Consider the space $C([0,1])$ of real valued continuous functions on $[0,1]$.
What are elementary arguments showing that the dimension of $C([0,1])$ as a real vector space is uncountable?
I look for an argument that could be understood by a first year student. So I don't want to use for instance the Baire category theorem establishing that every infinite dimensional Banach space has uncountable dimension or some Hilbert space methods.
$\frac{1}{x-a}$ for $a>1$ are linearly independent.
If $f(x) = \sum_{i=1}^n\frac{c_i}{x-a_i}=0$ for all $x\in[0,1]$, the argument can be finished in different ways,
(1) $f(x)$ is analytic outside $a_i$, so it has isolated zeroes unless $f(x)\equiv 0$, but if $c_i\not=0$, when $x\rightarrow a_i$, $f(x)\rightarrow\infty$, so $f(x)$ is not identically $0$.
(2) Multiply $\prod_{i}(x-a_i)$ on both sides, we get a polynomial $p(x)$ that vanishes on $[0,1]$, so it must be the zero polynomial for othwerwise it has at most finitely many zeroes. And if $c_i\not=0$, then let $x = a_i$, we have $p(a_i)= c_i\prod_{j\not=i}(a_i-a_j)\not=0$.
