# Can any spanning subset be reduced to a Hamel basis in infinite dimensions?

As per this question, it is possible to extend any l.i. subset of vectors $$A \subset V$$ of a vector space to a (Hamel) basis, $$B \subset V$$. Is it likewise possible to delete vectors from a spanning subset $$S\subset V$$ ($$span(S) = V$$)? The argument is trivial in finite-dimensions, but it's not clear to me that this is possible in infinite dimensions. Even a sketch of the proof would be greatly appreciated.

## 1 Answer

Let $$B$$ be a maximal independent subset of $$S$$, which exists by Zorn’s Lemma. Then we can show that $$span(B) = span(S) = V$$. For given $$s \in S$$, suppose $$s \notin span(B)$$. Then $$B \cup \{s\}$$ is a larger independent subset of $$S$$ than $$B$$ is; contradiction.

Also, note that we cannot prove this theorem without Zorn’s Lemma (or some statement of the equivalent axiom of choice). In fact, over ZF, the claim that every vector space has a basis is equivalent to Zorn’s Lemma. Clearly, your claim implies that all vector spaces have bases, as we can take $$S = V$$.

• Ah, so to be clear you're saying that Zorn's lemma is not just equivalent to saying that $span(S) = V$ has a basis, but it's also equivalent to saying that I can choose that basis as coming from elements of $S$?
– EE18
Commented Apr 4, 2023 at 19:38
• Yes. We just proved that Zorn’s Lemma implies we can pick a basis from $S$. Trivially, we can thus find some basis for $V$. Far less trivially, we can then use this principle to prove Zorn’s Lemma. Commented Apr 5, 2023 at 22:02