Every vector space has a basis Zorn's lemma can be used to show that every
vector space V has a
basis. 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
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
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 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
v1,
v2, ...,
vk ∈
B and
scalars a1,
a2, ...,
ak, not all zero, such that :a_1\mathbf{v}_1 + a_2\mathbf{v}_2 + \cdots + a_k\mathbf{v}_k = \mathbf{0}. Since
B is the union of all the sets in
T, there are some sets
S1,
S2, ...,
Sk ∈
T such that
vi ∈
Si for every
i = 1, 2, ...,
k. As
T is totally ordered, one of the sets
S1,
S2, ...,
Sk must contain the others, so there is some set
Si that contains all of
v1,
v2, ...,
vk. This tells us there is a linearly dependent set of vectors in
Si, contradicting that
Si 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
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.
Remark: While less common, it is possible to construct a basis somehow more directly using transfinite recursion, still assuming the axiom of choice. For that, see for example as a comparison.
Every nontrivial ring with unity contains a maximal ideal Zorn's lemma can be used to show that every nontrivial
ring R with
unity contains a
maximal ideal. Let
P be the set consisting of all
proper 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
r1 =
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.
A proof of Tychonoff's theorem Zorn's lemma implies the
ultrafilter lemma, which in turn implies
Tychonoff's theorem (together with the axiom of choice). However, it is also possible to prove Tychonoff's theorem directly from Zorn's lemma as follows (by using an ultrafilter implicitly). Let X_i be a family of
compact spaces, not necessarily Hausdorff. We shall show that a family F of closed subsets of X := \prod_i X_i with the
finite intersection property has nonempty intersection. Let P be the collection of all the families of subsets of X with the finite intersection property (not necessarily closed subsets). We let P be ordered by set inclusion. If C \subset P is a chain, then clearly the union of C has the finite intersection property; so, the union is in P. Thus, by Zorn's lemma, there is a maximal element M in P containing F. We shall show that the intersection of all the closed sets in M has nonempty intersection;
a fortiori, F has nonempty intersection. For each finite subset M' \subset M, since \cap M' is nonempty, \cap p_i(M') is nonempty for the projection p_i : X \to X_i. Thus, for each i, the set :N_i := \left \{ \overline{p_i(A)} \mid A \in M \right \} has nonempty intersection since it has the finite intersection property and X_i is compact. We note • If M' \subset M is a finite subset, \cap M' is in M. • For an open set U_i intersecting \cap N_i, we have p_i^{-1}(U_i) is in M. Indeed, (1) holds since M = M \cup \{ \cap M' \} by maximality, as the set on the right has the finite intersection property. Similarly, for a finite subset M' \subset M, since A = \cap M' is in M, U_i \cap p_i(A) \ne \emptyset or p_i^{-1}(U_i) \cap A \ne \emptyset. Thus, the set M \cup \{ p_i^{-1}(U_i) \} has the finite intersection property and (2) follows by the maximality of M. Now, by the axiom of choice, we can find an element m in \prod_i (\cap N_i). We claim this m is in each closed set A in M. Since A is closed, it is enough to show m is in the closure of A; i.e., each basic neighborhood of m intersects A. A basic neighborhood of m has the form \cap_{j=1}^r p_{i_j}^{-1}(U_{i_j}). But by the property (2) above, we have \left \{ A, p_{i_j}^{-1}(U_{i_j}) \mid j = 1, \dots, r \right \} \subset M, which gives the claim by the finite intersection property. \square
Remark: The ultrafilter lemma is strictly weaker than the axiom of choice; it is equivalent to the
boolean prime ideal theorem. On the other hand, somehow surprisingly, Tychonoff's theorem implies (thus is equivalent to) the axiom of choice. Hence, the use of AC above cannot be eliminated. (If X_i are compact Hausdorff, then the use of AC can be eliminated; indeed, each \cap p_i(M) above has exactly one point in that case.) An alternative (perhaps more standard) proof of Tychonoff's theorem is to first prove
Alexander's subbase lemma, but the proof of that lemma typically uses Zorn's lemma. == Equivalent formulations ==