Timeline for The unprecedented success of the “intersection” operator
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
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| Jan 4, 2014 at 18:10 | comment | added | fosco | Reading your answer was extremely useful; I was animated by a question which led me here looking for an abstraction which gives the "saturation" of a class with respect to a given property stable under intersection (which is of course a closure operation!). Nevertheless I'm left with a question now: "being infinite" is a property which allows you to define the smallest infinite set containing a given one, and yet it's absolutely non-stable under intersection. How comes this? | |
| Jul 27, 2010 at 3:47 | comment | added | Joel David Hamkins | Incidentally, when the property is preserved under arbitrary intersections and unions, then the following MO answer shows that it is exactly the same as being closed under pointwise images of a family of functions. mathoverflow.net/questions/11435/… | |
| Jul 27, 2010 at 3:33 | comment | added | T.. | Also, in all six of the examples of intersection-closed properties given in the question, the intersection was of substructures of a given structure. Under this interpretation it would be very interesting to see an $\forall$ property not closed under intersection. Being a field is an interesting example, since existence of inverses is a priori a $\forall \exists$ statement. However, direct sums of fields do have an equational presentation (by Birkhoff's theorem), and any subset of a field that is of this direct-sum form is a field, so the inverses axiom doesn't stop intersection-closure. | |
| Jul 27, 2010 at 3:27 | comment | added | Joel David Hamkins | I agree with that, although I would describe it differently. First order universal formulas used to define sets of points are of course closed under arbitrary intersection (or even just subsets). I was in contrast using a universal formula as a second order definition, to define a property of the relation appearing in it (linearity), and these are not necessarily closed under intersection. | |
| Jul 27, 2010 at 3:02 | comment | added | T.. | Joel, the $\forall$ property of being a linear order is closed in the usual sense, since it is expressed by equations on the $Z/2$-valued function on pairs that records the relation, such as $f(p,q) + f(q,p) = f(p,p)=1$. It satisfies the same intersection property as closed sets in topology: if $A$ and $B$ are substructures of $X$ having property $P$ (as defined in $X$ or inherited from it, e.g., both are closed sets in a given topology on $X$, or both are subsets of a linearly ordered set $X$) then so is $A \cap B$. Closed sets in different topologies need not have closed intersection. | |
| Jul 26, 2010 at 15:07 | comment | added | Joel David Hamkins | Here is an example of a $\forall$ property that is not closed under intersection: the intersection of two linear orders on a set is not necessarily a linear order. Rather, it is the partial order that the two linear orders have in common. But linearity is $\forall$-expressible, as $\forall p,q\, (p\leq q\vee q\leq p)$. | |
| Jul 26, 2010 at 11:39 | comment | added | Unknown | In Introduction to Topology by Bert Mendelson, around pp50's, it actually asks to prove the above notion of closure for closed sets. | |
| Jul 26, 2010 at 10:43 | history | answered | Joel David Hamkins | CC BY-SA 2.5 |