About the density character and the dimension of a Banach space

Question

The density character of a Banach space is the least size of a dense subset. By dimension I mean algebraic dimension.

I have three inquiries related to the title.

1. I would like to know if there is any relation between the dimension and the density character of a Banach space. In particular, if one parameter affects (or puts restrictions) in any way (on) the other.

2. Does the answer to inquiry 1. above change if we move to wider classes of objects, such as metric or even general topological spaces?

3. Given an infinite cardinal $\kappa$, does there always exist a Banach space of density character exactly $\kappa$?

Regarding this final inquiry, I would also like to know whether, given an infinite cardinal $\kappa$, there always exists a Banach space of cardinality exactly $\kappa$.

Answer 1

Partial answer for the first question. The main result of Toruńczyk's paper asserts that:

Every Fréchet space is homeomorphic to some Hilbert space.

Moreover, every Hilbert space has an orthonormal basis and hence is isometric with $l^2(X)$ for some set $X$. So the question reduces to calculation of the cardinality of $l^2(X)$ for a set $X$. For $X$ of large cadinality it seems to be $\mathrm{card}(X)^{\aleph_0}$ as it is indicated by OP's comment below this answer.

Answer for the third question assuming that the density character means the smallest cardinality of a dense subset.

Let $X = \mathbb{R}^{\oplus \kappa}$ be a direct sum of $\kappa$ copies of $\mathbb{R}$ equipped with norm

$$||(\alpha_k)_{k\in \kappa}|| = \sum_{k\in \kappa}|\alpha_k|$$

(the sum is finite, since almost all $\alpha_k=0$). Clearly $X$ is not complete. Let $\mathcal{X}$ be its completion with respect to $||-||$. Now since $\kappa \geq \aleph_0$, we derive that

$$\mathbb{Q}^{\oplus \kappa}\subseteq \mathcal{X}$$

is dense. Indeed, it is dense in $X$ and $X$ is dense in its completion $\mathcal{X}$. Therefore, $$d(\mathcal{X})\leq \kappa$$ On the other hand if $d(\mathcal{X}) < \kappa$, then there exists a topological base of $\mathcal{X}$ consisting of $<\kappa$ open balls. So there exists a topological base of $X$ consisting of $<\kappa$ open balls and thus $X$ has a dense subset of cardinality $<\kappa$. This is impossible.

Remark.

The result of Toruńczyk also implies that

Two non-locally compact Fréchet spaces are homeomorphic if and only if they have the same density.

Indeed, suppose that non-locally compact Fréchet spaces $Y_1$ and $Y_2$ have the same density. By the main result of Toruńczyk for each $i=1,2$ there exists a set $X_i$ such that $Y_i$ is homeomorphic to $l^2(X_i)$. Now $l^2(X_i)$ is non-locally compact and hence has density equal to $\mathrm{card}(X_i)$. In addition $l^2(X_1)$ have the same density as $l^2(X_2)$. Thus $\mathrm{card}(X_1)=\mathrm{card}(X_2)$ and therefore, $l^2(X_1)$ is isometric to $l^2(X_2)$. This implies that $Y_1$ and $Y_2$ are homeomorphic.

Answer 2

If $\kappa$ is an infinite cardinal, $C(\alpha D(\kappa))$, all real-valued continuous functions on the one-point compactification of a discrete space of size $\kappa$, in the supremum norm, is a Banach space of density (character) equal to $\kappa$. In metric spaces we can always just take $D(\kappa)$ itself as an example (for metric spaces weight equals density).