How does Rellich–Kondrachov not lead to a contradiction? I'm currently working through Lawrence Evan's PDE's textbook, and in it, he states the Rellich-Kondrachov theorem:
Assume $U$ is a bounded open set of $\mathbb R^n$, and $\partial U$ is $C^1$. Suppose $1 \leq p < n$. Then
$$W^{1,p}(U) \subset \subset L^p(U)$$
for each $1 \leq q < p^*$ where $p^*$ is the Sobolev conjugate of $p$.
I am currently failing to see how this does not contradict the fact that a linear space is compact if and only if it is finite dimensional. In particular the argument I have in mind is that, if $\overline{W^{1,p}(U)}$ is compact, then so is the unit ball in $W^{1,p}(U)$, being a closed subset of a compact set. But, $W^{1,p}(U)$ is clearly not finite dimensional.
Where does the above argument go wrong? Thanks!
Edit: After doing some further digging, it seems like the main cause of my confusion is that the topologist's definition of compact embedding and the analyst's definition are not equivalent, so I was under the impression that the theorem said something that it didn't.
 A: An embedding $X\subset Y$ being compact means that if we


*

*consider the image of the unit ball of $X$ under this embedding, and

*take its closure in the topology of $Y$,

*then we'll get a compact set in the topology of $Y$.


None of this speaks of the closed unit ball of $X$ being compact in the topology of $X$. And it will not be, unless $X$ is finite dimensional.
A: According to Evans (paraphrasing slightly):
Definition: Let $X,Y$ be Banach spaces with $X \subset Y$.  We say that $X$ is compactly embedded in $Y$, written $X \Subset Y$ provided both:


*

*There is a constant $C$ for which $\Vert x \Vert_Y \leq C \Vert x \Vert_X$ for all $x \in X$.

*Every bounded sequence in $X$ has a subsequence which converges in $Y$.
The qualifier "bounded" in (2) is important.  In writing $X \Subset Y$, we are not asserting that every sequence in $X$ has a subsequence which converges in $Y$.  That is: It does not follow that $\overline{X}$ (the closure of $X$ in the topology of $Y$) is compact (in the topology of $Y$).
