TheoremBase

Proof sketch: a basis is extracted from a spanning family, and the exchange lemma applied in both directions gives equal lengths.

Proof

If an nn-tuple (v1,…,vn)(v_{1},\dots,v_{n}) spans VV, then A Finite Spanning Family Contains a Basis (i) gives m≤nm\leq n and a strictly increasing map k:[m]→[n]k:[m]\to[n] such that the mm-tuple (vk(1),…,vk(m))(v_{k(1)},\dots,v_{k(m)}) is a basis of VV. This proves (i).

Now let (b1,…,bm)(b_{1},\dots,b_{m}) and (c1,…,cn)(c_{1},\dots,c_{n}) be bases of VV. By Finite Basis of a Vector Space (i) the first is linearly independent and the second spans VV, so m≤nm\leq n by The Steinitz Exchange Lemma (i). Exchanging the roles of the two bases gives n≤mn\leq m. Hence m=nm=n, which proves (ii).

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