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.
Best Answer
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.