How to show that $C[a,b]$ is infinite dimensional? How can we give a rigorous proof of the fact that the space $C[a,b]$ of all continuous real (or complex)-valued functions defined on a closed interval $[a,b]$, where $a$, $b$ are any two given real numbers such that $a<b$, is infinite-dimensional? 
We of course take the following norm: $$ ||x|| := \max_{a\leq t \leq b} |x(t)|$$ for any $x \in C[a,b]$, the vector addition and saclar multiplication being defined pointwise as usual. 
 A: Hint: this does not have anything to do with the norm. You can for instance observe that $C[a,b]$ contains the subspace of all polynomials, which is infinite-dimensional.
A: If the support of a continuous function on $[0,1]$ is $[k/n,(k+1)/n]$, then there you have a set of $n$ continuous functions, none of which is a linear combination of the others.  So the dimension is at least $n$.  For other intervals, it should be clear how to do the same thing.
A: I just discovered an alternate proof I wanted to share.
$C[a,b]$ with sup-norm is complete. Since any Banach space has either finite or uncountable basis and basis is not finite since $1,x,x^2,...$ are linearly independent functions, so the basis is uncountable. Further, the basis has cardinality less than the cardinality of $C[a,b]$ which is same as the cardinality of $\mathbb{R}$ so the basis has cardinality same as that of $\mathbb{R}$.  
A: Since C[a,b] is a function space, in every point of C[a,b] we can define a real valued function which is continuous in the interval [a,b] those functions will work as a basis for C[a,b]. As the point is infinite in number then C[a,b] is infinite dimensional.
