bio | website | jvanname.myweb.usf.edu |
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location | Nowhere, KS | |
age | 25 | |
visits | member for | 2 years, 7 months |
seen | 1 hour ago | |
stats | profile views | 2,804 |
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.
The government needs to refrain from giving students college scholarships and funding institutions of higher education because institutions of higher education do not do the public any good.
Oct 20 |
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Canonical functions in set theory and their applications
Also, the handbook of set theory page 99 refers to canonical functions as well along with other places. |
Oct 20 |
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Canonical functions in set theory and their applications
Canonical functions can also be found in Jech Ch. 24 Lem 24.5. |
Oct 13 |
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the meaning of “Cauchy filter” for an arbitrary topological group
The topological groups where the left uniformities and the right uniformities coincide are known as SIN (Small Invariant Neighborhood) groups. These are precisely the groups $G$ such that the identity $e$ has a basis consisting of sets $U$ such that $xUx^{-1}=U$ for all $x\in G$. |
Oct 13 |
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Homeo-Fixed point property
I suppose the fixed point property could also refer to $C^{r}$-maps on $C^{r}$-manifolds. |
Oct 12 |
answered | Homeo-Fixed point property |
Oct 12 |
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Why do roots of polynomials tend to have absolute value close to 1?
@Francois Dorais. I added to my answer some remarks that do explain why there should be rotational symmetry. Unfortunately it is not very rigorous at this point since the denominator could be near zero close to the contour. But I would agree that there is definitely more than meets the eye (just see Terry Tao's comments on my answer). |
Oct 12 |
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Why do roots of polynomials tend to have absolute value close to 1?
@Terry Tao. I have added a similar argument for why there should be rotational symmetry as well. Of course, this informal argument still suffers from the pesky fact that the denominator could be near zero. It seems like one could use the Poisson-Jensen Formula (for nearly arbitrary domains instead of simply the unit circle) for the pizza slice shaped domains to deduce rotational symmetry using $\log(|p(z)|)$ instead of $p'(z)/p(z)$. Is the usage of the Poisson-Jensen Formula for pizza slices the best way to formally rigorously deduce rotational symmetry or should different domains be used? |
Oct 12 |
revised |
Why do roots of polynomials tend to have absolute value close to 1?
added 1100 characters in body |
Oct 5 |
awarded | Guru |
Oct 3 |
awarded | Good Answer |
Oct 3 |
awarded | Nice Answer |
Oct 3 |
revised |
Why do roots of polynomials tend to have absolute value close to 1?
added 5 characters in body |
Oct 3 |
revised |
Does there exist a supersmooth non-polynomial function?
added 204 characters in body |
Oct 3 |
answered | Why do roots of polynomials tend to have absolute value close to 1? |
Oct 3 |
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Does there exist a supersmooth non-polynomial function?
@Pietro Majer. I asked the further question you suggested here mathoverflow.net/q/182423/22277. |
Oct 3 |
asked | Is every supersmooth function a local polynomial? |
Oct 2 |
reviewed | Approve suggested edit on Does there exist a supersmooth non-polynomial function? |
Oct 2 |
asked | Does there exist a supersmooth non-polynomial function? |
Sep 30 |
awarded | Explainer |
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 $x$. |