User apollo - MathOverflow

Answer by Apollo for Must a linearly ordered, separable space be metrizable?

2011-03-01T00:01:43Z

No. Take $[0,1]\times{0,1}$ with the lexicographic order. This gives a counterexample --- it is separable (for example $\mathbb{Q}\times{1}$ is a countable dense set), yet it is not metrizable. One way to see this is to notice that the subspace $[0,1]\times{1}$ (homeomorphic to the Sorgenfrey line) is not second-countable, hence not metrizable. The counter-example can also be viewed as an example of an Alexandroff "double-point" construction, which is an example of the general construction of "(special) resolution" (which is a nice technique for generating counterexamples).

(Edited to incorporate comments --- original answer was incorrect, citing Sorgenfrey line as a counterexample.)

Answer by Apollo for Why semigroups could be important?

2011-02-18T19:40:23Z

(Commutative) semigroups and their analysis shows up in the theory of misère combinatorial games. The "misère quotient" semigroup construction gives a natural generalization of the normal-play Sprague-Grundy theory to misere play which allows for complete analysis of (many) such games. (See http://miseregames.org/ for various papers and presentations.)

Answer by Apollo for How can I write down polynomial relations that define when a polynomial is a square?

2011-02-11T20:26:31Z

http://en.wikipedia.org/wiki/Square-free_polynomial gives a method for finding a square-free factorization of a polynomial (over characteristic zero field), ie $f=a_1\cdot a_2^2\cdots a_n^n$ where each $a_i$ is a square-free polynomial. Then $f$ is a perfect square iff $a_{2i+1}=1$ for each $i$.

Answer by Apollo for probability question regarding brownian motion

2011-02-10T22:00:04Z

A straightforward approach is to simply integrate the density of $X_t$ at time $a$ (which will be normally distributed with mean $\mu$ and variance $\sigma^2 a$) against the probability of hitting 0 conditional on the value at time $a$ (which is also known in closed-form). This will give you a messy integral (with an exponential multiplied by a cumulative-normal) but it should be reducible to a (messy sum of) bivariate cumulative normal(s).

The value we want to compute is $$\int_0^\infty \mathbb{P}[X_\xi>0 \text{ for } a\leq\xi\leq T\ |\ X_a=z] e^{-z^2/2\sigma^2T}\frac{dz}{\sigma\sqrt{T}\sqrt{2\pi}}$$ where I'm integrating the density at time $a$ for positive values against the non-hitting time.

The next step is to observe that the probability $\mathbb{P}[X_\xi>0 \text{ for } a\leq\xi\leq T\ |\ X_a=z]$ is equal to the probability $\mathbb{P}[X_\xi>-z \text{ for } 0\leq\xi\leq T-a]$ but this probability is equal to a difference of (basically) cumulative normals (it's just a hitting time computation for a (scaled) Brownian motion with drift). Then plug that formula into the above integral.

A quick calculation (might be wrong, so beware) gives me $$\mathbb{P}[X_\xi>-z \text{ for } 0\leq\xi\leq T-a] = \Phi\left[\frac{-z+\alpha (T-a)}{\sigma\sqrt{T-a}}\right] - e^{2\alpha z/\sigma^2}\Phi\left[\frac{z+\alpha (T-a)}{\sigma\sqrt{T-a}}\right] $$ where $\Phi[z]=\int_{-\infty}^z e^{-\xi^2/2}\frac{d\xi}{\sqrt{2\pi}}$ is the standard cumulative normal distribution function. (This follows from application of Girsanov to a reflection argument, a well-known result.)

Answer by Apollo for The difference between a sequential space and a space with countable tightness

2011-02-02T23:20:37Z

Some examples to expand Andreas' answer might be of interest: (it's too much to fit in a comment so I'm adding it as an answer, though I think Andreas' response is great)

any metric space will be Frechet-Urysohn (choose $x_n$ in $A$ within $1/n$ of $p$ ); (more generally, any first-countable space is F-U: just choose $x_n$ in the intersection of $A$ and $U_n$ where $\{U_n\}_n$ is a countable base at the limit point);
a sequential but not Frechet-Urysohn space is given by taking $((\omega+1)\times\omega)\cup\{*\}$ where each copy of $\omega+1$ has the usual topology and a base for $*$ consists of sets $A_{m,n}=\{(m,n)|m>M,n>N_m\}$ for $M,N_m\in\omega$ (ie cofinitely many elements of cofinitely many fibers) - then $*$ is in the closure of $\omega\times\omega$ but is not the limit of any sequence of points in $\omega\times\omega$ ; however, it is the limit of the sequence $x_n=(\omega,n)$ and each $x_n$ is the limit of a sequence of points from $\omega\times\omega$ ;
a countably-tight but not sequential space could be given by taking $(\omega\times\omega)\cup\{*\}$ where all points $(m,n)$ are open and a base for $*$ consist of sets $A_{M,N}=\{(m,n)|m>M, n>N_m\}$ for $M,N_m\in\omega$ - this space is trivially countably tight (it's countable) but is not sequential: $*$ is not the limit of any sequence in $\omega\times\omega$ (since we can always exclude any putative sequence converging to $*$ );
finally a non-countably-tight space is given by $\omega_1+1$ with the usual topology: $\omega_1$ (as a point) is in the closure of $\omega_1$ (as a set) but any countable subset of $\omega_1$ has bounded (countable) closure.

Answer by Apollo for Is there an uncountable, non-discrete, Hausdorff Toronto space?

2011-02-02T22:03:20Z

From "Open Problems in Topology" the following facts are known (described there as "folklore")

any Hausdorff non-discrete Toronto space is scattered with countably many isolated points
hence such a space must have derived length $\omega_1$ and be hereditarily separable, thus must be an $S$-space
this gives a way to have a model where there are no non-discrete Hausdorff Toronto spaces of size $\aleph_1$: assume $2^{\aleph_0}\neq2^{\aleph_1}$ and note that hereditary separability implies that the space has only $2^{\aleph_0}$ autohomeomorphisms while any Toronto space of size $\lambda$ must have $2^\lambda$ autohomeomorphisms

I have no idea what has been proven since then. It is also mentioned that questions concerning Toronto spaces with larger cardinalities and with stronger separation axioms is still open and gives for example the question "Are all regular (or normal) Toronto spaces of size $\aleph_1$ discrete?"

Answer by Apollo for Relating Category Theory to Programming Language Theory

2009-11-05T16:55:25Z

The connections I have seen are in the area of denotational semantics and type theory. For example, the objects can be the types: Integer, List(Integer), Integer+List(List(Integer)), etc. Functions (in the language) are arrows between the objects. (So the successor function is an arrow from Integer to Integer.) Then categorical constructions translate into programming language constructs ("List" is an endo-functor, for example). The existence of recursive functions is guaranteed by certain categorical properties of the category, etc.

Answer by Apollo for Derivate Bessel Function with respect to order

2009-11-04T23:59:24Z

Abramowitz and Stegun give a couple of special cases but don't give a general result. Starting from some of the integral or series representations and differentiating you can get a corresponding integral or series for the derivative, but I would guess that it's unlikely to simplify to a "known" function in the general case. An example they give is (for the spherical Bessel function ):

[ d/d\nu j_\nu(x) ]_{\nu=0} = (\pi/(2x))(Ci(2x)\sin x - SI(2x)\cos x)

They also give examples evaluated at and similar results for the case of the "other" spherical bessel .

Answer by Apollo for Is the long line paracompact?

2009-11-04T17:54:04Z

The Handbook of Set-Theoretic Topology is the bible for this kind of question. It has a chapter on the long-line and a chapter on questions of paracompactness of manifolds.

Answer by Apollo for Characterizations of non-wellfounded models?

2009-11-03T16:02:49Z

You may want to look at Peter Aczel, Jon Barwise, Non-Well-founded Sets where they describe doing set-theory replacing the Axiom of Foundation by the Anti-Foundation Axiom. There, certain kinds of non-well-founded sets are postulated to exist and they give means to determine if two sets are the "same" set. (For example given A={A} and B={C}, C={B}, is A=B? (I think the answer is yes, shown by a bi-simulation argument.)

Answer by Apollo for Discharging assumptions

2009-11-03T12:37:56Z

As I understand it, to discharging a premise or assumption is the opposite of introducing it: you absorb it (for example) into the antecedent of an implication --- this means that it is no longer an assumption. A trivial example:

P 1. Assume P

P 2. From 1

P->P 3. Discharging 1

Thus I have concluded that P->P without any assumptions (iow |- P->P). If we didn't discharge the assumption, we would have P|-P

Comment by Apollo

2012-10-04T14:12:20Z

They're all countable metric spaces with no isolated points. See at.yorku.ca/p/a/c/a/25.htm

Comment by Apollo

2012-10-04T13:50:40Z

They're homeomorphic.

Comment by Apollo

2011-03-23T23:09:56Z

Solovay's model is a model of ZF where every set of reals is measurable. (Assuming the existence of an inaccessible cardinal.) Thus, ZF without choice does not allow you to build a non-measurable set.

Comment by Apollo

2011-03-10T23:42:50Z

Thanks. I'd be surprised if there was a faster method (based on the need (maybe) for very powerful large cardinals (at least more than PRA) to prove facts about the periodicity of the top row) to compute arbitrary entries...

Comment by Apollo

2011-03-09T22:52:55Z

Do you have a reference you could give for the algorithm? Thanks,

Comment by Apollo

2011-03-01T00:24:37Z

Actually, now that I think about it, yes it is separable: $[0,1]\times\{0,1\}$ ordered lexicographically.

Comment by Apollo

2011-03-01T00:19:25Z

The Sorgenfrey line is a subspace of a linear order with the order topology, but yes, it is not an order topology

Comment by Apollo

2011-02-23T15:56:05Z

I would recommend using Sobol' sequences as they tend to give better results in my experience. Second, using low-discrepancy sequences you do not want to use methods like Box-Muller to generate normals from the uniforms since, as pointed out by Tim, the points in the low-discrepancy sequences are not independent. Simply invert the cumulative normal density function (the Ziggurat method is an efficient method for this).

Comment by Apollo

2011-02-11T23:13:23Z

This question was also asked on math.stackexchange.com/questions/21601/… (seems a better fit there).

Comment by Apollo

2011-02-10T22:44:33Z

A useful fact for simplifying the resulting integral is that $\int_{-\infty}^\lam \phi(u)\Phi(\alpha+\beta u)du = \Psi[\lambda,\frac{\alpha}{\sqrt{1+\beta^2}};\frac{-\beta}{\sqrt{1+\beta^2}}]$ where $\Psi[x,y;\rho]$ is the bivariate normal cumulative distribution function. This is a straightforward bit of algebra (expand the inner cumulative normal as an integral and play with it until you get the standard bivariate).

Comment by Apollo

2011-02-10T22:34:31Z

We're integrating along the distribution of $X$ at time $a$ and multiplying by the conditional probability (given our location at time $a$) that we make it further to time $T$ without hitting zero. If we're already below zero then this probability is $0$ (so the lower bound of the integral starts at $0$) if we're above zero then we just need to keep the minimum of the remaining path above zero.

Comment by Apollo

2011-02-10T22:18:03Z

Added a little more detail.

Comment by Apollo

2011-02-04T19:56:31Z

From the 7th status report: in G. Gruenhage, J.T. Moore, Countable Toronto spaces, Fund. Math. 163 (2) (2000) 143–162, they show that there is an $\omega$ -Toronto space, where an $\alpha$ -Toronto space is a scattered space of Cantor-Bendixson rank $\alpha$ which is homeomorphic to each of its subspaces of rank $\alpha$ . They constructed countable $\alpha$ -Toronto spaces for each $\alpha<\omega$ . Gruenhage also constructed consistent examples of countable $\alpha$ -Toronto spaces for each $\alpha<\omega_1$ .

Comment by Apollo

2010-05-10T14:09:00Z

This argument doesn't work. The set of countable subsets of a cardinal $\kappa$ will be of size $\kappa^\omega$ which will be equal to $2^\omega \kappa$ for $\kappa$ with uncountable cofinality (and hence $>\kappa$ for any $\kappa<2^\omega$. For $\kappa$ of countable cofinality $\kappa^\omega>\kappa$. So in all of these cases your construction fails. However, for $\kappa$ of uncountable cofinality you will get a $\sigma$-algebra of cardinality $\kappa$.

Comment by Apollo

2010-04-19T14:12:21Z

David- No problem. Thanks for posting the scan --- it's nice to have an e-copy (mine is a 20 year old xerox...) Cheers