I read that If X is an infinite-dimensional Banach space then every compact operator cannot be Also Bijective. But why? I know that if dim(X) < inf then the space of compact operator is equal to the space of linear and bounded operator, but in which way this imply the thesis?
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There are definitely bounded linear operators which are not compact, so your comment is wrong. – Aug 11 '21 at 22:58
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Not if the space is finite dimensional space – maru0032 Aug 11 '21 at 22:59
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@maru0032 In infinite dimensions, the identity operator is not compact; there are many non-compact operators! – Aug 11 '21 at 23:01
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Yes, in infinite dimension yes, but Not in finite! But in why way this imply that cannot be exist a compact bijective operator? – maru0032 Aug 11 '21 at 23:03
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@maru0032 Ah, you edited your post to correct it. But now the second sentence of your post is irrelevant--if you care about infinite dimensional spaces, knowing about finite dimensional spaces doesn't help. – Aug 11 '21 at 23:04