bio  website  jvanname.myweb.usf.edu 

location  Nowhere, Antarctica.  
age  24  
visits  member for  2 years, 1 month 
seen  14 mins ago  
stats  profile views  2,112 
I do mathematics.
9h

comment 
Which complete lattices arise as images of the Galois connections induced by binary relations?
For finite lattices, the collection of all join (meet)irreducible sets is the smallest join (meet)dense subset, so that will be a canonical choice. More generally, if a complete lattice has no infinite chains, then the join(meet)irreducible sets forms the smallest join(meet)dense subset. However, I can only think of canonical choices of joindense subsets for certain specialized classes of complete lattices. I therefore cannot think of a canonical joindense subset in general. 
10h

revised 
Which complete lattices arise as images of the Galois connections induced by binary relations?
added 1568 characters in body 
10h

answered  Which complete lattices arise as images of the Galois connections induced by binary relations? 
Apr 15 
answered  Conjugation Quandles and… “QuandleGroups”? From quandles to Groups 
Apr 9 
comment 
Conjugation Quandles and… “QuandleGroups”? From quandles to Groups
A good reference for LDmonoids that give monoid conjugations is the book Braids and SelfDistributivity by Patrick Dehornoy. I will probably give a full blown answer to this question tomorrow when I have more time. 
Apr 9 
comment 
Conjugation Quandles and… “QuandleGroups”? From quandles to Groups
The notion of an LDmonoid (LD stands for leftdistributive) is an algebra $(M,\cdot,1,\wedge)$ where $(M,\cdot,1)$ is a monoid and $\wedge$ is an operation that acts like a conjugation on the monoid $(M,\cdot,1)$. In fact, if $(M,\cdot,1)$ is a group and $(M,\cdot,1,\wedge)$ is a LDmonoid, then the operation $\wedge$ is precisely conjugation. However, the operation $\wedge$ generally does not give a quandle operation as in the case of Laver tables. The operation $\wedge$ satisfies the $x\wedge(y\wedge z)=(x\wedge y)\wedge(x\wedge z)$, but generally not the other quandle identities. 
Apr 1 
comment 
Is every Heyting algebra the Lindenbaum algebra of an intuitionistic first order theory?
I explained why quantifier elimination works in this theory. 
Apr 1 
revised 
Is every Heyting algebra the Lindenbaum algebra of an intuitionistic first order theory?
added 131 characters in body 
Apr 1 
answered  Is every Heyting algebra the Lindenbaum algebra of an intuitionistic first order theory? 
Mar 29 
comment 
Does there exist a supercompactness theorem?
Even though Mohammad Golshani has given an affirmative answer that I accepted, I conjecture that there are other characterizations of supercompact cardinals that involve some sort of compactness. In particular, I conjecture that there is some purely combinatorial "compactness" theorem that characterizes supercompact cardinals. 
Mar 29 
accepted  Does there exist a supercompactness theorem? 
Mar 28 
comment 
extending $\sigma$complete boolean homomorphism
For each nonc.c.c algebra $A$, we can set $C$ to be any complete c.c.c. Boolean algebra and the same result with the same argument will work except that instead of having a $\sigma$complete ultrafilter on $\aleph_{1}$, we would have a $\sigma$complete $\sigma$saturated ideal on $\aleph_{1}$ which is still impossible. In other words, the image $C$ does not have to be trivial. 
Mar 28 
asked  Does there exist a supercompactness theorem? 
Mar 28 
answered  extending $\sigma$complete boolean homomorphism 
Mar 26 
answered  Structures that turn out to exhibit a symmetry even though their definition doesn't 
Mar 24 
revised 
Is the countable intersection of residual sets in [0,1] with Hausdorff dimension 1 of full Hausdorff dimension?
added 342 characters in body 
Mar 24 
answered  Is the countable intersection of residual sets in [0,1] with Hausdorff dimension 1 of full Hausdorff dimension? 
Mar 20 
awarded  Yearling 
Mar 14 
revised 
measure zero in R but not in R^2
added 122 characters in body 
Mar 14 
answered  measure zero in R but not in R^2 