Why Does a Vector Space Have a Basis (Module Theory)

Module and vector space

First we recall some backgrounds. Suppose A is a ring with multiplicative identity 1A. A left module of A is an additive abelian group (M,+), together with an ring operation A×MM such that

for x,yM and a,bA. As a corollary, we see (0A+0A)x=0Ax=0Ax+0Ax, which shows 0Ax=0M for all xM. On the other hand, a(xx)=0M which implies a(x)=(ax). We can also define right A-modules but we are not discussing them here.

Let S be a subset of M. We say S is a basis of M if S generates M and S is linearly independent. That is, for all mM, we can pick s1,,snS and a1,,anA such that

and, for any s1,,snS, we have

Note this also shows that 0MS (what happens if 0MS?). We say M is free if it has a basis. The case when M or A is trivial is excluded.

If A is a field, then M is called a vector space, which has no difference from the one we learn in linear algebra and functional analysis. Mathematicians in functional analysis may be interested in the cardinality of a vector space, for example, when a vector space is of finite dimension, or when the basis is countable. But the basis does not come from nowhere. In fact we can prove that vector spaces have basis, but modules are not so lucky.

Examples of non-free modules

First of all let’s consider the cyclic group Z/nZ for n2. If we define

which is actually m copies of an element, then we get a module, which will be denoted by M. For any x=k+nZM, we see nk+nZ=0M. Therefore for any subset SM, if x1,,xkM, we have

which gives the fact that M has no basis. In fact this can be generalized further. If A is a ring but not a field, let I be a nontrivial proper ideal, then A/I is a module that has no basis.

Following Z/nZ we also have another example on finite order. Indeed, any finite abelian group is not free as a module over Z. More generally,

Let G be a abelian group, and Gtor be its torsion subgroup. If Gtor is non-trival, then G cannot be a free module over Z.


Next we shall take a look at infinite rings. Let F[X] be the polynomial ring over a field F and F[X] be the polynomial sub-ring that have coefficient of X equal to 0. Then F[X] is a F[X]-module. However it is not free.

Suppose we have a basis S of F[X], then we claim that |S|>1. If |S|=1, say PS, then P cannot generate F[X] since if P is constant then we cannot generate a polynomial contains X with power 1; If P is not constant, then the constant polynomial cannot be generate. Hence S contains at least two polynomials, say P10 and P20. However, note X2P1F[X] and X2P2F[X], which gives

Hence S cannot be a basis.

Why does a vector space have a basis

I hope those examples have convinced you that basis is not a universal thing. We are going to prove that every vector space has a basis. More precisely,

Let V be a nontrivial vector space over a field K. Let Γ be a set of generators of V over K and SΓ is a subset which is linearly independent, then there exists a basis of V such that SBΓ.

Note we can always find such Γ and S. For the extreme condition, we can pick Γ=V and S be a set containing any single non-zero element of V. Note this also gives that we can generate a basis by expanding any linearly independent set. The proof relies on a fact that every non-zero element in a field is invertible, and also, Zorn’s lemma. In fact, axiom of choice is equivalent to the statement that every vector has a set of basis. The converse can be found here.

Proof. Define

Then T is not empty since it contains S. If T1T2 is a totally ordered chain in T, then T=i=1Ti is again linearly independent and contains S. To show that T is linearly independent, note that if x1,x2,,xnT, we can find some k1,,kn such that xiTki for i=1,2,,n. If we pick k=max(k1,,kn), then

But we already know that Tk is linearly independent, so a1x1++anxn=0V implies a1==an=0K.

By Zorn’s lemma, let B be the maximal element of T, then B is also linearly independent since it is an element of T. Next we show that B generates V. Suppose not, then we can pick some xΓ that is not generated by B. Define B=B{x}, we see B is linearly independent as well, because if we pick y1,y2,,ynB, and if

then if b0 we have

contradicting the assumption that x is not generated by B. Hence b=0K. However, we have proved that B is a linearly independent set containing B and contained in S, contradicting the maximality of B in T. Hence B generates V.

The Big Three Pt. 1 - Baire Category Theorem Explained

About the ‘Big Three’

There are three theorems about Banach spaces that occur frequently in the crux of functional analysis, which are called the ‘big three’:

  1. The Hahn-Banach Theorem
  2. The Banach-Steinhaus Theorem
  3. The Open Mapping Theorem

The incoming series of blog posts is intended to offer a self-read friendly explanation with richer details. Some basic analysis and topology backgrounds are required.

First and second category

The term ‘category’ is due to Baire, who developed the category theorem afterwards. Let X be a topological space. A set EX is said to be nowhere dense if E has empty interior, i.e. int(E)=.

There are some easy examples of nowhere dense sets. For example, suppose X=R, equipped with the usual topology. Then N is nowhere dense in R while Q is not. It’s trivial since N=N, which has empty interior. Meanwhile Q=R. But R is open, whose interior is itself. The category is defined using nowhere dense set. In fact,

  • A set S is of the first category if S is a countable union of nowhere dense sets.
  • A set T is of the second category if T is not of the first category.

Baire category theorem (BCT)

In this blog post, we consider two cases: BCT in complete metric space and in locally compact Hausdorff space. These two cases have nontrivial intersection but they are not equal. There are some complete metric spaces that are not locally compact Hausdorff.

There are some classic topological spaces, for example Rn, are both complete metric space and locally compact Hausdorff. If a locally compact Hausdorff space happens to be a topological vector space, then this space has finite dimension. Also, a topological vector space has to be Hausdorff.

By a Baire space we mean a topological space X such that the intersection of every countable collection of dense open subsets of X is also dense in X.

Baire category states that

(BCT 1) Every complete metric space is a Baire space.

(BCT 2) Every locally compact Hausdorff space is a Baire space.

By taking the complement of the definition, we can see that, every Baire space is not of the first category.

Suppose we have a sequence of sets {Xn} where Xn is dense in X for all n>0, then X0=nXn is also dense in X. Notice then X0c=nXnc, a nowhere dense set and a countable union of nowhere dense sets, i.e. of the first category.

Proving BCT 1 and BCT 2 via Choquet game

Let X be the given complete metric space or locally Hausdorff space, and {Xn} a countable collection of open subsets of X. Pick an arbitrary open subsets of X, namely A0 (this is possible due to the topology defined on X). To prove that nVn is dense, we have to show that A0(nVn). This follows the definition of denseness. Typically we have

A subset A of X is dense if and only if AU for all nonempty open subsets U of X.

We pick a sequence of nonempty open sets {An} inductively. With An1 being picked, and since Vn is open and dense in X, the intersection VnAn1 is nonempty and open. An can be chosen such that

For BCT 1, An can be chosen to be open balls with radius <1n; for BCT 2, An can be chosen such that the closure is compact. Define

Now, if X is a locally compact Hausdorff space, then due to the compactness, C is not empty, therefore we have

which shows that A0Vn. BCT 2 is proved.

For BCT 1, we cannot follow this since it’s not ensured that X has the Heine-Borel property, for example when X is the Hilbert space (this is also a reason why BCT 1 and BCT 2 are not equivalent). The only tool remaining is Cauchy sequence. But how and where?

For any ε>0, we have some N such that 1N<ε. For all m>n>N, we have AmAnAN, therefore the centers of {An} form a Cauchy sequence, converging to some point of K, which implies that K. BCT 1 follows.

Applications of BCT

BCT will be used directly in the big three. It can be considered as the origin of them. But there are many other applications in different branches of mathematics. The applications shown below are in the same pattern: if it does not hold, then we have a Baire space of the first category, which is not possible.

R is uncountable

Suppose R is countable, then we have

where xn is a real number. But {xn} is nowhere dense, therefore R is of the first category. A contradiction.

Suppose that f is an entire function, and that in every power series

has at least one coefficient is 0, then f is a polynomial (there exists a N such that cn=0 for all n>N).

You can find the proof here. We are using the fact that C is complete.

An infinite dimensional Banach space B has no countable basis

Assume that B has a countable basis {x1,x2,} and define

It can be easily shown that Bn is nowhere dense. In this sense, B=nBn. A contradiction since B is a complete metric space.

The series

Since there is no strong reason to write more posts on this topic, i.e. the three fundamental theorems of linear functional analysis, I think it’s time to make a list of the series. It’s been around half a year.