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Let $\alpha$ be a cardinal number with $\alpha\geq c:= \operatorname{card}(\Bbb R).$ Is there a Banach space $X$ which satisfies $\operatorname{card}(X)= \alpha?$ With many thanks for your answers.

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    $\begingroup$ Two people downvoted this well formulated and polite question: that was a really stupid and mean move, especially after the system's warning "Be nice" to this new contributor. $\endgroup$ – Georges Elencwajg Jan 2 at 11:09
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    $\begingroup$ Yes. Consider the space $X$ of maps $\sigma:\alpha \rightarrow \mathbb{ R}$ with countable support, which are square summable. You can equip $X$ with a sufficiently complete norm as follows $\Vert \sigma \Vert = \Sigma \{ \sigma(\gamma)^2: \gamma \in supp(\sigma)\}$. ( $X$ is a linear space when equiped the pointwise addition and scalar mult.) $\endgroup$ – Not Mike Jan 2 at 11:20
  • $\begingroup$ I just want to know if this problem is open or it solve previously? $\endgroup$ – Ali Bayati Jan 2 at 11:23
  • $\begingroup$ If I had to guess, I'd guess this was solved close to a 100 years ago. $\endgroup$ – Asaf Karagila Jan 2 at 11:24
  • $\begingroup$ Are there any references in this area? Why we have $card( l^2(\sigma))=card (\sigma)$ when $card( \sigma) \geq c?$ $\endgroup$ – Ali Bayati Jan 2 at 11:27
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The answer is no.

Suppose that $\alpha=\beth_\omega$, where $\beth_0=\aleph_0$ and $\beth_{\alpha+1}=2^{\beth_\alpha}$, with supremum at limits. Then $\alpha>\frak c$.

If $X$ is a Banach space of cardinality $\alpha$, its dimension over $\Bbb R$ is $\alpha$ as well, fix a basis of size $\alpha$ and let $X_n$ be the closed span of the first $\beth_n$ vectors in the basis. Since taking closure uses only countable sequences, and $$\beth_n^{\aleph_0}\leq\beth_n^{\beth_n}=2^{\beth_n}=\beth_{n+1}$$ it follows that $X_n$ is a closed subspace of size at most $\beth_{n+1}$. In particular, it has empty interior, since any open ball must have size $|X|$. But it is also easy to see that $\bigcup X_n=X$, which contradicts the Baire Category Theorem.

This can be generalized to all $\alpha$ such that $\alpha^{\aleph_0}>\alpha$. And indeed that is the only limitation. If $|S|=\alpha$ such that $\alpha^{\aleph_0}=\alpha$, then $\ell^\infty(S)$ has size $\alpha$ and a natural Banach space structure.

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