Prove that $(C([0,1]),\lVert \cdot \rVert_\infty)$ is infinite dimensional Let $C([0,1])$ equipped with $\lVert \cdot \rVert_\infty$ be set of all functions continuous on $[0,1]$. Prove that $C([0,1])$ is not finite dimensional.
There is a theorem which states that a normed vector space is finite dimensional if and only if its closed unit ball $\{v\in V: \lVert v \rVert =1   \}$ is compact. In our case, $S=\{f\in C([0,1]): \lVert f \rVert_\infty =1   \}$. We know that a subset of a vector space is compact if and only if it is closed and bounded. Clearly, $S$ is bounded. So all is needed to be proved here is that $S$ cannot be closed to establish that $(C([0,1]),\lVert \cdot \rVert_\infty)$ is not finite dimensional.
One of the approaches that I've tried is this:

If $f_0$ is a limit point of $S$ then $\exists \{f_n\}\subset S$ such that $\{f_n\}$ converges to $f$. In this case, one would have to prove that somehow $\lVert f_0 \rVert_\infty \ne 1$. This can be done by providing a counterexample that $\exists f_n\subset S$ converging to $f_0$ where $\lVert f_0 \rVert_\infty \ne 1$. I came up with the sequence of functions $f_n=e^{-nx}$, which supposedly converges to the zero function. The only problem is: what if $x=0$? Then $f_n$ converges to $1$.

I would appreciate some advice.
 A: Your argument is not gonna work because $S$ is closed in any normed space. The point is not to show it's not closed, but to show it's not compact: as addressed in my comment, the equivalence between compactness and closedness-boundedness is only a property of finite dimensional vector spaces. (Edit: as spotted by @Mariano Suárez-Álvarez, we need to assume first that the vector space is complete.)
To prove non-compactness you might want to use the following (equivalent) characterisation of compactness in any metric space: if $K$ is compact, then any sequence in it has a subsequential limit in it. Now pick $\{e^{in2\pi t}\}\subset S$, and you should be able to show it has no subsequential limit at all.

Or, for a much more straightforward approach, just see @carmichael561's comment: $\{1,x,x^2,\cdots\}$ are linear independent, but no finite-dimensional vector space admits an infinite set of linearly independent vectors.
A: The norm is completely irrelevant for this exercise, it's just linear algebra.
Let $p_n(x) = x^n$, so that $p_n \in C([0,1])$. Suppose that $f := \sum a_k p_k = 0$ for some coefficients $(a_0, a_1, \dots)$ (where only a finite number are nonzero). Then $a_k = k! \cdot f^{(k)}(0)$, simply by differentiating the polynomial and evaluating at $x = 0$. Since $f$ is actually the zero function, all its derivatives vanish, therefore $f^{(k)}(0) = 0$ and thus $a_k = 0$.
In other words, as soon as a we pick a linear combination of the $p_k$ which vanish, then all the coefficients must vanish. By definition, this means that the family $(p_0, p_1, \dots)$ is linearly independent. It is also infinite, thus $C([0,1])$ is infinite-dimensional, as otherwise any infinite family would be linearly dependent.

Now, something more interesting is that the dimension of $C([0,1])$ is not even countable. Indeed, it is a Banach space, and infinite-dimensional Banach spaces never have a countable basis (by the Baire category theorem). For this the norm is really relevant.
A: Let $f_n$ linearly interpolate the points $(0,0), ({1 \over n+1},0), ({1\over 2}({1 \over n}+ { 1\over n+1}),1), ({1 \over n},0),(1,0)$.
Suppose $\sum_n \alpha_n f_n = 0$, then be evaluating at  ${1\over 2}({1 \over k}+ { 1\over k+1} )$, we see that $\alpha_k = 0$, hence the $f_n$ are
linearly independent.
