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Let $B = l_\infty$ be the Banach space of bounded sequences and the operator $$A: (x_1,x_2,x_3,\dots) \mapsto (x_1,x_2/2,x_3/3,\dots).$$ This operator is bounded. Therefore its image $B':=A(l_\infty)$, with the norm inherited from $l_\infty$, is a Banach space and $A:B\to B'$ is a bounded bijection. By the bounded inverse theorem, $A^{-1}$ is bounded, which is obviously not true.

What is the mistake in this reasoning?

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 When you claim $A(l_\infty)$ is a Banach space you have not said which norm you are putting on it. This is a natural question to think of while learning functional analysis, but it is in my opinion not at the right level for MO. mathoverflow.net/faq#whatnot – Yemon Choi Jan 26 2011 at 19:04
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 Here's a hint: the mistake is in the sentence starting "Therefore". – Mark Meckes Jan 26 2011 at 19:13
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 Continuous images do not have to be closed, and it is instructive to check that directly in your case. – Yemon Choi Jan 26 2011 at 19:25
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 No: $(x_n)$ is not a Cauchy sequence. E.g. if $x_n = (x^{(n)}_m) = (n^{1/2}\delta_{n,m})$ then $y_n = (n^{-1/2}\delta_{n,m})$ which converges to $0$ for the sup norm. Here $\delta_{n,m}$ is the Kronecker delta. If you want more help, ask this on math.stackexchange.com – Matthew Daws Jan 26 2011 at 20:14
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 Injective continuous maps can still move things that were far away close to each other. So just because points in the image are close together, it doesn't mean the corresponding pre-images had to be close together – Yemon Choi Jan 26 2011 at 21:00
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closed as too localized by Yemon Choi, Bill Johnson, Andres Caicedo, Andrew Stacey, Mariano Suárez-Alvarez Jan 26 2011 at 21:41

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