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A Banach space of (Hamel) dimension $\kappa$ exists if and only if $\kappa^{\aleph_0}=\kappa$.

How will we prove the converse implication. One sided implication for Hilbert Space is proved in question: Can you equip every vector space with a Hilbert space structure?

If we don't assume Axiom of Choice, and we have a Banach space with (Hamel Basis B existence given). Will it be true $B^\Bbb N$ equinumerous with $B$?

Note: $B^\Bbb N$ is not empty as $B$ is specified.

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No, this is not true.

If $D$ is a Dedekind finite set with a Dedekind finite power set, then $\ell_1(D)$ is a Banach space which has a Hamel basis which is also a Schauder basis, and every linear operator from $\ell_1(D)$ to a normed space is continuous.

But if $D$ is Dedekind finite, then $|D|^{\aleph_0}>|D|$. So it suffices to assume that an infinite Dedekind finite set like that exists. Which is of course consistent with the failure of choice.

See also:

Brunner, Norbert "Garnir's dream spaces with Hamel bases." Arch. Math. Logik Grundlag. 26 (1987), no. 3-4, 123–126.

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  • $\begingroup$ Why Hamel Basis will be Schauder Basis. Schauder Basis are always countable while normed linear space has uncountable Hamel Basis. I think you want to say Hamel Basis has cardinality same as D $\endgroup$ – Sushil Jun 27 '15 at 11:02
  • $\begingroup$ And I might be rude in asking questions and questions only. But If I assume axiom of countable choice atleast. What will be your conclsion. $\endgroup$ – Sushil Jun 27 '15 at 11:06
  • $\begingroup$ (1) If you just assume a normed space, then it is perfectly possible to find one with a countable dimension. (2) Read the paper by Brunner. (3) If we assume countable choice, I don't know what happens, I imagine it might still be consistent that a counterexample holds, but it is probably going to be much more difficult to come by. $\endgroup$ – Asaf Karagila Jun 27 '15 at 11:26
  • $\begingroup$ But proof given here: math.stackexchange.com/q/1340333/168520 I don't see were we using CC $\endgroup$ – Sushil Jun 27 '15 at 11:42
  • $\begingroup$ Not every normed space is a Banach space. I never claimed that there is a Banach space of countable dimension. $\endgroup$ – Asaf Karagila Jun 27 '15 at 11:46

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