bio | website | jvanname.myweb.usf.edu |
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location | Nowhere, KS | |
age | 25 | |
visits | member for | 2 years, 6 months |
seen | 17 hours ago | |
stats | profile views | 2,457 |
I am interested in to varying degrees ordered sets, general topology, point-free topology, set-theory, universal algebra, and model theory. Most of my mathematics research involves dualities that are similar to Stone duality.
Sep 6 |
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Morphisms associated to measured spaces
One should obtain an equivalence of categories if we restrict this category to just the $\sigma$-algebras of the form $(2^{X},\mathcal{M})$ where $2^{X}$ is the Cantor cube and $\mathcal{M}$ is the Baire $\sigma$-algebra on $2^{X}$. This should follow from the fact that $\mathcal{M}$ is freely generated as a $\sigma$-complete Boolean algebra by all half cubes $\{f\in 2^{X}|f(x)=1\}$ where $x$ ranges over all elements in $x$. |
Sep 6 |
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Morphisms associated to measured spaces
The morphisms between measured spaces $(X,\mathcal{M},\mathcal{I}),(Y,\mathcal{N},\mathcal{J})$ where $(X,\mathcal{M})$ is a $\sigma$-algebra and $\mathcal{I}$ is a $\sigma$-complete ideal should simply be the functions $f:X\rightarrow Y$ where $f^{-1}[R]\in\mathcal{M}$ for each $R\in\mathcal{N}$ and where $f^{-1}[R]\in\mathcal{I}$ whenever $R\in\mathcal{I}$. |
Sep 6 |
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Morphisms associated to measured spaces
Of course, I do not believe that we obtain an equivalence of categories between the categories mentioned above unless one restricts the category of measured spaces to an appropriate full subcategory. |
Sep 6 |
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Morphisms associated to measured spaces
Also, in my new paper "Representations of algebras in varieties generated by infinite primal algebras", I consider a generalization of the category of filters mentioned in Andreas Blass' paper. This category can be used to characterize up-to-equivalence the pro-completion of the category of sets and more general categories. |
Sep 6 |
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Morphisms associated to measured spaces
If I understand the question correctly, then the paper "Two closed categories of filters" by Andreas Blass matwbn.icm.edu.pl/ksiazki/fm/fm94/fm94115.pdf discusses the category of filters which is a special case of the category which you want. |
Sep 4 |
answered | Comparing ideals in posets |
Aug 28 |
awarded | Nice Answer |
Aug 28 |
revised |
What is the maximal number of distinct values of the product of n permuted ordinals
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Aug 28 |
answered | What is the maximal number of distinct values of the product of n permuted ordinals |
Aug 27 |
awarded | Nice Answer |
Aug 22 |
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Ostaszewski space's construction Lemma
You don't need to use one of the special metrization theorems since we are dealing with locally compact spaces. In particular, a locally compact space is metrizable if and only if it can be written as a disjoint union of second countable spaces (this little result follows easily from the fact that metric spaces are paracompact and every paracompact locally compact space is a disjoint union of $\sigma$-compact spaces). Also, it does not seem too difficult to show that this space is a disjoint union of second countable spaces using the fact that $X$ space is locally compact and metrizable. |
Aug 20 |
awarded | Good Answer |
Aug 6 |
awarded | set-theory |
Aug 5 |
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non-Borel set which intersects every compact in a Borel set
I cannot immediately think of a a specific example of a non-Borel subset of $\omega_{1}$. However, the fact that $\mathcal{M}$ is a proper $\sigma$-algebra follows from the fact that the club filter is not an ultrafilter since no non-principal $\sigma$-complete ultrafilters appear until we reach the first measurable cardinal. In fact, Solovay's theorem states that $\omega_{1}$ can be partitioned into $\aleph_{1}$ stationary sets (and each element of this partition is non-Borel). |
Aug 4 |
answered | non-Borel set which intersects every compact in a Borel set |
Jul 29 |
awarded | Pundit |
Jul 29 |
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Escape the zombie apocalypse
If the zombies can strategize, then regardless of how slow the zombies, walk the zombies can easily form a circle around you of a sufficiently large radius before you make it to the boundary of the circle so that whenever you cross from the inside of the circle to the outside of the circle, the zombies will be in a $d$ distance of you. All the zombies have to do is form a sequence of circles of radius say $2^{n}$ for all $n$ and the zombies are simply commanded to walk to the boundary of the nearest circle. |
Jul 28 |
answered | Locally compact Hausdorff space that is not normal |
Jul 27 |
revised |
The word problem of the free left distributive algebra on one generator
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Jul 27 |
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The word problem of the free left distributive algebra on one generator
Yes. These results can be proven in ZFC alone, and the Handbook of Set Theory outlines ZFC proofs of these results even though the proofs are not very set theoretical. |