The unprecedented success of the “intersection” operator - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T18:15:22Z http://mathoverflow.net/feeds/question/33366 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/33366/the-unprecedented-success-of-the-intersection-operator The unprecedented success of the “intersection” operator To be cont'd 2010-07-26T07:48:32Z 2010-08-06T16:40:05Z <p>You might think that the title is an overstatement of a well-known fact but it is the best title I can come up with for the wonders the intersection operator does in some fields of math. </p> <p>Recently,(on summer vacation) I was studying one subject after another and after changing about three subjects, I began to notice that in all these the set-theoretic intersection operator always carried over some property of the parent sets to the one obtained after the intersection.</p> <p>To summarize briefly: </p> <blockquote> <p>Let $A$ and $B$ be two sets, say with property P. Then, $A \cap B$ has property P.</p> </blockquote> <p>Evidences, some trivial:</p> <h2>Topology</h2> <ul> <li><p>The intersection of two open sets is open.</p></li> <li><p>The intersection of two closed sets is closed.</p></li> <li>The nonempty intersection of two subspaces of a metric space is a metric space.</li> </ul> <p>......and so forth.</p> <h2>Algebra</h2> <ul> <li><p>The intersection of two subspaces of a vector space is a vector.</p></li> <li><p>The intersection of two subgroups of a group is a group(w.r.t the same binary operation and clearly the intersection is between the underlying sets).</p></li> <li>The intersection of two sub-fields of a field is a field.</li> </ul> <p>......the list continues.</p> <p>The third subject was Graph Theory, but I haven't yet come across the notion of intersection.</p> <p>Now I would like to ask <s>whether this trend always holds or</s> whether there is some underlying principle each discipline abides by when using the notion of intersection. <s>Is there any property deviating from this trend?</s> What are the reasons for the ubiquity of the quoted property? </p> http://mathoverflow.net/questions/33366/the-unprecedented-success-of-the-intersection-operator/33368#33368 Answer by Dylan Wilson for The unprecedented success of the “intersection” operator Dylan Wilson 2010-07-26T08:00:35Z 2010-07-26T08:00:35Z <p>So it's pretty clear, I think, that this trend doesn't always hold. For example, the intersection of two simply connected subspaces of a space does not have to be simply connected. (For example, the intersection of hemispheres on the sphere is a circle). </p> <p>I think that the reason it holds in all the cases you mention except the case of open sets, is that the property is somehow a "closure" property. Closed sets are closed under limits (or nets, more correctly), subobjects of algebraic objects are closed under some operations, etc. The pervasiveness you see is due to the fact that "closure" type properties behave well under intersection. The open set example, is, I think, a red herring. There, notice that only finite intersections maintain the property in question, and that this behavior with respect to intersections is part of the definition... Probably to avoid having too few open sets. </p> <p>Anyway, I'm not sure what else there is to say about this question- it's rather vague, so I don't even know if this is the kind of answer you wanted.</p> http://mathoverflow.net/questions/33366/the-unprecedented-success-of-the-intersection-operator/33373#33373 Answer by T. for The unprecedented success of the “intersection” operator T. 2010-07-26T08:42:52Z 2010-07-26T09:08:46Z <p>When property P is universal ($\forall ...$) it is likely to correspond to closed sets, and thus be preserved under intersection. Examples: axioms of a group, ring, field, directed graph; having symmetry under a given group. </p> <p>However, if P is existential ($\exists ...$) it corresponds to open sets and is more likely to be preserved by unions (or products), not intersections. Examples: being algebraically closed, having at least 53 elements. (Well, algebraic closure is $\forall \exists$ so of course it is even more complicated. But falling out of the pure $\forall$ class it fails the intersection property.)</p> <p>The first situation is possibly more common because we want structures to satisfy some, well, structural properties. Properties expressed by equations usually correspond to closed sets.</p> <p>To some extent this is formalized in Birkhoff's theorem On equational presentations. Any book on Universal Algebra will discuss it.</p> <p>Also, the sample of concepts is biased, because definitions that become standard are often selected for their useful formal properties. Concepts not having stability under intersection (or union, or inheritance by sub- or super-structures) are less likely to be used.</p> http://mathoverflow.net/questions/33366/the-unprecedented-success-of-the-intersection-operator/33380#33380 Answer by Joel David Hamkins for The unprecedented success of the “intersection” operator Joel David Hamkins 2010-07-26T10:43:28Z 2010-07-26T10:43:28Z <p>There is a general sense in which any property that is closed under arbirtrary intersection is exactly a closure property.</p> <p>To explain what I mean, suppose that $X$ has property $P$ and that the collection of subsets of $X$ with property $P$ is closed under arbitrary intersection. Then I claim that there is a function $cl$, a <a href="http://en.wikipedia.org/wiki/Closure_operator" rel="nofollow">closure operator</a>, defined on subsets of $X$ such that $A\subset cl(A)=cl(cl(A))$ and $A\subset B\to cl(A)\subset cl(B)$, for which the sets with property $P$ are exactly the sets $A$ that are closed with respect to $cl$, meaning that $A=cl(A)$. </p> <p>To see this, simply let $cl(A)$ be the intersection of all $B$ with property $P$ such that $A\subset B$. </p> <p>Apart from this, there many examples of natural properties that are not closed under intersection. </p> <ul> <li>The intersection of two groups is not necessarily a group. (e.g. when they are not subgroups of a larger group.) </li> <li>Same for almost any other type of algebraic structure. </li> <li>The intersection of two nonempty sets may not be nonempty.</li> <li>The intersection of two ultrafilters on a set is not necessarily an ultrafilter.</li> <li>The intersection of two maximal ideals in a ring is not necessarily a maximal ideal.</li> <li>The intersection of two unbounded subsets of the plane may not be unbounded. </li> <li>etc.</li> </ul> http://mathoverflow.net/questions/33366/the-unprecedented-success-of-the-intersection-operator/33383#33383 Answer by yatima2975 for The unprecedented success of the “intersection” operator yatima2975 2010-07-26T10:56:06Z 2010-07-26T15:52:32Z <p>Mulling this over, I thought: "Compared to what?". So another question is "Why are more properties closed under intersection than under union"? While that may be arguable, there's something to be said for looking at it (e.g. a union of two non-trivial subgroups of the same subgroup is not a subgroup).</p> <p>My intuition is that it has to do with the fact that, constructively, $(P \rightarrow X \wedge Y) \Leftrightarrow (P \rightarrow X) \wedge (P \rightarrow Y)$ but $(P \rightarrow X \vee Y) \nLeftrightarrow (P \rightarrow X) \vee (P \rightarrow Y)$. Here $P$ restricts the properties $X$ and $Y$ to an interesting 'sub-universe'. </p> <p>I'm sure someone with more knowledge of lattice or even category theory can restate this more succinctly.</p> http://mathoverflow.net/questions/33366/the-unprecedented-success-of-the-intersection-operator/33406#33406 Answer by alephomega for The unprecedented success of the “intersection” operator alephomega 2010-07-26T15:17:14Z 2010-07-26T15:17:14Z <p>You might want to check the definition of a filter: <a href="http://en.wikipedia.org/wiki/Filter_%28mathematics%29" rel="nofollow">http://en.wikipedia.org/wiki/Filter_%28mathematics%29</a></p> <p>I think the definition of a filter exemplifies the "intersection property" you talk about. </p>