2008-03-09 14:36MATH 880 PROSEMINAR JT SMITHDIMENSIONALITY PROOF SKETCH SPRING 2008Here is a sketch of the proof that any two bases of a vector space are equinumerous. Icouldn’t find a single proof that works for both finite and infinite dimensionality. Thissketch combines Van der Waerden’s proof for the finite case with Zariski & Samuel’s forthe infinite case. The originals are shown at the end of this document.1. For n 0 let Φ(n) denote the following proposition about a vector space V:a. For every independent A f V with n elements, and every subset B f V, if A fB– then there is an injection f : A 6 B such that A c (B – Rng f )——————— = B–.b. I’ll prove œn Φ(n) by recursion on n, following Van der Waerden [1937] 1953,101–102.c. Φ(0) is trivial, since n = 0 implies A = f = φ.d. Assume Φ(n).e. To prove: Φ(n + 1).f. Given an independent A f V with n + 1 elements, and a subset B f V such thatA f B–.g. To find: an injection f : A 6 B such that A c (B – Rng f )——————— = B–.h. There exists an element a 0 A.i. Let Ar = A – { a}, so thatj. Ar has n elements,k. and Ar f A f B–.l. By (d) there is an injection e : Ar 6 B such that Ar c (B – Rng e )——————— = B–.m. By (f,h), a 0 B–.n. By the finiteness principle, a 0 E– for some finite E f Ar c (B – Rng e); choose Eas small as possible.o. If E 1 (B – Rng e) = φ, then a 0 Ar—, contrary to the independence of A. Thusthere exists b 0 E 1 (B – Rng e).p. a ó E – {b}———– by the minimality of E.q. a 0 E– = (E – {b}) c { b}——————— by (n).r. By the exchange principle, b 0 (E – {b}) c {a}———————.s. Consider f = e c {<a, b>}. t. f : A 6 B injectively because e is injective, a ó Dom e, and b ó Rng e.u. Every element of B depends on Ar c (B – Rng e) by (l),v. and thus on Ar c (B – Rng f ) c {b} by (s),w. and thus on Ar c (B – Rng f ) c (E – { b}) c { a}= Ar c (B – Rng f ) c { a} = A c (B – Rng f ) by (i,n,o,r).x. Therefore B– f A c ( B – Rng f )——————— . The reverse inclusion follows from the assumptionA f B–.y. T h u s f satisfies the requirements of (g).2. Corollaries of Van der Waerden’s theorem:a. If A is independent and finite and B spans V, then #A # #B.Page 2 DIMENSIONALITY PROOF SKETCH2008-03-09 14:36b. If A is independent and B is finite and spans V, then #A # #B.i. Proof: Apply (a) to every finite subset of A: they must all have cardinals # #B.Were A infinite, it would have a countably infinite subset by the axiom of choice,and thus subsets of every finite cardinality. Thus A must be finite, and (a)yields the result.c. If A, B are bases of V and one is finite, then #A = #B.i. If B is finite, use (b) to get #A # #B, then interchange A, B to get the reverseinequality.3. Proposition, following Zariski and Samuel 1958–1960, volume 1, 99.a. Suppose A spans V and B is a basis of V.b. To prove: #B # (#A) ω.c. Let E be the family of all nonempty finite subsets of B.d. By the finiteness principle, (œ x 0 A)(› S 0 E )[x 0 S–].e. By the axiom of choice, there is a choice function E 0 Xx 0A{ S 0 E : x 0 S–}.i. Zariski and Samuel presented an involved construction of E without using theaxiom; but since they used it elsewhere, there’s little point to that.f. Since x 0 Ex— for each x 0 A, A f ^x 0 AEx———–. Since A spans V, so does ^x 0 AEx. g. Since Ex f B for each x 0 A, ^x 0 AEx f B. Since B is a basis, ^x 0 AEx = B.h. #B = #(^x 0 AEx) # Σx 0 A(#Ex) # Σx 0 Aω = (#A)ω.i. There is probably a tacit application of the axiom of choice in the second inequal-ity here.4. Corollaries of Zariski and Samuels’s proposition:a. If A is infinite and spans V, and B is a basis of V, then #B # #A.i. Proof: (#A) ω = #A.b. If A spans V, B is a basis of V, and B is infinite, then so is A and #B # #A.c. If A and B are bases of V, then #A = #B.i. Proof: if A or B is finite, use (2c); otherwise use (4a or 4c).DIMENSIONALITY PROOF SKETCH Page 32008-03-09 14:36Van Der Waerden [1937] 1953, 101–102Page 4 DIMENSIONALITY PROOF SKETCH2008-03-09 14:36Zariski and Samuel 1958–1960, volume 1,