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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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Sushil
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  • As noted in this answer: http://math.stackexchange.com/a/1100952/151552, this is Lemma 2 in the paper "Badly incomplete normed linear spaces" by A.H. Kruse. – PhoemueX Jun 27 '15 at 09:05
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    In case you do not have access to Kruse's article from Springer website, it is useful to know that older issues of Mathematische Zeitschrift are freely available from GDZ: http://gdz.sub.uni-goettingen.de/dms/load/toc/?IDDOC=8487 – Martin Sleziak Jun 27 '15 at 10:47

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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.

Asaf Karagila
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  • 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 – Sushil Jun 27 '15 at 11:02
  • 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. – Sushil Jun 27 '15 at 11:06
  • (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. – Asaf Karagila Jun 27 '15 at 11:26
  • But proof given here: http://math.stackexchange.com/q/1340333/168520 I don't see were we using CC – Sushil Jun 27 '15 at 11:42
  • Not every normed space is a Banach space. I never claimed that there is a Banach space of countable dimension. – Asaf Karagila Jun 27 '15 at 11:46
  • Is it not true: 'No infinite-dimensional normed linear space with a countable Hamel basis can be complete' – Sushil Jun 27 '15 at 12:03
  • Yes, it is true. How does that contradict what I said? Is every normed space complete? – Asaf Karagila Jun 27 '15 at 12:05
  • in your answer you said: ℓ1(D) is a Banach space which has a Hamel basis which is also a Schauder basis, – Sushil Jun 27 '15 at 12:06
  • And I said to that, read the paper. – Asaf Karagila Jun 27 '15 at 12:07