# Prove that the real vector space consisting of all continuous, real-valued functions on the interval $[0,1]$ is infinite-dimensional.

Prove that the real vector space consisting of all continuous, real-valued functions on the interval $[0,1]$ is infinite-dimensional.

Clearly it's infinite dimensional, because if you consider say $P (\mathbb{F})$ on $[0,1]$, then there are an infinite amount of continuous real-valued functions on the interval, but how do I prove this?

• You can exhibit an infinite linearly independent collection of continuous functions. To use $1,x,x^2,x^3,\dots$ look up Vandermonde matrix (or determinant, I forget which) in Wikipedia. Oct 8 '13 at 19:34
• The fact that a vector space has infinitely many elements doesnt mean that its dimension is infinite as well. Consider $\mathbb R$ which clearly has infinitely many elements, but has dimension one. Hint: Show that the subset of polynomials forms an infinite-dimensional subspace. Oct 8 '13 at 19:35

Recall that if $V$ is a finite-dimensional vector space, then each subspace of $V$ is also finite-dimensional. So if $V$ contains an infinite-dimensional subspace, then it is infinite-dimensional. As you point out, $C[0,1]$ (the space of continuous real-valued functions on $[0,1]$) has the space $P(\mathbb{R})$ of real polynomials as a subspace. If you can show that $P(\mathbb{R})$ is not finite-dimensional, then you're done.

The plethora of answers already submitted are more than sufficient to answer the question. Interestingly, one may actually show that the collection of smooth functions on any smooth manifold forms an infinite dimensional vector space. Since $C^{\infty}(M) \subseteq C(M)$ this would also prove your result.

To prove that $C^\infty(M)$ is infinite dimensional requires a bit of work, but is actually quite intuitive. To show that a vector space $V$ is infinite dimensional, it is sufficient to show that for any $k \in \mathbb N$, one may find $k$-linearly independent vectors. Fix an arbitrary chart $\phi: U \subseteq M \to \mathbb R^n$ and take $k$-distinct points $\{x_i\}_{i=1}^k$ in $\phi(U)$. By Hausdorffness of $\mathbb R^n$, one may find open neighbourhoods $X_i$ of each $x_i$ and consequently $\epsilon_i$ such that $B_i:=\overline{B_{\epsilon_i}(x_i)} \subseteq X_i$. If $f_i:\mathbb R^n \to \mathbb R$ are bump functions on each $B_i$ and $W_i = \phi^{-1}(B_i)$, define the continuous functions $$g_i: M \to \mathbb R, \qquad g_i(x) = \begin{cases} f_i(\phi(x)) & x \in W_i \\ 0 & x \notin W_i \end{cases}.$$ These may be extended to smooth functions $\hat g_i$ on $M$ such that $g_i|_{W_i} = \hat g_i|_{W_i}$, and these $\hat g_i$ (of which there are $k$) are easily seen to be a linearly independent set in $C^\infty(M)$.

• I'm afraid this may be a bit above my current knowledge. It seems like an interesting answer though! Oct 8 '13 at 19:51
• Well done! Very well done indeed! +1! Oct 8 '13 at 19:54
• The answer seems complicated because it is framed in the language of manifolds. Do you know what bump functions are? The idea is that for any $k \in \mathbb N$ you can find $k$ compact disjoint sets $\{X_i\}_{i=1}^k$ in $[0,1]$. The bump functions on these $X_i$ are all linearly independent. Since you could do this for any $k$, you must be infinite dimensional. Edit: The intuition behind the manifold proof is then to just do this in a local chart. Oct 8 '13 at 19:54
• I see. Well, I have learned about manifolds yet, but when I start to I will keep this answer in mind. Oct 8 '13 at 20:46

The fact that there are infinite amounts of continuous functions is not enough. In $\mathbb{R}^2$ there are infinite amounts of vectors but it's still finite dimensional. One way to prove this is to exhibit an infinite set of continuous functions that are linearly independent. As a suggestion, consider polynomials.

Consider the subspace of $C([0, 1], \Bbb R)$ generated by the polynomials with real coefficients, i.e. $\Bbb R[x]$. It is infinite dimensional since the functions $1, x, x^2, . . .$ are linearly independent. How so? A linear relation $\sum_0^n c_n x^n = 0$ affirms the the polynomial $\sum_0^n c_n x^n$ vanishes identically; thus $c_i = 0$ for $0 \le i \le n$. This shows the $x^i$, infinite in number, are linearly independent, so $R[x]$ is an infinite dimensional subspace of $C([0, 1], \Bbb R)$ which, having an infinite dimensional subspace, is itself of infinite dimension.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!