bio  website  jvanname.myweb.usf.edu 

location  Nowhere, Antarctica.  
age  24  
visits  member for  2 years 
seen  8 hours ago  
stats  profile views  2,104 
I do mathematics.
1d

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 
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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?
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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?
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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
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Mar 14 
answered  measure zero in R but not in R^2 
Mar 13 
answered  Rational points in the Alexandroff line 
Mar 9 
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Connectedness properties of groups of homeomorphisms
@Ludolila. Have you read about the mapping class group? Essentially the mapping class group of a space is the group of autohomeomorphisms modulo the subgroup of autohomeomorphisms isotopic to the identity. In particular, the mapping class group is the collection of all path components in $Aut(X)$. 
Mar 8 
answered  Characterizing Inf and Sup sets 