Let $S \subset X, Y$ be normed spaces over $K$. An operator $A:S \to Y$ is called compact if:
- $A$ is continuous
- $A$ transforms bounded set into relatively compact sets i.e. if $(c_n)$ is a bounded sequence in $\mathbb{S}$, $\exists (c_n') \subset (c_n)$ s.t. $(Ac_n')$ is convergent in Y
Why should it be obvious that all continuous operators $A$ in finite dimensional space i.e. $R^n$ are compact?