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bio website jvanname.myweb.usf.edu
location Nowhere, Antarctica.
age 24
visits member for 2 years
seen 8 hours ago

I do mathematics.


1d
answered Conjugation Quandles and… “Quandle-Groups”? From quandles to Groups
Apr
9
comment Conjugation Quandles and… “Quandle-Groups”? From quandles to Groups
A good reference for LD-monoids that give monoid conjugations is the book Braids and Self-Distributivity 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… “Quandle-Groups”? From quandles to Groups
The notion of an LD-monoid (LD stands for left-distributive) 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 LD-monoid, 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 non-c.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
Mar
13
answered Rational points in the Alexandroff line
Mar
9
comment 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