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Answers for “为什么 \ell^p(\mathbb{N}) 上的恒等算子不是紧算子?”

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Every linear operator between finite-dimensional normed spaces is compact because it is automatically continuous (equivalently bounded), which makes the image of the unit ball bounded. Taking the closure of this bounded image yields a closed and bounded set, and in finite-dimensional spaces, closed and bounded sets are compact.

Conditions: The operator is linear.; The domain and codomain are finite-dimensional normed spaces.