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Let $S \subset X, Y$ be normed spaces over $K$. An operator $A:S \to Y$ is called compact if:

  1. $A$ is continuous
  2. $A$ 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?

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You can answer this in several ways, drawing upon the mainstream structural results in finite-dim. spaces, e.g. the Bolzano-Weierstrass property that every bounded sequence has a convergent subsequence, and then noting that continuous operators preserve sequential convergence.

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