10,349 reputation
12444
bio website jvanname.myweb.usf.edu
location Gotham, NY
age 25
visits member for 3 years, 3 months
seen 12 hours ago

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 mathematics research involves dualities that are similar to Stone duality.


Jun
24
accepted How long can a cycle of antichains in a finite partial order be?
Jun
24
comment Example of a collection of metacompact spaces with non-metacompact box-product
See also mathoverflow.net/q/209661/22277.
Jun
24
revised Example of a collection of metacompact spaces with non-metacompact box-product
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Jun
24
answered Example of a collection of metacompact spaces with non-metacompact box-product
Jun
22
comment Which topological properties are preserved under taking box products?
An Alexandrov space usually refers to a space such that the intersection of arbitrarily many open sets is open. This is a weaker property than saying that every open set is clopen. In fact, the spaces where every open set is clopen are precisely the spaces whose $T_{0}$-reflection is discrete.
Jun
20
comment Which topological properties are preserved under taking box products?
Dominic van der Zypen. The mistake began when you said "we can assume that $N((x_{i})_{i\in I})=\prod_{i\in I}N_{i}(x_{i})$." Here the $N_{i}(x_{i})$ depends on the entire function $(x_{i})_{i\in I}$ instead of the individual point $x_{i}$. Even though the notion of a $D$-space is a covering property like paracompactness, the notion of a $D$-space is hardly preserved under any constructions and is difficult to work with.
Jun
19
comment Which topological properties are preserved under taking box products?
Dominic van der Zypen. That same paper that you referenced claims that not even finite products of $D$-spaces are $D$-spaces. In general, the notion of a $D$-space is not very well behaved.
Jun
19
asked Which topological properties are preserved under taking box products?
Jun
18
revised Compact open topology on $\mathrm{Homeo}(X)$
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Jun
18
comment $T_2$-space $X$ with $X\cong \text{Aut}(X)$
The most natural structure to put on sets of continuous functions from a non-locally compact space $X$ to a space $Y$ is not a topological structure but a convergence structure. Of course, if $X$ is a uniform space, then the uniformity of uniform continuity on $X^{Y}$ is a fairly natural uniformity to put on such function spaces, but this uniformity induces the compact open topology whenever $X$ is a compact space.
Jun
18
comment $T_2$-space $X$ with $X\cong \text{Aut}(X)$
The compact open topology on $\text{Aut}(X)$ only seems natural for compact spaces; the compact-open topology only behaves the way it is supposed to for locally compact spaces $X$. Furthermore, there are locally compact spaces where the compact-open topology on $\text{Aut}(X)$ is generally not a topological group since the operation $f\mapsto f^{-1}$ is generally not continuous mathoverflow.net/q/58690/22277. However, for compact spaces $X$, the space $\text{Aut}(X)$ however is a topological group.
Jun
3
revised Cohen algebra and $\mathcal P(\omega)/ \mathrm{fin}$
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Jun
3
revised Cohen algebra and $\mathcal P(\omega)/ \mathrm{fin}$
added 164 characters in body
Jun
3
answered Cohen algebra and $\mathcal P(\omega)/ \mathrm{fin}$
Jun
1
comment Set of ideals of the set of finite subsets of $\mathbb{N}$
If $X$ is a join-semilattice, then $\mathcal{I}(X)$ is an algebraic lattice, and if $L$ is an algebraic lattice, then the set $K(L)$ of compact elements in $L$ is a join-semilattice. If $L$ is an algebraic lattice, then $L\simeq\mathcal{I}(K(L))$, and if $X$ is a join-semilattice, then $X\simeq K(\mathcal{I}(X))$. The lattice $P(\mathbb{N})$ is an algebraic lattice and $K(P(\mathbb{N}))=F(\mathbb{N})$. Therefore, $\mathcal{I}(F(\mathbb{N}))=I(K(P(\mathbb{N})))\simeq P(\mathbb{N})$. See newton.case.edu/papers/algfca.pdf for a proof of this duality.
Jun
1
comment Set of ideals of the set of finite subsets of $\mathbb{N}$
Good answer. A more high-level way to see this isomorphism is through the duality between join-semilattices and algebraic lattices. Algebraic lattices are described in detail in the book A Course in Universal Algebra by Burris and Sankappanavar. Of course, the duality between join-semilattices and algebraic lattices is a part of a larger framework of dualities between ordered sets and complete lattices.
May
23
awarded  Nice Question
May
22
comment How long can a cycle of antichains in a finite partial order be?
domotorp. For a simple connected example better than $|X|+1$, take the tree $X=\{0,00,000,01,011,0111\}$ ordered by extension of strings, then the cycle in $\mathcal{A}_{X}$ containing $\emptyset$ has length 13.
May
21
revised How long can a cycle of antichains in a finite partial order be?
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May
21
accepted When is a formula preserved under taking factors in a reduced product or the stalk in a Boolean product?