Proving that every (possibly infinite) span contains a basis in a vector space.

Question

The book "A First Course in Algebra" says

In a finite dimensional vector space, every finite set of vectors spanning the space contains a subset that is a basis.

All that is fine. But what about a span having an infinite number of vectors? Surely that too must contain the basis!! An example is $\{\overline{i},\overline{j},\overline{k}+r\overline{i}\},\forall r\in\Bbb{R}$. Can we select all linearly independent vectors in this infinite span, and then prove it is the basis? Is this a sound mathematical technique? The proof given for finite spans does not seem to suggest this.

Thanks in advance!

Answer 1

Given that $V$ is finite-dimensional one can argue as follows: Take a basis $(e_i)_{1\leq i\leq n}$ of $V$. Each of the $e_i$ is a finite linear combination of vectors from the spanning set $S$. It follows that there is a finite subset $S_0\subset S$ that spans $V$. From your "First Course in Algebra" you can then conclude that $S_0$ contains a basis, and so does $S\supset S_0$.

Answer 2

It is certainly true, but not a trivial corollary like I'd originally thought. Typically the development of the notion of "dimension" goes something like this:

Talk about linear combinations, independence, and spanning sets. Show all linearly independent spanning sets have the same cardinality, call this the dimension of the space and call those linearly independent spanning sets bases. Prove various properties bases have, namely unique representation of vectors.

From this standpoint, it would be "obvious" that any infinite spanning set must contain a basis but maybe surprisingly clunky to actually argue rigorously. I suspect the proof of the finite case probably "throws away" vectors until you've got an independent set. With infinities, you can throw away an infinite number of elements an infinite number of times and still be left with an infinite set (start with the whole numbers and throw away all multiples of every prime except 3: you're left with a set containing $\{3^n : n \in \mathbb{N} \}$).

I would try this argument for the infinite case:

Let $S$ be an infinite spanning set for the vector space $V$, where $V$ has dimension $n< \infty$. Then $S$ must contain a linearly independent set of size $n$, otherwise it couldn't span $V$ but only a proper subspace. By definition, the linearly independent spanning set whose cardinality is the dimension of the space is a basis for $V$.

I don't write textbooks though, so my "proof" may not be valid for the book's structure: it matters how you've defined everything, and what theorems you already have. Not owning the book, I'm not sure what route the author has taken to reach this point.