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Is the normed vector space $\{0\}$ considered a Banach space? I am asking this question because Cartan's Differential Calculus implicitly assumes that the identity operator has norm 1 in Section 1.9.

However with the norm of $f$, defined as $$ \|f\| = \sup_{\|x\|\leq 1} \|f(x)\|,$$ the norm of the identity operator is $0$ for the trivial vector space.

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It certainly should be considered a Banach space. The kernel of a linear operator is a Banach space. A closed linear subspace of a Banach space is a Banach space. These statements become much more clumsy if you exclude $0$. On the other hand, you sometimes have to exclude the zero space in some statements and proofs, but this is hardly ever interesting or difficult. – Martin Jan 13 '13 at 9:46
Yes, it's a Banach space, but if considering it as such makes the definition of a norm stupid then just consider the definition to apply for a nonzero space. – KCd Jan 13 '13 at 9:46
Well, it IS a Banach space, there is nothing to argue about... – Kofi Jan 13 '13 at 10:48
The Banach spaces book by Megginson pays particular attention to zero-dimensional spaces throughout the text, pointing out how certain natural statements are false for them. – user53153 Jan 13 '13 at 15:18

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