Is $\mathbb N$ a Banach space with the norm $|x-y|$ from $\mathbb R$? I think is Banach space because there is no convergent sequence that is not constant after some $N$. Then all limit points are in the space. But I am not sure.
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6Have you looked up the definition of Banach space? – Jonas Meyer May 29 '13 at 14:49
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@JonasMeyer Yes. I forgot to regard the structure it have. How stupid of me. – blue May 29 '13 at 15:00
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4Not only is it a real question, it even displays some work and has a multiply-upvoted and accepted answer. While there’s no pressing need at this point to reopen the question, the closure is bizarre, to put it politely. – Brian M. Scott May 29 '13 at 19:08
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I agree. This ain't even a PSQ; it's just short. – Alexander Gruber May 29 '13 at 19:46
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It surely is a complete metric space, your proof is correct. Another justification of this would be that it is a closed subspace of the Banach space $(\Bbb R,|\cdot|)$. But it has no linear structure, so it's not a Banach space.
Olivier Bégassat
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No, a Banach space has to be a vector space over $\mathbb R$. This is not true for $\mathbb N$ because in general $n\in \mathbb N$ does not mean that $\lambda n\in \mathbb N$ for every $\lambda \in\mathbb R$.
Tim
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