Editing Zorn's lemma (section)

=== Every vector space has a basis ===

Zorn's lemma can be used to show that every [[vector space]] ''V'' has a [[basis (linear algebra)|basis]].<ref>{{cite web |last1=Smits |first1=Tim |title=A Proof that every Vector Space has a Basis |url=https://www.math.ucla.edu/~tsmits/115B/115A/Zorn's%20Lemma.pdf |access-date=14 August 2022}}</ref>

If ''V'' = {'''0'''}, then the empty set is a basis for ''V''. Now, suppose that ''V'' ≠ {'''0'''}. Let ''P'' be the set consisting of all [[linear independence|linearly independent]] subsets of ''V''. Since ''V'' is not the [[zero vector space]], there exists a nonzero element '''v''' of ''V'', so ''P'' contains the linearly independent subset {'''v'''}. Furthermore, ''P'' is partially ordered by [[subset|set inclusion]] (see [[inclusion order]]). Finding a maximal linearly independent subset of ''V'' is the same as finding a maximal element in ''P''.

To apply Zorn's lemma, take a chain ''T'' in ''P'' (that is, ''T'' is a subset of ''P'' that is totally ordered). If ''T'' is the empty set, then {'''v'''} is an upper bound for ''T'' in ''P''. Suppose then that ''T'' is non-empty. We need to show that ''T'' has an upper bound, that is, there exists a linearly independent subset ''B'' of ''V'' containing all the members of ''T''.

Take ''B'' to be the [[union (set theory)|union]] of all the sets in ''T''. We wish to show that ''B'' is an upper bound for ''T'' in ''P''. To do this, it suffices to show that ''B'' is a linearly independent subset of ''V''.

Suppose otherwise, that ''B'' is not linearly independent. Then there exists vectors '''v'''<sub>1</sub>, '''v'''<sub>2</sub>, ..., '''v'''<sub>k</sub> ∈ ''B'' and [[scalar (mathematics)|scalars]] ''a''<sub>1</sub>, ''a''<sub>2</sub>, ..., ''a''<sub>k</sub>, not all zero, such that
:<math>a_1\mathbf{v}_1 + a_2\mathbf{v}_2 + \cdots + a_k\mathbf{v}_k = \mathbf{0}.</math>
Since ''B'' is the union of all the sets in ''T'', there are some sets ''S''<sub>1</sub>, ''S''<sub>2</sub>, ..., ''S''<sub>k</sub> ∈ ''T'' such that '''v'''<sub>i</sub> ∈ ''S''<sub>i</sub> for every ''i'' = 1, 2, ..., ''k''. As ''T'' is totally ordered, one of the sets ''S''<sub>1</sub>, ''S''<sub>2</sub>, ..., ''S''<sub>k</sub> must contain the others, so there is some set ''S''<sub>i</sub> that contains all of '''v'''<sub>1</sub>, '''v'''<sub>2</sub>, ..., '''v'''<sub>k</sub>. This tells us there is a linearly dependent set of vectors in ''S''<sub>i</sub>, contradicting that ''S''<sub>i</sub> is linearly independent (because it is a member of ''P'').

The hypothesis of Zorn's lemma has been checked, and thus there is a maximal element in ''P'', in other words a maximal linearly independent subset ''B'' of ''V''.

Finally, we show that ''B'' is indeed a basis of ''V''. It suffices to show that ''B'' is a [[linear span|spanning set]] of ''V''. Suppose for the sake of contradiction that ''B'' is not spanning. Then there exists some '''v''' ∈ ''V'' not covered by the span of ''B''. This says that ''B'' ∪ {'''v'''} is a linearly independent subset of ''V'' that is larger than ''B'', contradicting the maximality of ''B''. Therefore, ''B'' is a spanning set of ''V'', and thus, a basis of ''V''.