In my notes on functional analysis it mentions that $C([0,1]),\ell^p$ and, $\ell^\infty$ are normed vector spaces, and gives some examples of norms that we can define on them. However, it then simply states that these three spaces are infinite-dimensional normed vector spaces.

The only thing mentioned in my notes so far is in relation to finite-dimensional vector spaces, namely, that a vector space is finite-dimensional if it has a finite basis.

My question(s): how is it exactly that one understands these spaces to be infinite-dimensional; what does it mean to say that they are infinite-dimensional and how do they differ from an example of a finite-dimensional vector space, say, $\mathbb R^n$. Going on what I know about finite-dimensional spaces, is it simply then that an infinite-dimensional space has an infinite basis? How would one visualize this? Can anybody show me why the examples I gave above are indeed infinite-dimensional?

  • 2
    $\begingroup$ This might be helpful math.stackexchange.com/questions/630142/… $\endgroup$
    – user223391
    Commented Aug 12, 2015 at 8:01
  • 1
    $\begingroup$ I would say that an infinite-dimensional space has no finite basis, rather than has an infinite basis. Both are correct but the second is way trickier. $\endgroup$ Commented Aug 12, 2015 at 8:01

2 Answers 2


Given the axiom of choice, every vector space has a basis (though it will be a very unnatural basis), and you are correct that infinite-dimensional vector spaces are exactly those where the basis is infinite. But this kind of basis (often called a Hamel basis) is rather useless and impossible to visualize.

So, a more concrete way of thinking about it might be that in an infinite-dimensional vector space, you can exhibit infinitely many vectors $v_1, v_2, v_3, \ldots$ that are all linearly independent; no (finite) linear combination of vectors is zero. Equivalently, $v_n$ is not a linear combination of $v_1, v_2, \ldots, v_{n-1}$ for any $n$. In particular, this means $\oplus_{i=1}^\infty \mathbb{R}$ (the set of infinite sequences of real numbers where all but finitely many terms are zero) is algebraically a subspace of every infinite-dimensional vector space over $\mathbb{R}$.

How do such vector spaces differ from finite-dimensional vector spaces? Many things break. For example:

  • Some linear maps do not have any eigenvalues.

  • Some linear maps are not continuous; you end up having to restrict to Bounded operators which are basically, operators which behave nicely with the norm on your vector space.

  • The Dual space of the dual space of $V$, $(V^*)^*$, might not equal $V$.


If $V$ is a vector space of finite dimension $n$, any collection of $n + 1$ vectors is linearly dependent. Using this property you can easily show that the vector spaces in the first posting are not finite-dimensional by proving the existence of a linearly independent set of arbitrary size.

For example, let's take a look at $C([0, 1])$. For every natural number $n$ you can construct a set of functions ${f_1, \ldots, f_n}$ that is linearly independent. Just define $f_i$ to be $0$ on $[0, \frac{2i - 2}{2n}] \cup [\frac{2i}{2n}, 1]$, $1$ at $\frac{2i - 1}{n}$ and (affine) linear between $\frac{2i - 2}{2n}$, $\frac{2i - 1}{2n}$ and $\frac{2i}{2n}$. It is easy to see that this family of functions is linearly independent.

  • $\begingroup$ Surely you mean linearly dependent in your first sentence? $\endgroup$ Commented Aug 12, 2015 at 9:17
  • 1
    $\begingroup$ @JeremyJeffreyJames Yes, I will correct it. $\endgroup$
    – Dominik
    Commented Aug 12, 2015 at 9:18
  • $\begingroup$ Thanks for your answer, it certainly adds to and clarifies the answer below. $\endgroup$ Commented Aug 12, 2015 at 9:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .