# Existence of a Banach space of arbitrary cardinal number $\alpha\geq card( \Bbb R)$

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.

• 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. – Georges Elencwajg Jan 2 at 11:09
• 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.) – Not Mike Jan 2 at 11:20
• I just want to know if this problem is open or it solve previously? – Ali Bayati Jan 2 at 11:23
• If I had to guess, I'd guess this was solved close to a 100 years ago. – Asaf Karagila Jan 2 at 11:24
• Are there any references in this area? Why we have $card( l^2(\sigma))=card (\sigma)$ when $card( \sigma) \geq c?$ – Ali Bayati Jan 2 at 11:27

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.