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I am reading a book about functional analysis and have a question:

Let $X$ be a infinite-dimensional Banach-space and $A:X \rightarrow X$ a compact operator. How can one show that $A$ can not be surjective?

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    What is your preferred definition of "compact operator"? Proofs might look quite different depending on which definition you use.2011-10-19
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    Jonas has given an answer, but you can also remember that in the context of Banach spaces or more generally complete metrizable topological vector spaces, surjective (continuous) linear maps are automatically *open*. Thus you would have $$c B_X\subset A(B_X)$$ where $c$ is a positive real number and $B_X$ is the unit ball in $X$. The left hand side has compact closure iff $X$ is finite dimensional by a theorem of Riesz, while the right hand side has compact closure by definition of $A$ being compact. So $X$ must be finite dimensional for $A$ to be compact and surjective.2011-10-19
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    @Olivier: Please add that as an answer.2011-10-19
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    @t.b. ok, done.2011-10-19
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    What about the case when $A\colon X\rightarrow Y$, when $X\neq Y$ in general?2015-12-08
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    Then there can be surjection for sure. Take $Y = \mathbb R$ for example and a non zero functional $f: X\to \mathbb R$. @charlestoncrabb2016-09-01

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