Monthly Archives: October 2013

A vector space of uncountable dimension

Let V=\{ f: {\bf N}\rightarrow R\} be the real vector space of all functions from the natural numbers to the reals. Then V has uncountable dimension. To see this, for each a>0, let f_a be the function such that f_a(n)=a^n.

Claim: The f_a are linearly independent.

Proof: It suffices to show that any finite collection \{f_{a_i}\}_{i=1}^n with a strictly increasing sequence of a_i are linearly independent.

We prove this by induction on n, the fact being clear for n=1. Suppose \sum_{i=1}^nc_if_{a_i}=0 with n>1. The equation means


for all m. Thus, \sum_{i=1}^{n-1}c_i(a_i/a_n)^m+c_n=0 for all m. Now let m\rightarrow \infty. This shows that c_n=0. Thus, by induction, all c_i=0.

We have displayed an uncountable linearly independent collection of functions in V. Now, let \{b_i\}_{i\in I} be a basis for V. For each f_a there is then a unique finite set I(a)\subset I of indices such that f_a can be written ¬†as a linear combination with non-zero coefficients of \{b_i\}_{i\in I(a)}. The linear span of any given finite set of the b_i is finite-dimensional. Hence, for any finite subset S\subset I, there are at most finitely many a such that I(a)=S. That is, the map a\mapsto I(a) is a finite-to-one map from the positive reals R_{>0} to the finite subsets of I. Hence, the set of finite subsets of I must be uncountable. But then I itself must be uncountable. (I leave it as an exercise to show that the set of finite subsets of a countable set is itself countable. You should really write out the proof if you’ve never done it before.)

I might point out that before the tutorials, I was a bit confused myself. That is, the first bit about the f_a‘s being an uncountable linearly independent set is rather easy. However, I started hesitating: Still, why can’t there be a countable set of elements in terms of which we can express all of them? After all, the set of coefficients we can use for the expressions is uncountable… So think through again clearly: how is this resolved above?

As a final remark, note that this proves that V is not isomorphic to R[x]. This is perhaps the first example you’ve seen where you can prove that two vector spaces of *infinite* dimensions are not isomorphic by simply counting the dimensions and comparing them.

Linear independence of polynomials

One of the exercises this week asked for a proof of linear independence for the set

\{x^i\}_{i\in {\bf N}}

inside the polynomials R[x] with real coefficients. However, note that the polynomials here are regarded as *functions* from R to R. Thus, it amounts to showing that if

c_0+c_1x+\cdots c_nx^n=0

as a function, then all c_i have to be zero. This does require proof. One quick way to do this is to note that all polynomial functions are differentiable. And if

f(x)=c_0+c_1x+\cdots c_nx^n

is the zero function, then so are all its derivatives. In particular,


for all i. But f^{(i)}(0)=i!c_i. Thus, c_i=0 for all i.

One possible reason for confusion is that there is another ‘formal’ definition of R[x] by simply identifying a polynomial with its sequence of coefficients. That is, you can think of an element of R[x] as a function f:N \rightarrow R that has *finite support* in that f(i)=0 for all but finitely many i. With this definition, the polynomial x^i becomes identified with the function e_i that sends i to 1 and everything else to zero. If you take this approach, the linear independence also becomes formal. But in this problem, you are defining R[x] as a function in its variable. This of course is the natural definition you’ve been familiar with at least since secondary school.

Here are two questions:

1. If you think of two polynomials f and g as functions from N to R with finite support, what is a nice way to write the product fg?

2. What is the advantage of this formal definition?

Mathematics in Society

After several conversations recently about the social status of mathematics, I thought I’d put here links to two short essays I wrote on this.

1. An exchange on Mathoverflow

2. Mathematical Vistas