Density character of $\ell_\infty(\kappa, S)$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T12:33:59Z http://mathoverflow.net/feeds/question/95135 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/95135/density-character-of-ell-infty-kappa-s Density character of $\ell_\infty(\kappa, S)$ Jan Veselý 2012-04-25T09:15:41Z 2012-04-25T13:57:18Z <p>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 $\kappa&lt;|S|$). 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 $2^\omega \cdot \kappa$?</p> http://mathoverflow.net/questions/95135/density-character-of-ell-infty-kappa-s/95155#95155 Answer by Joel David Hamkins for Density character of $\ell_\infty(\kappa, S)$ Joel David Hamkins 2012-04-25T12:33:27Z 2012-04-25T13:57:18Z <p>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.</p> <p>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 $|S|^\kappa$.</p> <p>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 $\kappa\geq 2^\omega$.</p> <p>(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 <em>strictly less than</em> $\kappa$, 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 $|S|^{\lt\delta}$)</p>