# Can we reduce infinitely linearly dependent spanning set to an independent set while keeping spanning?

I'm quite unfamilar to infinite-dimensional case. Let $V$ be an infinite-dimensional vector space, and let $S$ be a subset of $V$. If $S$ spans $V$ and $S$ is linearly dependent, can we know that $S$ can be reduced to a linearly independent spanning set? If so, how to prove it? Does it require AC?

• Probably a maximal linearly independent subset of $S$ will still span $V$. – Daniel Schepler Sep 15 '17 at 17:00

## 1 Answer

This statement is in fact stronger than the claim that every vector space has a basis, which requires the full axiom of choice. So the answer is yes, but only with choice.

• What is the full axiom of choice mean? – Eric Sep 15 '17 at 16:59
• There are some weaker forms of choice, generally involving the ability to "choose" on some particular set. You can see the axiom of countable choice, or the axiom of dependent choice. However, in ZF, the assertion that every vector space has a basis is exactly equivalent to the (full-strength) axiom of choice. So not only will you need "something like" choice to prove your claim, but your claim would in turn imply the axiom of choice. – AJY Sep 15 '17 at 17:03
• Thanks. Very very clear. Learned a lot from your answer! By the way, do you know where can I find the proof of my claim? I think there are few linear algebra books have these. – Eric Sep 15 '17 at 17:10
• I'd look for a proof of every vector space having a basis. I imagine you'll have to change maybe five words to turn it into a proof of your claim. – AJY Sep 15 '17 at 17:11