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For a set A, we are incrementally building a basis for A. Suppose some oracle keeps giving us new vectors from A that are not spanned by our partially-built basis. There must be some theorem that tells us this constructive procedure of adding new linearly independent vectors will always terminate after the same number of vectors are added (no matter what vectors that oracle gives us), and all sets constructed this way must form a basis for A. What is the theorem? Is it in Axler?

user3180
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    I'm not sure what "same number" would mean in an infinite-dimensional context. Certainly it makes sense for finite-dimensional $A$, in which case the proof that any two bases are of the same cardinality is standard. Hint: Try to show that for $V$ a vector space, if $U = {u_1, \ldots, u_m }$ is a set of linearly independent vectors and $W = { w_1, \ldots, w_n }$ spans $V$, then $m \leq n$ and ${ u_1, \ldots, u_m, w_{m+1}, \ldots, w_n }$ spans $V$ by induction. (This is the Steinitz exchange lemma.) – Jake Lai Jul 27 '23 at 20:44
  • I assume that the oracle should be able to answer that there are no more such vectors. Otherwise, there is no such vector space. In general, might need to query the oracle a transfinite number of times. So, assuming the axiom of choice you can prove that there is some maximal linearly independent set $A$ such that the oracle stops. I don't know if in that book, but see here. The other part is showing that all such maximal sets have the same cardinality. See here – NDB Jul 27 '23 at 21:13
  • You could name the oracle Hartogs. – blargoner Jul 27 '23 at 21:51
  • @JakeLai How do we know that this constructive procedure forms a basis? Certainly, the procedure tells us our partial basis is linearly independent. How do we know that by picking additional linearly independent vectors, until the oracle tells us there are no more such vectors, we are guaranteed to span our set A after a finite number of vectors picked? And yes lets assume A is finite vector space. – user3180 Jul 27 '23 at 23:14
  • Steinitz exchange lemma. – lonza leggiera Jul 28 '23 at 05:44

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