This theorem is from Rudin book which he says that obvious, but I'm quite confused how to prove it completely. Hope someone can help me clarify.
Let $X$, $Y$ be Banach spaces, If $T \in B(X,Y)$ and dim $R(T) \lt \infty$, then $T$ is compact
Because dim $R(T) \lt \infty$, $R(T)$ is closed in $Y$. Then $R(T)$ is complete because $Y$ is complete. Following from open mapping theorem, we conclude that $T$ is an open mapping from $X$ onto $R(T)$. Because $\overline{T(U)}$ is closed, we need to prove that it is bounded. But $U$ is bounded, then $T(U)$ is bounded, so $\overline{T(U)}$ is bounded. Therefore $\overline{T(U)}$ is compact (without using dim $R(T) \lt \infty$). So what did I do wrong? Where to use the statement dim $R(T) \lt \infty$. Thanks
You use the fact that the range is finite dimensional to conclude that it is closed. In general it may not be closed! I would prove it like this:
Let $\{x_n\}$ be a sequence in the unit ball of $X$. Then $T(x_n)$ is bounded: For any $n\in \Bbb{N}$, we have $$\|T(x_n)\| \leq \|T\| \|x_n\| \leq M$$
for some constant $M$ because $T$ is bounded and $\{x_n\}$ is in the unit ball. Furthermore $T(x_n)$ is contained in a finite dimensional subspace. Now use the fact that a bounded sequence in a finite dimensional complete normed vector space has a convergent subsequence.