User peter krautzberger - MathOverflow

Answer by Peter Krautzberger for Are there q-filters which are not ultrafilters?

2011-09-15T15:08:07Z

Another way to get a Q-filter that is not an ultrafilter is to destroy a Q-point using an $\omega^\omega$-bounding forcing, e.g., Grigorieff forcing with a selective ultrafilter.

Grigorieff forcing destroys the selective ultrafilter, but because Grigorieff forcing with a P-filter is $\omega^\omega$-bounding, the (formerly selective) filter will remain a Q-filter.

Destroying the P-filter-property

2011-05-31T16:03:26Z

It is known that if a forcing notion is proper, then every P-filter will generate a P-filter in the generic extension (see, e.g., Shelah, Proper and Improper Forcing, VI.5)

On the other hand, if we start collapsing cardinals, we can destroy the P-filter property. For example, making a base of a P-point countable will add a sequence of elements from the filter (namely, a complete enumeration of that base) that serves as a counterexample for the P-filter property in the extension.

So my question is:

Are there examples of "nicer" (e.g., not collapsing cardinals) forcing notions that destroy P-filters in this way, i.e., add a sequence in the filter that the filter cannot decide?

More spectacularly, is there maybe a forcing notion that could preserve a P-point as an ultrafilter while destroying the P-filter property?

EDIT: As Martin Goldstern pointed out, I should add that I'm interested in filters on $\omega$.

How "much" does (Grigorieff) forcing destroy an ultrafilter?

2011-04-25T22:56:44Z

Introduction. I recently revisited Shelah's model without P-points and I was wondering how "badly" Grigorieff forcing destroys ultrafilters, i.e., what kind of properties can survive the destruction of the "ultra"ness.

An example. Given a free (ultra)filter $F$ on $\omega$, Grigorieff forcing is defined as $$ G(F) := \{ f:X \rightarrow 2: \omega \setminus X \in F \},$$ partially ordered by reverse inclusion. A simple density argument shows that "$G(F)$ destroys $F$", i.e., the filter generated by $F$ in a generic extension is not an ultrafilter (the generic real being the culprit).

Of course, there are many forcing notions that specifically destroy ultrafilters (also, Bartoszynski, Judah and Shelah showed that whenever there's a new real in the extension, some ground model ultrafilter was destroyed).

My question is:

If $F$ is destroyed, how far away is $F$ from being the ultrafilter it once was?

Maybe a more positive version: Which properties of $F$ can we destroy while preserving others?

This might seem awfully vague, so before you vote to close let me explain what kind of answers I'm hoping for.

Positive answers.
- If the forcing is $\omega^\omega$-bounding and $F$ is rapid, then $F$ will still be rapid. That's a very clean and simple preservation.
- In Shelah's model without P-points, all ground model Ramsey ultrafilters stop being P-points but "remain" Q-points.
"Minimal" answers. Is it possible that $F$ together with the generic real generates an ultrafilter, i.e., there are only two ultrafilters extending $F$? For Grigorieff forcing, I'd expect this needs at least a Ramsey ultrafilter. But maybe other forcings have this property?
Negative answers. Say $F$ is a P-point; can $F$ still be extended to a P-point? Shelah tells us that forcing with the full product $G(F)^\omega$ denies this. Is it known whether $G(F)$ already denies this? Do other forcing notions allow this?

I know there is a lot of literature on preserving ultrafilters (mostly P-points, I think) but I'm more interested in the case where the ultrafilter is actually destroyed. But I'd welcome anything that sheds light on this.

PS: community wiki, of course.

Answer by Peter Krautzberger for How "much" does (Grigorieff) forcing destroy an ultrafilter?

2011-05-31T15:43:54Z

I wanted to add two comments that I received in 'meatspace'. I hope this isn't too inappropriate.

If $F$ is a P-filter, and the forcing is proper, than $F$ generates a P-filter in the extension.
If $F$ is a Q-filter, i.e., every finite-to-one map becomes injective on a set in $F$, and the forcing is $\omega^\omega$-bounding, then $F$ generates a Q-filter in the extension.

Proofs of these facts can be found, e.g., in Shelah, Proper and Improper Forcing, Chapter VI, Section 4 and 5 resp.

One more example from myself.

If $F$ is an idempotent filter then $F$ will remain an idempotent filter in any forcing extension. In particular, if $F$ is an idempotent ultrafilter, it will still extend to an idempotent ultrafilter.

Answer by Peter Krautzberger for Is every p-point ultrafilter Ramsey?

2011-04-20T01:31:42Z

Another small addendum to Andres's and Andreas's answers.

It is also consistent that the answer to your question is yes.

Shelah has constructed a model of ZFC in which there exists (up to isomorphism) exactly one p-point -- and that p-point is, in fact, selective. This construction is Section XVIII.4 in Shelah, Proper and Improper Forcing .

Answer by Peter Krautzberger for Spaces of filters

2011-04-14T14:57:47Z

Some partial answers taken from Neil Hindman, Dona Strauss, "Algebra in the Stone-Čech compactification", chapter 21 (aptly named "Other Semigroup Compactifications").

The most general is probably Theorem 21.31:

If $X$ is discrete, $Y$ compact, $g: \beta X \rightarrow Y$ continuous and onto, then $Y$ is isomorphic to a space of filters on $X$ (simply intersect the preimages of points in $Y$)
In particular, every compactification of $X$ can be viewed as a space of filters.

If you're interested in algebraic aspects, there's section 21.3 of the book.

If $(X,\cdot)$ is a semigroup (not necessarily discrete, but completely regular), "nice" semigroup compactifications such as the AP and WAP compactifications, i.e., the (maximal) topological and semitopological semigroup compactifications respectively, are very interesting objects. They also have nice descriptions as filters.

Answer by Peter Krautzberger for What are some reasonable-sounding statements that are independent of ZFC?

2010-07-04T21:38:29Z

One of my favourites is

"Three clouds cover the plane"

where a subset $A \subseteq \mathbb{R}^2$ is a cloud around $a$ if every line through $a$ has a finite intersection with $A$.

This is due to Péter Komjáth; see http://www.cs.elte.hu/~kope/p28.ps.

In fact, three clouds cover the plane if and only if CH is true.

If the continuum is at most $\aleph_n$, then you can cover the plane with $n+2$ clouds ~~(whether the reverse holds is open)~~ (see comments).

Answer by Peter Krautzberger for Dynamical systems, minimal sets and the Axiom of Choice

2010-12-15T23:47:06Z

This is somewhat of a non-answer to the second question (I do need a little bit of AC) so I apologize. But I thought it might be worthwhile for its generality.

The uncountable example I had in mind is $\beta \mathbb{N}$, the set of ultrafilters on the natural numbers, i.e., the Stone–Čech compactification of the discrete space $\mathbb{N}$.

$(\beta \mathbb{N},+)$ is a semigroup (with + extending the usual addition on $\mathbb{N}$ to $\beta \mathbb{N}$ in such a way that addition with a fixed right hand side is continuous)
The shift $s(p)= 1+ p$ makes $(\beta \mathbb{N},s)$ a dynamical system (because addition with natural numbers is left-continuous, too)
Its minimal systems are exactly the minimal left ideals of the semigroup
Its cardinality is $2^{2^{\aleph_0}}$ (i.e., the size of the power set of the reals)
The minimal left ideals are universal minimal systems for discrete time, so one minimal left ideal induces minimal systems everywhere

To prove all this, if I'm not mistaken, you require "only"

The ultrafilter lemma for (the power set of) $\mathbb{N}$ so that you actually get the space (the ultrafilter lemma is strictly weaker than AC)
An application Zorn's Lemma to find a minimal left ideal

So that's not "a lot" of AC to get one minimal system everywhere (which is why I thought it'd be worthwhile).

On the other hand, if you assume, e.g., ZF+AD you do not find any free ultrafilters on $\mathbb{N}$, so no $\beta \mathbb{N}$, and I have no idea what the dynamics look like then...

Answer by Peter Krautzberger for Tools for long-distance collaboration

2010-12-14T18:00:50Z

I'm a little late to the party, and a lot of my favorite tools have been mentioned, but since Willie Wong asked for testimonials...

Meta: don't expect it to be 'just as easy'! If you suggest to use some of these tools, make sure your collaborator understands that this requires effort (at least initially) and a different routine.
audio/video: video calls are cheap and easy to use -- perfect for all those hand-waving arguments. I mostly use skype (I tried tokbox for multi-user conferencing but never used it frequently). Also, as mentioned by Beth, skype is good enough to broadcast a (small) blackboard. If you have real camcorder, you can broadcast using vlc, justin.tv or ustream for higher resolution video (this comes with lag, so keep some other audio solution). Also great to hook people up to seminars btw.
Online whiteboards with tablets! The combination of online whiteboards and tablets is my favorite since it can be set up almost anywhere. I personally use scriblink and dabbleboard extensively; scriblink is more reliable and uses less bandwidth, but dabbleboard has fancier technology (shape recognition, upload documents as background). For more privacy, there's also jarnal which can connect across the net, but I could never get it to work through firewalls. There are also all-in-one tools like dimdim -- but they were always too general for my purpose (and had problems with flash under linux). As already mentioned by Michael, whiteboards only make sense with a tablet of some sorts (I was happy with a wacom bamboo (cheap), but a tabletpc is even better (I have an HP TM2 running ubuntu)) Of course, you can type text on online whiteboards (scriblink even does a little TeX), but there are better tools for plain text.
Remote Desktops Sometimes I also like to connect the desktops, i.e., allowing one side to fully access the other side's desktop. I usually do this via teamviewer (connects through firewalls), but vnc, rdp are good, too. The advantage: all your programs are there! Anything you can do on your computer, you can do together. E.g., Xournal, OneNote for tablet-scribbling, gummi or latexian for live-previewed LaTeX, pdf-viewers for collaborative document browsing etc.
LaTeX (in the cloud) If you just want to scribble some TeX, there are many wikis with $\LaTeX$ support. I use Tiddlywiki with the mathsvg-plugin a lot these days -- a single html/javascript file, portable, fast TeX. Combined with a cloud service like dropbox or box.net and you can keep everything up-to-date in real time.

Answer by Peter Krautzberger for Social Reading Platform for Math or LaTeX texts

2010-12-05T19:25:54Z

This is far from an answer, but if no answer is found I thought a partial answer might be useful.

Since the last point (switching annotations on and off) seems to me the most difficult, Google Wave with its time slider comes to mind. Unfortunately, LaTeX support was extremely bad in my experience. But since the Apache foundation might take over the development there's at least a future.
On the other hand Tiddlywiki has most of the features you're looking for. Above all, it is very hackable with tons of plugins (excellent LaTeX plugins like MathSVGPlugin and jsmath). Of course, its technology (javascript, single file) does not scale well but it might offer some ideas. I use it in several projects with a small number of users. There's a wonderful copy of the Tractatus as an example for its abilities.

EDIT: Dec 14, 2010 I recently remembered this questions when I came across a tool for Wordpress.

http://digress.it/ a plugin for comments per paragraph.
There's a number of LaTeX plugins, e.g. http://wordpress.org/extend/plugins/mathjax-latex/
There's a plugin for comment rating http://wordpress.org/extend/plugins/comment-rating/

That seems to be the best base for a multi-user approach -- a couple of more plugins could turn it into a social reading platform.

Answer by Peter Krautzberger for What is a satisfactory way to format definitions in Latex?

2010-11-19T16:22:17Z

Since Colin's comment indicates that this is not about $\LaTeX$ and some answers have given good specific advice I want to throw in a more abstract answer:

make sure you write markup

worry more about structuring your content
and realize a typesetting for your own purposes (say according to established typesetting/layout rules for screen reading, website design, epaper or good old printouts, whichever you prefer to read your own stuff with)
BUT do it in such a way that anyone with access to the source (e.g. journal, website) can easily modify the layout (e.g. in $\LaTeX$ make sure redefining your environment is easy, maybe even via options for your own sty file, for a website use good css)

This is not as hard as it sounds -- you just have to overcome the urge to control your layout and focus on your content.

Answer by Peter Krautzberger for Sexy vacuity ....

2010-11-13T20:18:22Z

$\bigcap \emptyset = V$

Unfortunately, I have read more than one philosophical comment on the "set theoretic depth" of this logical triviality.

Answer by Peter Krautzberger for Hahn-Banach without Choice

2010-11-12T17:55:59Z

Since I can't (yet) leave a comment to Andres Calceido's answer.

A link to another paper by Luxemburg with a proof of the Hahn-Banach Theorem using ultrapowers http://www.ams.org/bull/1962-68-04/S0002-9904-1962-10824-6/S0002-9904-1962-10824-6.pdf

Answer by Peter Krautzberger for Awfully sophisticated proof for simple facts

2010-11-08T22:18:36Z

Every finite semigroup contains an idempotent element.

You can nuke this problem using a theorem by Ellis that every compact, semi-topological semigroup contains an idempotent (which uses Zorn's Lemma).

Answer by Peter Krautzberger for Best tablet computer for mathematics

2010-09-01T22:32:23Z

I recently got an HP tm2 -- very affordable (compared to Thinkpads, Latitudes and motion computing slates), good specs, wacom digitizer+multitouch, 5h+ regular battery life.

Productivity is great using Xournal under linux (kubuntu, almost flawless hardware support). Xournal runs on windows, too, btw.

During a conference recently Xournal was brilliant -- infinite paper with infinite zoom, free rearranging, shape recognition etc. makes it easy to keep good notes (except that, well, it's still my handwriting...). The battery life of the tm2 lasted an entire conference day (with dimmed screen and wifi off).

You might want to check out the gottabemobile blog. It is an good source on mobile computing, with a lot of reviews on note taking software for tablet pcs and the iPad.

Generally speaking, pressure sensitive touch technology (like wacom or n-trig) gives better results for handwriting. But some iPad apps have features to compensate for that.

Being on a budget, a graphics tablet is a very good alternative for taking notes. I used to use one extensively before, especially for online whiteboards (like scriblink and dabbelboard) during phone conversations.

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

2010-08-12T17:07:09Z

Just a partial answer.

For $\beta \mathbb{N}$ (the set of all ultrafilters on $\mathbb{N}$ with the Stone topology) it is not hard to see that a sequence converges iff it is eventually constant. Hence any subset of $\beta \mathbb{N}$ is sequentially open -- and of course, $\beta \mathbb{N}$ is not discrete, so it cannot be sequential. Similarly, for the ultrafilters on $\mathbb{R}$.

If I recall correctly, this 'trivial sequential convergence' holds in all extremally disconnected spaces -- this should be an exercise in the book 'Rings of continuous functions' by Gillman and Jerison (there is also a PDF/TeX-file with all exercise solutions freely available on the web somewhere).

Also, I could be wrong, but I think $\beta \mathbb{N}$ is not countably tight since its remainder is not (since there exist weak P-points). Maybe somebody else can confirm or reject.

Answer by Peter Krautzberger for Maximal ideals and ultrafilters

2010-08-01T14:29:34Z

To get the actual answer out of the way: the usual definition of ideal implies that any ideal contains the empty set -- an ideal $I$ (on a set $X$ / the power set of $X$) is non-empty, closed under taking subsets and under taking finite unions (and of course $I\subseteq P(X)$). The first two should convince you that it contains the empty set.

The notion of filter is dual -- non-empty, closed under taking finite intersections and supersets. Ideals correspond to filters by mapping each element $A\in I$ to its complement $X\setminus A$. Maximal ideals correspond to maximal filters.

Just in case this question was only due to a confusion of ideals and filters, let me add:

A proper ideal by definition does not contain the 'full' set $X$ (e.g. in your example $X = { 1,2,3 }$). Similarly, a proper filter does not contain the empty set by definition. The 'improper' cases of these definitions coincide -- both the improper filter and the improper ideal are just the full power set (as is clear from being closed under subsets and supersets respectively).

Usually, filter/ideal means proper filter/ideal, but for notational or technical convenience, it sometimes seems nice to allow the improper case -- for example in the Stone-Cech compactification of the natural numbers, $\beta \mathbb{N}$, proper filters (on $\mathbb{N}$) correspond to closed non-empty subsets and the improper filter to the empty set. But I think the general preference (as Joel David Hamkins pointed out in the comments) is not to do this since no convenience outweighs the confusion caused by the improper case.

In your example and (as mentioned by Robin Chapman) for any finite set $X$ , the maximal (proper) filters (or ultrafilters) are the principal filters, i.e. those of the form $\dot{x} = { A \subseteq X:\ x \in A } $ for some $x\in X$. To see this just partition $X$ into singletons -- a finite partition by assumption on $X$ -- every maximal filter contains exactly one part of the partition. The maximal ideals are again the dual.

Comment by Peter Krautzberger

2011-07-10T15:24:39Z

Wouldn't you rather look at the power of an arbitrary element and cycles therein?

Comment by Peter Krautzberger

2011-06-14T16:57:48Z

Martin, thanks again. I was really hoping for a positive answer, but hank you for sharing these two partial ones.

Comment by Peter Krautzberger

2011-06-06T20:55:34Z

Thank you, Martin.

Comment by Peter Krautzberger

2011-06-04T19:28:23Z

Martin, thanks! I hope you don't mind that I'll wait a bit to see if more (partial) answers turn up.

Comment by Peter Krautzberger

2011-06-04T19:10:03Z

Yes, I mean filters on $\omega$. I'll add that to my question.

Comment by Peter Krautzberger

2011-05-14T21:03:55Z

@Peter McNamara: I think Adobe Reader does not show back/forward buttons as default and I don't use it myself. Try editing the toolbar and look for back/forward buttons. Also, try alt+right, alt+left (on windows) or whatever combination your browser uses to navigate. And if all fails, google.

Comment by Peter Krautzberger

2011-05-13T17:09:40Z

@André Most modern PDF-viewers come with a "back in the document" button that does what you are asking for; e.g., acrobat reader, okular, skim etc.

Comment by Peter Krautzberger

2011-05-10T17:07:49Z

My last comment was silly -- of course the P-point has to be a Ramsey ultrafilter to begin with because if we don't kill it, all its ground-model RK-predecessors cannot be killed either. In other words, the P-point should better be minimal/selctive/Ramsey from the start to end up the unique (and minimal) P-point. Thanks to Andreas for the correction.

Comment by Peter Krautzberger

2011-05-09T02:27:55Z

Thank you for this answer! It's a wonderful general argument that I can keep in mind.

Comment by Peter Krautzberger

2011-04-26T21:38:42Z

I just looked at Shelah's construction and I think it's worth pointing out that Shelah really needs (or at least thinks he needs) a Ramsey ultrafilter to start with, i.e., the preservation lemma requires the "surviving" P-point to be selective.

Comment by Peter Krautzberger

2011-04-20T18:20:50Z

Doh! Thanks Tanmay! I shouldn't ask questions after midnight...

Comment by Peter Krautzberger

2011-04-20T03:59:21Z

Andres, is there an easy argument why a unique p-point must be selective? Also, it could be interesting to the OP to point out that you can pick any Ramsey ultrafilter in a CH model and have it survive to become the unique one in an extension. But then again, selective ultrafilters all look the same (in the sense of what Andreas calls 'complete combinatorics'). I would agree that the model is more involved though I have heard good arguments that this is mostly due to the presentation.

Comment by Peter Krautzberger

2011-04-06T01:26:22Z

A gist of the argument: show that Grigorieff forcing with a non-meagre P-filter is proper and $\omega^\omega$-bounding; understand why those properties are preserved under proper countable support iterations; prove the key combinatorial argument: the reals added by the full $\omega$-product of Grigorieff forcing with a P-point prohibits that P-point from being "resurrected" in any further $\omega^\omega$-bounding forcing extension; now do the iteration killing any P-point that comes along.

Comment by Peter Krautzberger

2011-04-06T00:49:03Z

The link in the question is very good though. Wolfgang Wohofsky's Diplomarbeit is (imho) the most accessible presentation of Shelah's model without P-points. Assuming some experience with iterated forcing, a good introduction to iterated proper forcing would be Martin Goldstern's excellent "Tools for your forcing construction" to be found at info.tuwien.ac.at/goldstern/papers/tools.ps

Comment by Peter Krautzberger

2011-03-14T17:12:26Z

Felix, could you try to format your text a little better? I'm not sure I understand correctly what specific question you are asking. It seems to me you want something along the lines of "How do different logical set ups of a theory of the reals (e.g., classical, constructive) effect the type of an irrational number, i.e., statements about a single number?" I'm not an expert, but this might be less likely to be closed as "not a real question" (the pun would be wonderful, though)