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Let $S$ be an uncountable set. Consider the subspace $\ell_\infty(\kappa, S)$ of $\ell_\infty(S)$ formed by all functions with support of cardinality at most $\kappa$ (here $\kappa<|S|$). Certainly, $\ell_\infty\subset \ell_\infty(\kappa, S)$, whence the density character of $\ell_\infty(\kappa, S)$ is at least $2^\omega$. Can we specify it? Is it for example $2^\omega \cdot \kappa$?

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up vote 6 down vote accepted

I understand you to mean that $\ell_\infty(S)$ is the collection of bounded functions from $S\to\mathbb{R}$, under the sup norm, and $\ell_\infty(\kappa,S)$ consists of those functions which take value $0$ outside a set of size at most $\kappa$, a fixed infinite cardinal less than $|S|$. The density character is the size of the smallest dense subset of this space under the sup norm metric.

I claim that the density character is exactly $|S|^\kappa$. It can be no larger than this, because the number of possible supports is $|S|^\kappa$, and for each support, there are $|\mathbb{R}|^\kappa=2^\kappa$ many functions having that support, leading to $|S|^\kappa\cdot 2^\kappa=|S|^\kappa$ many functions altogether in $\ell_\infty(\kappa,S)$. Conversely, the density character is at least this large. To see this, consider the set of all functions $f:S\to \{0,1\}$ with support of size $\kappa$. There are precisely $|S|^\kappa$ many such functions, and any two of these functions have distance $1$. So any dense set must have elements close to each of these functions, and so must have size at least $|S|^\kappa$.

Regarding the $2^\omega\cdot\kappa$ proposal, note that $|S|^\kappa$ is at least $2^\kappa$, which is strictly larger than $2^\omega\cdot\kappa$ when $\kappa\geq 2^\omega$.

(Lastly, if I may, allow me to suggest a small modification to your notation. It might be better to consider $\kappa\leq|S|$ and define that $\ell_\infty(\kappa,S)$, or perhaps denote it $\ell_\infty({\lt}\kappa,S)$, to be the collection of functions in $\ell_\infty(S)$ with support of size strictly less than $\kappa$, rather than at less-than-or-equal-to $\kappa$. This way of doing it includes your case, by replacing $\kappa$ with $\kappa^+$, but also allows one to consider the case of limit cardinals, such as the functions with support less than $\aleph_\omega$, which is not possible to express in your notation. A similar argument to the above shows that $\ell_\infty({\lt}\delta,S)$ has density character $|S|^{\lt\delta}$)

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In the third paragraph of your answer, you don't really mean to say that $2^\kappa$ is always strictly larger than $2^\omega\cdot\kappa$. Suppose, for example, that $2^\omega=2^{\omega_1}=\omega_2$ and $\kappa=\omega_1$. – Andreas Blass Apr 25 '12 at 13:43
Oops, you're right. I have now corrected it. – Joel David Hamkins Apr 25 '12 at 13:58
A perfectly good answer. Thank you. – Jan Veselý Apr 25 '12 at 15:25

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