About the density character and the dimension of a Banach space 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.


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*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.

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

*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 $\kappa$.
 A: 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.
A: 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). 
