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If you two have inner product spaces and one is complete, and there is an isomorphism between the two spaces, is the other space also complete?

Or do we absolutely require equivalence of norms?

michael_faber
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    What do you mean by isomorphism here? Just algebraic, i.e., a linear bijection? – Harald Hanche-Olsen May 01 '13 at 18:52
  • @HaraldHanche-Olsen Yeah, that's right. – michael_faber May 01 '13 at 18:55
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    For a concrete example, note that $\ell^1$ is contained in $\ell^2$, but is not complete for the $\ell^2$ norm. And since $(x_n)\longmapsto (x_n/n^2)$ injects $\ell^2$ into $\ell^1$, Schroeder-Bernstein for vector spaces says that $\ell^1$ and $\ell^2$ are isomorphic as vector spaces. – Julien May 01 '13 at 19:15

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You have to take norms into account. Consider an infinite dimensional Banach space $X$. Take an infinite, linearly independent (in the algebraic sense) sequence of vectors $(x_n)$ in $X$. Create a linear map $X\to X$ by mapping $x_n$ to $nx_n$, and expand to a linear isomorphism. This map is clearly not bounded.

The expansion requires some transfinite induction, however: Typically, expand $(x_n)$ to an algebraic (a.k.a. Hamel) basis of $X$, then map each new basis vector to itself.

Harald Hanche-Olsen
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  • I should add that this argument really requires the axiom of choice (AC), so you can never give an explicit example. I am pretty sure that there exists an extension of ZF without AC in which all linear isomorphisms between Banach spaces are Banach space isomorphisms. But I am not totally sure, which is why I mention this in a comment and not in the answer proper. – Harald Hanche-Olsen May 01 '13 at 19:08