\subseteq B$, then $r(B\cup\{x\})=r(B)$.
4. A maximal independent subset $J$ of a finite $A$ has $|J|=r(A)$. Also $r(A\cup B)\le r(A)+r(B)$.
5. If finite independent sets $J,K$ satisfy $|J|<|K|$, some $x\in K\setminus J$ makes $J\cup\{x\}$ independent.

Proof. Repeated application of (R2), starting from (R1), proves the bounds in part 1. If $r(A)=|A|$ and $T\subseteq A$, then

$$|A|=r(A)\le r(T)+|A\setminus T|\le |T|+|A\setminus T|=|A|.$$

Both inequalities must be equalities, so $r(T)=|T|$. This proves part 2; its converse follows by taking $T=A$. It also proves that subsets of independent sets remain independent.

For part 3 it suffices to enlarge $A$ by one element $y$. If $r(A\cup\{y\})=r(A)$, apply (R3) to $x,y$. Otherwise (R2) and integer-valuedness give $r(A\cup\{y\})=r(A)+1$, and

$$r(A\cup\{y\})\le r(A\cup\{x,y\}) \le r(A\cup\{x\})+1=r(A)+1.$$

Again equality holds. Adding the finitely many elements of $B\setminus A$ one at a time proves part 3. Consequently, if each element of a finite set $T$ individually leaves the rank of $A$ unchanged, adjoining all of $T$ leaves it unchanged: before adjoining each new element, apply part 3 to the current enlargement of $A$.

Choose a maximal independent $J\subseteq A$; a maximal member exists because $A$ is finite and $\varnothing$ is independent. If $x\in A\setminus J$, then $J\cup\{x\}$ is dependent. Parts 1 and 2 imply $r(J\cup\{x\})=r(J)=|J|$. The preceding consequence of part 3 gives $r(A)=r(J)=|J|$.

To obtain subadditivity, choose such a finite base $K$ of $B$. Every $b\in B$ leaves $r(K)$ unchanged. Part 3 says that it also leaves $r(A\cup K)$ unchanged. Adjoining the elements of $B$ therefore gives

$$r(A\cup B)=r(A\cup K)\le r(A)+|K|=r(A)+r(B).$$

Finally, if no element of $K\setminus J$ augments the independent set $J$, every element of $K$ leaves $r(J)$ unchanged. Hence $r(J\cup K)=r(J)=|J|$, whereas monotonicity gives $r(J\cup K)\ge r(K)=|K|>|J|$. This contradiction proves part 5.

For an ordered tuple, define $I$ to be one precisely when its entries are distinct and their set is independent, and zero otherwise. In particular $I()=1$ for the empty tuple. For $m\ge0$ the source's equation (11) is

$$I(y_1,\ldots,y_m)I(x_1,\ldots,x_{m+1}) \le \sum_{j=1}^{m+1}I(y_1,\ldots,y_m,x_j).$$

If the left side is zero, this follows from nonnegativity. If it is one, part 5 gives an entry $x_j$ outside the first tuple that augments its independent set, so at least one term on the right is one. This includes $m=0$. Repeated entries are not silently treated as an independent tuple. $\square$

The infinite-cardinality augmentation theorem requires a separate representative-selection argument. It is not obtained merely by using an infinite cardinal in the finite proof above.