Is this basic function space compact?

Question

Let $A=L^2(X)$ be the space of square integrable functions on a compact Euclidean space $X$. If we equip $A$ with the usual 2-norm, is $A$ compact?

Edit: And if we restrict AA by adding the assumption that the functions are totally bounded, i.e. the supremum norms of all the functions are bounded by a constant?

Answer 1

No. If $X=[0,1]$, then the sequence $\{e^{2\pi inx}\}_{n\in\mathbb{Z}}$ has no convergent subsequence.

Answer 2

$d(x,y)=\|x-y\|$ is an unbounded metric. Any metric on a compact space is bounded.