Where is the mistake in my reasoning?:
Let X be a separable metric space, then for every $p\in [1,\infty)$ and for every borel measure $\mu$ on $X$: $L^p_{\mu}(X)$ is separable. Therefore by a thoeroem of Willard $L^p_{\mu}(X)$ is homeomorphic to a totally counded metric space $Y$.
Moreover, since $L^p_{\mu}(X)$ is complete and homeomorphism preserve completness then $Y$ is a complete and totally bounded metri cspace. Whence $Y$ must be compact. Since homeomorphism preserve compactness, then $L^p_{\mu}(X)$ is compact.
But clearly this is not the case since it is not sequentially compact.
My question is where did my reasoning go wrong? Is it that total boundedness is not preserved by homeomorphism; even then I didnt really use that....
Homeomorphisms, in general, preserve neither completeness nor total boundedness. Consider $\Bbb R$ and $(0,1)$.