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

Question

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.

Answer 1

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.