bio | website | jvanname.myweb.usf.edu |
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location | Gotham, NY | |
age | 25 | |
visits | member for | 3 years, 3 months |
seen | 12 hours ago | |
stats | profile views | 3,417 |
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
added 1663 characters in body |
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 $\mathcal P(\omega)/ \mathrm{fin}$
added 164 characters in body |
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 |
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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?
edited body |
May 21 |
accepted | When is a formula preserved under taking factors in a reduced product or the stalk in a Boolean product? |