Editing Zorn's lemma (section)

== Example applications ==

=== 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''.

=== Every nontrivial ring with unity contains a maximal ideal ===

Zorn's lemma can be used to show that every nontrivial [[ring (mathematics)|ring]] ''R'' with [[Unital ring|unity]] contains a [[maximal ideal]].

Let ''P'' be the set consisting of all ''proper'' [[ideal (ring theory)|ideals]] in ''R'' (that is, all ideals in ''R'' except ''R'' itself). Since ''R'' is non-trivial, the set ''P'' contains the trivial ideal {0}. Furthermore, ''P'' is partially ordered by set inclusion. Finding a maximal ideal in ''R'' is the same as finding a maximal element in ''P''.

To apply Zorn's lemma, take a chain ''T'' in ''P''. If ''T'' is empty, then the trivial ideal {0} is an upper bound for ''T'' in ''P''. Assume then that ''T'' is non-empty. It is necessary to show that ''T'' has an upper bound, that is, there exists an ideal ''I'' ⊆ ''R'' containing all the members of ''T'' but still smaller than ''R'' (otherwise it would not be a proper ideal, so it is not in ''P'').

Take ''I'' to be the union of all the ideals in ''T''. We wish to show that ''I'' is an upper bound for ''T'' in ''P''. We will first show that ''I'' is an ideal of ''R''. For ''I'' to be an ideal, it must satisfy three conditions:

# ''I'' is a nonempty subset of ''R'',
# For every ''x'', ''y'' ∈ ''I'', the sum ''x'' + ''y'' is in ''I'',
# For every ''r'' ∈ ''R'' and every ''x'' ∈ ''I'', the product ''rx'' is in ''I''.

'''#1 - ''I'' is a nonempty subset of ''R''.'''

Because ''T'' contains at least one element, and that element contains at least 0, the union ''I'' contains at least 0 and is not empty. Every element of ''T'' is a subset of ''R'', so the union ''I'' only consists of elements in ''R''.

'''#2 - For every ''x'', ''y'' ∈ ''I'', the sum ''x'' + ''y'' is in ''I''.'''

Suppose ''x'' and ''y'' are elements of ''I''. Then there exist two ideals ''J'', ''K'' ∈ ''T'' such that ''x'' is an element of ''J'' and ''y'' is an element of ''K''. Since ''T'' is totally ordered, we know that ''J'' ⊆ ''K'' or ''K'' ⊆ ''J''. [[Without loss of generality]], assume the first case. Both ''x'' and ''y'' are members of the ideal ''K'', therefore their sum ''x'' + ''y'' is a member of ''K'', which shows that ''x'' + ''y'' is a member of ''I''.

'''#3 - For every ''r'' ∈ ''R'' and every ''x'' ∈ ''I'', the product ''rx'' is in ''I''.'''

Suppose ''x'' is an element of ''I''. Then there exists an ideal ''J'' ∈ ''T'' such that ''x'' is in ''J''. If ''r'' ∈ ''R'', then ''rx'' is an element of ''J'' and hence an element of ''I''. Thus, ''I'' is an ideal in ''R''.

Now, we show that ''I'' is a ''proper'' ideal. An ideal is equal to ''R'' [[if and only if]] it contains 1. (It is clear that if it is ''R'' then it contains 1; on the other hand, if it contains 1 and ''r'' is an arbitrary element of ''R'', then ''r''1 = ''r'' is an element of the ideal, and so the ideal is equal to ''R''.) So, if ''I'' were equal to ''R'', then it would contain 1, and that means one of the members of ''T'' would contain 1 and would thus be equal to ''R'' – but ''R'' is explicitly excluded from ''P''.

The hypothesis of Zorn's lemma has been checked, and thus there is a maximal element in ''P'', in other words a maximal ideal in ''R''.