bio | website | jvanname.myweb.usf.edu |
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location | Gotham, NY | |
age | 25 | |
visits | member for | 3 years, 4 months |
seen | yesterday | |
stats | profile views | 3,492 |
I am interested in to varying degrees ordered sets, general topology, point-free topology, Boolean algebras, set-theory, universal algebra, and model theory. Most of my previous mathematics research involves dualities that are similar to Stone duality. I am currently interested in using forcing extensions to study objects such as Boolean algebras and frames in the ground model. Furthermore, I am also interested in self-distributive algebras with operations of arbitrary finite arity and their relations with algebras of rank-into-rank elemetary embeddings and other mathematical structures.
Jul 28 |
answered | Baire Category Theorem for complete uniform spaces |
Jul 28 |
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Are there discontinuities in the large cardinal hierarchy?
Andres Caicedo. Perhaps you should give an answer elaborating on your comment. Since your comment has already been highly upvoted, I am sure that the people on MO with an affinity towards set theory would appreciate your answer. |
Jul 28 |
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Are there discontinuities in the large cardinal hierarchy?
Andreas Blass. It seems like I completely overlooked the compactness theorem when posting this question. I was afraid of something like that. |
Jul 23 |
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Minimum regular open set containing a given set in a T0 Alexandrov topological space
I should mention that $\overline{A},A^{\circ}$ are definable in $(X,A,\leq)$, so the least regular open set containing $A$, namely $(\overline{A})^{\circ}$ is definable in $(X,A,\leq)$, so if $A$ is a definable subset of $X$, then so is $(\overline{A})^{\circ}$. I am unsure if anything more substantial can be said about the definability of the minimum regular open set containing $A$ though with the way the question is asked so far. |
Jul 23 |
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Minimum regular open set containing a given set in a T0 Alexandrov topological space
Ken Y. I see. You are talking about the minimum regular open set containing a set while I was thinking of simply the minimum non-empty regular open set. I apologize for causing confusion. |
Jul 23 |
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Which combinations of normality, separability, and paracompactness do complex manifolds possess?
If I understand this answer correctly, this would give a Riemann surface, but every Riemann surface is paracompact, so I do not think this construction would work. |
Jul 23 |
asked | Are there discontinuities in the large cardinal hierarchy? |
Jul 20 |
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Minimum regular open set containing a given set in a T0 Alexandrov topological space
Ken Y. Thanks for clarifying. I should have read your question more carefully since you mentioned "the" minimum regular open instead of "a". However, this means that the only regular open set in $X$ is $X$ itself. In other words, the separative quotient of $X$ is must be a one element set. This is only possible if every two elements are comparable. i.e. if $X$ is a downwards directed set. |
Jul 20 |
asked | Which combinations of normality, separability, and paracompactness do complex manifolds possess? |
Jul 20 |
answered | “Abnormal” manifold |
Jul 20 |
answered | Is the long line paracompact? |
Jul 20 |
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Minimum regular open set containing a given set in a T0 Alexandrov topological space
I think you mean to say minimal regular open sets rather than minimum regular open sets. The minimal regular open sets are the minimal elements in the separative quotient of your partially ordered set (if they exist of course). |
Jul 18 |
answered | Is an open subset of a compact subset of a Hausdorff locally convex TVS paracompact? |
Jul 15 |
answered | Posets isomorphic to their endomorphism poset |
Jul 15 |
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Posets isomorphic to their endomorphism poset
I must also mention that if $\mu$ is the strong limit cardinal with $\mu\leq|P|\leq 2^{\mu}$, then $\mu$ is not a weakly compact cardinal since if $\mu\rightarrow(\mu)^{2}_{3}$, then $|P|$ would have a well-ordered or anti-well ordered chain or an antichain of cardinality $\mu$ which is a contradiction. |
Jul 13 |
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Why is the set-theoretic principle $\diamondsuit$ called $\diamondsuit$?
Avshalom. My source is Nicky Oppenheimer, the director of DeBeers and richest man in South Africa who made all of his money from diamonds, who said "Diamonds are intrinsically worthless, except for the deep psychological need they fill." The story I gave about diamonds is well known. |
Jul 13 |
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Why is the set-theoretic principle $\diamondsuit$ called $\diamondsuit$?
Asaf Karagila. The phrase "A diamond is forever" is simply a very successful slogan from the diamond industry to make people buy diamonds. Also, diamonds only appear rare because a certain organization has a near monopoly on diamonds and diamond mines and they want us to think they are rare. In reality, they have piles and piles of diamonds. |
Jul 13 |
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Lattice homomorphism from ${\cal Id}(L)$ onto $L$
In fact, the Stone-Cech compactification of a completely regular frame $L$ is a subframe of $Id(L)$: If $L$ is a completely regular frame, then let $R(L)⊆Id(L)$ be the subframe consisting of all ideals $I\subseteq L$ such that if $x∈I$ then there is some $y∈I$ with $x≪y$. Then the mapping $f_{|R(L)}:R(L)\rightarrow L$ is the point-free analogue to the Stone-Cech compactification mapping. |
Jul 13 |
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Lattice homomorphism from ${\cal Id}(L)$ onto $L$
I must add that for distributive lattices (and clearly only for distributive lattices), the condition that finite meets distribute over joins of ideals if and only if finite meets distribute over arbitrary joins. i.e. if $L$ is a complete lattice, then $f:Id(L)→L$ is a lattice homomorphism if and only if $L$ satisfies $x\wedge\bigvee_{i\in I}y_{i}=\bigvee_{i\in I}(x\wedge y_{i})$, and this infinite distributivity law states that $L$ is a frame. If $L$ is a frame, then $f$ is a frame homomorphism. Furthermore, $Id(L)$ is a compact frame and the right adjoint of $f$ is a dense localic embedding. |
Jul 13 |
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Lattice homomorphism from ${\cal Id}(L)$ onto $L$
Let me correct and undelete the comments then. |