Spaces of filters - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T04:46:07Z http://mathoverflow.net/feeds/question/61659 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/61659/spaces-of-filters Spaces of filters Abel Stolz 2011-04-14T07:03:15Z 2012-04-23T03:26:32Z <p>This question arose more from curiosity than from an actual problem. There are situations when you embed some space $X$ in a set of filters on $X$, which inherits properties of $X$ or has even better properties. The prominent examples are (and in fact the only ones I am aware of):</p> <p>(1) $X$ is a topological space and we consider the Stone-Cech compactification $\beta X$, which can be constructed from the set of all ultrafilters on $X$. We gain the obvious advantage of compactness.</p> <p>(2) $X$ is a uniform space, and the completion of $X$ can be constructed from a set of Cauchy filters on $X$.</p> <p>In some examples algebraic structures can be preserved. For example when $X$ is a a SIN topological group, the completion is a complete topological group.</p> <p>The question is:</p> <blockquote> <p>What other examples with an interesting application of spaces of filters like in the above examples are there?</p> </blockquote> <p>For a motivation consider the Stone-Cech compactification $\beta \mathbb{N}$ of the natural numbers, which has a natural structure of a (noncommutative) monoid with a multiplication, which is only left continuous. Nevertheless it can be used to sho interesting results, for example about IP-sets. (http://en.wikipedia.org/wiki/IP_set)</p> <p>(I hope this question is not too vague, in order to qualify as a real question.)</p> http://mathoverflow.net/questions/61659/spaces-of-filters/61670#61670 Answer by Juris Steprans for Spaces of filters Juris Steprans 2011-04-14T09:02:34Z 2011-04-14T09:02:34Z <p>The ultrafilters are, of course, measures on the natural numbers and can therefore be considered as elements of the dual space of $\ell^\infty$. A generalization of the semigroup construction on $\beta\mathbb{N}$ is the Arens product on the double dual of a Banach algebra. Many of the concepts that arise in the study of $\beta\mathbb{N}$ have analogues in this context.</p> http://mathoverflow.net/questions/61659/spaces-of-filters/61709#61709 Answer by Peter Krautzberger for Spaces of filters Peter Krautzberger 2011-04-14T14:57:47Z 2011-04-14T15:09:20Z <p>Some partial answers taken from Neil Hindman, Dona Strauss, "Algebra in the Stone-Čech compactification", chapter 21 (aptly named "Other Semigroup Compactifications"). </p> <p>The most general is probably Theorem 21.31: </p> <ul> <li>If $X$ is discrete, $Y$ compact, $g: \beta X \rightarrow Y$ continuous and onto, then $Y$ is isomorphic to a space of filters on $X$ (simply intersect the preimages of points in $Y$)</li> <li>In particular, every compactification of $X$ can be viewed as a space of filters.</li> </ul> <p>If you're interested in algebraic aspects, there's section 21.3 of the book.</p> <p>If $(X,\cdot)$ is a semigroup (not necessarily discrete, but completely regular), "nice" semigroup compactifications such as the AP and WAP compactifications, i.e., the (maximal) topological and semitopological semigroup compactifications respectively, are very interesting objects. They also have nice descriptions as filters.</p> http://mathoverflow.net/questions/61659/spaces-of-filters/61803#61803 Answer by Greg Graviton for Spaces of filters Greg Graviton 2011-04-15T10:40:27Z 2011-04-15T10:40:27Z <p>I would argue that, morally, any kind of topological completion can be described as a space of filters. In other words, interesting spaces of filters are equivalent to interesting completions and the question reduces to "What are other interesting examples of completions?". I don't know any good examples for the latter, though.</p> <p>Consider an ultrafilter $p$ and write $A \in p$ as "$p \in A$" for a moment. Then, the ultrafilter axioms read</p> <ol> <li>$p\not\in \emptyset$</li> <li>$p \in A \wedge A\subseteq B \implies p \in B$ </li> <li>$p \in A \wedge p\in B \implies p \in A\cap B$ </li> <li>$p \in A \vee p\in A^c$</li> </ol> <p>In other words, ultrafilters are just an axiomatization of the notion of "point".</p> <p>Now, completing a space means adding points. But since points can be described by ultrafilters, it is no surprise that a completion can be described as a collection of ultrafilters, or a quotient thereof.</p> <p>In any case, that's how I like to think about ultrafilters. Another point of view with the same effect would be the observation that in order to call a space $Y$ a completion of $X$, it should be compact, which immediately gives a surjective morphism</p> <p>$$\beta X \to Y$$</p> <p>allowing us to write $Y$ as a quotient of a space of ultrafilters.</p> http://mathoverflow.net/questions/61659/spaces-of-filters/69160#69160 Answer by Benjamin Steinberg for Spaces of filters Benjamin Steinberg 2011-06-30T02:46:32Z 2011-06-30T02:46:32Z <p>You can view the profinite completion of a group as the space of ultrafilters on the Boolean algebra of finite unions of cosets of finite index subgroups. A similar thing is true for semigroups. For example, points of the free profinite monoid are ultrafilters of regular languages.</p> http://mathoverflow.net/questions/61659/spaces-of-filters/94902#94902 Answer by Joseph Van Name for Spaces of filters Joseph Van Name 2012-04-23T03:26:32Z 2012-04-23T03:26:32Z <p>Let $X$ be a separating proximity space. Then the Smirnov compactification of $X$ can be defined in terms of filters. A filter $F$ on a proximity space $X$ is said to be a round filter if for each $R\in F$, there is an $S\in F$ with $S\preceq R$. An end is a round filter $F$ such that if $A\preceq B$, then $B\in F$ or $X\setminus A\in F$. It can be shown that the ends are precisely the maximal round filters. The Smirnov compactification of $X$ is simply the collection of all ends. See [1] for details on proximity spaces and the Smirnov compactification. In my research, I used the Smirnov compactification in finding the maximal ideal space of $L^{\infty}(\mu)$ for any measure $\mu$.</p> <p>One can generate other examples of spaces of ultrafilters by looking at zero-dimensional spaces. Recall that a space is zero-dimensional if it has a basis consisting of clopen sets. The advantage of looking at zero-dimensional spaces is that the ultrafilters are generally ultrafilters on Boolean algebras. If $X$ is a zero-dimensional space, then the Banaschewski compactification of $X$ is simply the collection of all ultrafilters on the Boolean algebra of clopen sets of $X$. A space $X$ is said to be strongly zero-dimensional if whenever $Z_{1},Z_{2}\subseteq X$ are disjoint zero sets, then there is a clopen set $C$ with $Z_{1}\subseteq C,Z_{2}\subseteq C^{c}$. For example, every zero-dimensional Lindelof space is strongly zero-dimensional. One can show that a space $X$ is strongly zero-dimensional if and only if $\beta X$ is zero-dimensional. It turns out that the Banaschewski compactification of zero-dimensional space $X$ is the Stone-Cech compactification if and only if $X$ is strongly zero-dimensional. </p> <ol> <li>Naimpally, S. A., and B. D. Warrack. Proximity Spaces. Cambridge [Eng] : University, 1970.</li> </ol>