User todd eisworth - MathOverflow

Answer by Todd Eisworth for If $\kappa \rightarrow (\alpha)^r_2$ holds for every $r\in \omega$, then is $\kappa$ an $\alpha$-Erdős cardinal?

2013-05-31T12:51:54Z

The two properties are not the same.

If $\kappa$ is weakly compact, then $\kappa\rightarrow(\kappa)^r_\lambda$ for any $r<\omega$ and $\lambda<\kappa$. These cardinals are compatible with $V=L$.

A cardinal $\kappa$ for which $\kappa\rightarrow(\kappa)^{<\omega}_2$ is called a Ramsey cardinal, and these cardinals imply that $0^\sharp$ exists.

Answer by Todd Eisworth for Question about Shelah's version of "Shooting a club" found in PIF

2013-05-09T20:58:31Z

Sorry this is sketchy, but I hope it helps!

A Cohen condition $p$ of length $n$ is going to determine $\langle C_m:m<n\rangle$ . We know there's going to be an $i<\omega$ such that $q_i$ extends both $C_{n-1}$ and a member of $N[G_P]\cap A$. Extend $p$ to a Cohen condition $p'$ with $p'(n)=i$. If $r^*$ is any Cohen generic extending $p'$, then running the construction with $r^*$ guarantees that $C_n$ extends a member of $N[G_P]\cap A$. So a density argument tells us the construction is forced to work.

The point is that in Shelah's construction, $m(n)$ is going to be equal to $r^*(n)$ very often because $r^*$ is Cohen.

Answer by Todd Eisworth for Why $z \in \overline{A}$?

2013-02-25T15:54:14Z

I may be missing something because this is quite elementary, but:

Since $Z$ is regular, for any open neighborhood $U$ of $z$ there is an open $V$ such that $z\in V$ and $\overline{V}\subseteq U$ (apply regularity to $z$ and the complement of $U$). Do this countably many times and take intersection, and you'll find that for any open neighborhood $U$ of $z$ there is a closed $G_\omega$-set $P$ such that $z\in P\subseteq U$.

If follows easily that if $z$ is not $G_\omega$-separated from $A$ (that is, if (b) fails), then $z$ must be in the closure of $A$.

Answer by Todd Eisworth for Disjoint stationary sets that reflect

2012-11-06T20:24:01Z

In the presence of large cardinals, one can (or rather Shelah can...) force the answer to be "NO" in a very strong sense. The place to look is Section 7 of Chapter X of Proper and Improper Forcing.

In particular, Theorem 7.4 shows that assuming the consistency of 2 supercompact cardinals, one can force that for any regular $\kappa>\omega_1$, any stationary subset of $S^\kappa_{\aleph_0}$ contains a closed copy of $\omega_1$.

This implies the answer to your question is no by the following argument:

Step 1: If $\kappa>\aleph_1$ is regular and $A$ reflects at all uncountable limit ordinals below $\kappa$, then so does $A\cap S^\kappa_{\aleph_0}$ (where $S^\kappa_\tau$ is the set of ordinals less than $\kappa$ of cofinality $\tau$).

Proof: Let $A_0= A\cap S^\kappa_0$, and let $A_1= A\setminus A_0$. $A_1$ cannot reflect at ordinals of cofinality $\omega_1$, and so it must be the case that $A_0$ reflects at all ordinals of cofinality $\omega_1$. But then $A_0$ also reflects at any place where $S^\kappa_{\aleph_1}$ reflects as well, and so $A_0$ reflects at all ordinals of uncountable cofinality below $\kappa$.

Step 2:
Assume we are in a model like that obtained by Shelah. If $\kappa$ is a regular cardinal greater than $\aleph_1$ and $A$ is a stationary subset of $S^\kappa_{\aleph_0}$. We know $A$ contains a closed copy $C$ of $\omega_1$, and if we set $\delta=\sup(C)$ then $\delta$ is an ordinal of cofinality $\omega_1$ where $A$ reflects but $\kappa\setminus A$ does not. In particular, no stationary subset disjoint to $A$ can reflect at $\delta$, hence there is no way to get your "$B"$.

Edit:

A "no" answer to your question at $\omega_2$ is equiconsistent with the existence of a Mahlo cardinal.

As Joel mentioned in (an earlier version of) his answer, one can build $A$ and $B$ in $\omega_2$ from a $\square_{\omega_1}$-sequence. The failure of $\square_{\omega_1}$ implies that $\aleph_2$ is Mahlo in $L$ (Credited to Jensen on page 453 of Jech's "Set Theory"; I don't know a better reference.)

On the other hand, Theorem 7.1 in Chapter XI (page 576) of Proper and Improper forcing tells us that from a Mahlo cardinal, we can force ZFC+GCH + "every stationary subset of $S^{\omega_2}_{\omega}$ contains a closed copy of $\omega_1$, which we argued above gives a "No" answer.

Note that what Shelah is really showing is the consistency of the following statement:

"If $S$ is a stationary subset of $S^{\omega_2}_{\omega}$ that reflects at every member of $S^{\omega_2}_{\omega_1}$, then $S^{\omega_2}_{\omega}\setminus S$ is non-stationary,"

while the original question is equivalent to asking of $S^\kappa_\omega$ can be partitioned into two disjoint stationary sets, each of which reflects at every ordinal in $S^\kappa_{\omega_1}$.

Answer by Todd Eisworth for Start with a topological group, take the meet of the two uniformities, and take the topology. Is the result again a topological group? [xpost from math.SE]

2012-10-19T16:22:58Z

This is most definitely not my field of expertise (so be kind!), but Section 1.8 of the book "Topological Groups and Related Structures, an introduction to topological algebra" by Arhangel'skii and Tkachenko deals with these sorts of questions.

The book is available online at SpringerLink if you have access. Theorem 1.8.15 deals with something called the Roelcke uniformity, and if I'm reading the result correctly, proves that it is compatible with the topology on the group, and also the finest uniformity on the group coarser than the left and right uniformities.

I hope this is helpful! I can edit with better information after my colleague Vladimir Uspenskij gets out of class, as he's an expert on this.

Edit: I asked Uspenskij about this, and his quote was something like "In general, the meet of two uniformities is something horrible, but in topological groups we get the nice Roelcke uniformity."

Answer by Todd Eisworth for Chain conditions in quotients of power sets

2012-06-08T18:34:16Z

I'll add in that Shelah has used pcf theory to investigate related questions. Typically these results are tucked away inside long papers dealing with other questions, but I know that the last section of [Sh:410] explicitly deals with ``strongly almost disjoint families", and characterizes their existence in terms of pcf.

For example, if $\aleph_0<\kappa\leq\kappa^{\aleph_0}<\lambda$, then the existence of a family of $\lambda^+$ sets in $[\lambda]^{\kappa}$ with pairwise finite intersection is equivalent to a ``pcf statement''.

I'm not sure which version of the paper to link to, as the published version has been reworked a few times. I THINK that the most recent version is here:

Sh:410

Answer by Todd Eisworth for closed set and z-ultrafilter on normal space

2012-03-22T00:39:52Z

No: think of what happens with $\omega_1$ in the usual topology. This is certainly normal (even hereditarily normal), and since every real-valued continuous function on $\omega_1$ is eventually constant, the co-bounded sets form a $z$-ultrafilter. Now let $W$ be the set of countable limit ordinals.

Answer by Todd Eisworth for Uncountable family of infinite subsets with pairwise finite intersections

2012-02-23T18:47:38Z

This all hinges on what you mean by constructive. An easy way to get such a family is to proceed as follows:

Put your countable set $X$ in bijective correspondence with the collection of finite sequences of 0s and 1s.

For every every subset $A$ of the natural numbers, let $\chi_A:\mathbb{N}\rightarrow\{0,1\}$ be the characteristic function of $A$, and let $S_A$ be the collection of finite sequences of the form $\chi_A|\{0,\dots,n\}$ for $n\in\mathbb{N}$.

Each $S_A$ is infinite, and if $A$ and $B$ are distinct subsets of $\mathbb{N}$, then $S_A\cap S_B$ is finite (once we get past the first difference between $A$ and $B$, the characteristic functions disagree). Now use your bijection to pull things back inside your original set $X$. This gives you a family of size $2^{\aleph_0}$ enjoying the property you want.

I don't know if this is constructive enough or not!

Answer by Todd Eisworth for Readings for an honors liberal art math course

2012-02-19T19:04:53Z

I've used Underwood Dudley's "Readings for Calculus" with such honors students, and I think it was successful. Don't let the title fool you, there are very few "math problems" in the book, and lots and lots of discussion questions about the nature of mathematics, its history, and its place in the world.

Answer by Todd Eisworth for A question about local connectedness in metric spaces

2012-01-19T21:19:38Z

There is a "folklore" counterexample. Peter Nyikos gives the construction here (see the last paragraph for the compactness)

Devlin's "Constructibility" as a resource

2011-10-10T19:44:39Z

It is fairly well-known among set-theorists that Keith Devlin's 1984 book "Constructibility" has flaws in its initial development of fine structure theory. (See Lee Stanley's review 1 of the text for the Journal of Symbolic Logic, for example.) I've had the book on my shelf for twenty years now, and although there is much in there that I find interesting, the fact that I know there are some errors in it means that I've been reluctant to invest a lot of time working through it. So, this brings me to my questions for the experts:

How badly do these flaws mar the rest of the book? Is the damage localized to the initial development of properties of the the J-hierarchy, or is it much more widespread?

Of particular interest to me are the following questions:

1) Is Devlin's treatment of the Covering Lemma for L on solid ground?

2) What about his treatment of morasses?

I know that there are other sources for this material, but I've always appreciated Devlin's writing style.

Answer by Todd Eisworth for Examples of forcing arguments which require an assumption in the ground model about the sizes of the power sets?

2011-11-15T16:48:47Z

There are some odd examples coming from PCF theory, where one begins with assumptions about pcf structures that are not known to be consistent (even if one assumes consistency of large cardinals), and then do some forcing in order to make things happen.

For example, consider the following conjecture of Shelah:

"If $\mathfrak{a}$ is a set of regular cardinals greater than $|\mathfrak{a}|$, then ${\rm pcf}(\mathfrak{a})$ cannot have a weakly inaccessible accumulation point."

An affirmative answer to the above conjecture implies \begin{equation} {\rm cf}(\prod{\rm pcf}\mathfrak{a}, <)={\rm cf}(\prod\mathfrak{a},<) \end{equation} for every set of regular cardinals $\mathfrak{a}$ with $|\mathfrak{a}|<\min\mathfrak{a}$. On the other hand, if the above conjecture fails for some $\mathfrak{a}$, one can do some mild forcing and get the consistency of \begin{equation} {\rm cf}(\prod{\rm pcf}\mathfrak{a}, <)\neq{\rm cf}(\prod\mathfrak{a}, <). \end{equation}

Shelah has many arguments of this sort: he's got a whole family of pcf conjectures that are "linearly ordered" (none of which are known to be consistent) and he uses forcing to show that many other natural questions are (essentially) equivalent to one of his basic conjectures.

For a more substantial example, consider the question of whether every compact Hausdorff space can be partitioned into two pieces, neither of which contains a copy of the Cantor set. Assuming a supercompact cardinal, Shelah is able to force the existence of counterexample, but the resulting model satisfies CH. He is able to prove, however, that the existence of such a counterexample in a model where $2^{\aleph_0}=\aleph_2$ implies some weird pcf behavior that is not known to be consistent. Moreover (and this part speaks to your original question), if one assumes there is a model where this weird pcf behavior holds, then one can force the existence of a counterexample together with the continuum being $\aleph_2$. So in this case, the topological question is essentially equivalent to a question about pcf. (This is from his paper [Sh:682].)

Edit: I know your original question asked about assumptions on sizes of power sets, but I interpreted things a little broader to include assumptions about cardinal arithmetic.

Answer by Todd Eisworth for Books you would like to read (if somebody would just write them...)

2011-10-25T20:06:22Z

"Cardinal Arithmetic: The New Corrected Edition (including index)" by Saharon Shelah...

Answer by Todd Eisworth for A model of CH +$\lnot \diamondsuit$

2011-10-23T14:28:56Z

"The easiest way" to produce a model of CH in which $\diamondsuit$ is false is to start with a model of GCH and then do a countable support iteration of length $\omega_2$ killing off a potential $\diamondsuit$ sequence at each stage.

The forcing for doing this is straightforward: supposing $\langle A_\alpha:\alpha<\omega_1\rangle$ is a sequence with $A_\alpha\subseteq \alpha$ , we force the existence of an uncountable $X\subseteq\omega_1$ such that $X\cap\alpha\neq A_\alpha$ for all $\alpha$. This is done by viewing conditions as telling us initial segments of $X$.

Iterating the above forcing $\omega_2$ times while using bookkeeping to make sure we kill of any potential $\diamondsuit$ sequence actually works.

BUT

There's a whole lot of work involved in proving that the iteration doesn't add new reals. This is where one must wrestle with the Shelah machinery of $\mathbb{D}$-completeness and $<\omega_1$-properness

Answer by Todd Eisworth for continuous function

2011-10-20T13:54:52Z

The criterion for "EVERY continuous map from $D$ to $[0, 1]$ has a continuous extension to $X$" is that any two disjoint zerosets in $D$ have disjoint closures in $X$. You can find this in Chapter 6 of Gillman and Jerison's classic "Rings of Continuous Functions". They also consider the "local problem" of continuously extending a single map at length in some of the exercises, e.g. given $f:D\rightarrow Y$ (not necessarily $Y=[0, 1]$) Exercise 6G characterizes the largest subspace of $X$ to which $f$ can be continuously extended in terms of $z$-filters.

Answer by Todd Eisworth for Substructure Argument for Chain Conditions

2011-10-19T20:34:15Z

A typical argument for showing that a notion of forcing is ccc will invoke the $\Delta$-system lemma at some point, and this result has a slick proof using elementary submodels (see Lemma 24.24 of "Discovering Modern Set Theory II" by Just and Weese, or Example 2.1 of Dow's "An introduction to applications of elementary submodels to topology".

One can usually combine these two steps into a single argument to prove that a notion of forcing is ccc using elementary submodels, although I can't point to a specific published example of this at the moment. I'll see if I can find a reference for you where someone does this.

Edit: The best reference is Ramiro's answer!

Answer by Todd Eisworth for Can we collapse $\omega_1$ to $\omega$ without adding a dominating real?

2011-10-12T21:43:34Z

Hi Noah, I'm adding an answer because this wouldn't fit in a comment. Unless I'm making a silly mistake, the situation is different with respect to adding an escaping real:

Let $P$ be the collection of finite partial functions from $\omega$ to $\omega_1$ as usual, and let $\dot f$ be a $P$-name for the generic surjection from $\omega$ onto $\omega_1$.

Given $g:\omega\rightarrow\omega$ in the ground model, and $n<\omega$, consder the set $D(g, n)$ of conditions $p$ such that for some $m>n$,

$m$ is in the range of $p$,
the domain of $p$ is an initial segment of $\omega$, and
the least $k$ for which $p(k)=m$ is greater than $g(m)$.

This set is dense in $P$ for each $g$ and $n$, and so the real $h$ in the extension defined by setting $h(m)$ equal to the least $k$ such that $\dot f(k)=m$ is not bounded by a ground model real. (So essentially, we are "inverting" the surjection on the initial segment $\omega$ of its range)

Edit: Even simpler, if we define a real $h$ in the extension by setting $h(n)=\dot f(n)$ if $\dot f(n)<\omega$, and $h(n)=0$ otherwise, then $h$ is Cohen over the ground model, hence unbounded.

Answer by Todd Eisworth for Generalizations of pcf theory

2011-10-11T15:34:45Z

I don't know if anyone has looked at such things systematically, but I know Shelah has made use of structures of this form at various times. The examples which follow are just what I can remember off-hand; I know there's more buried in his work, but this is where I remembered seeing such a construction:

1) Clause $(\gamma)$ on page 1641 of [Sh:589]:

Applications of PCF theory. J. Symbolic Logic 65 (2000), no. 4, 1624–1674.

(It's available on JSTOR here; the Arxiv version looks like an older iteration)

2) The second proof of his Revised GCH theorem in [Sh:460], in particular Claim 2.6.

The generalized continuum hypothesis revisited. Israel J. Math. 116 (2000), 285–321.

(Available on SpringerLink here, again the Arxiv version is a bit dated)

I don't think that you get anything really new as far as pcf theory by doing this. Rather, it's just that sometimes such a structure is the right way to organize things.

Answer by Todd Eisworth for Bounds on $\max \mathrm{pcf}(A)$ if $\Pi A$ is big

2011-10-11T14:59:17Z

Shelah proved that it is consistent that GCH holds below $\aleph_\omega$ , while $2^{\aleph_\omega}=\aleph_{\omega+\alpha+1}$ for any countable ordinal $\alpha$ you care to choose. (See Theorem 36.5 of Jech's book, for example).

In such a model, ${\rm max pcf}(A)=\aleph_{\omega+\alpha+1}$ as well. Now if you add $\aleph_{\omega_1+1}$ Cohen reals (which has no effect on the pcf structure) you end up with a model where

$|\prod A| = \aleph_{\omega_1+1}$ , and
${\rm max pcf}(A)=\aleph_{\omega+\alpha+1}$ .

So ${\rm max pcf} (A)$ could potentially be any successor cardinal below $\aleph_{\omega_1}$.

(Of course, it's still unknown if $\aleph_{\omega_1}\leq{\rm max pcf }(A)$ is possible, so this is the best answer we can hope for given our current knowledge.)

I don't know the answer to your "evens and odds" question, but certainly you can split $A$ up into two disjoint pieces whose "gap" is as large as possible:

Let $\tau$ denote ${\rm max pcf}(A)$ , and suppose $\aleph_{\omega+1}<\tau$ .

We know there exists an unbounded $B\subseteq A$ such that $\prod B$ contains a scale (mod finite) of length $\aleph_{\omega+1}$ . This implies ${\rm max pcf}(B)=\aleph_{\omega+1}$ .

The set $A\setminus B$ cannot be in the ideal $J_{<\tau}[A]$ (otherwise, we contradict ${\rm max pcf}(A)=\tau$), and so we must conclude ${\rm max pcf}(A\setminus B)=\tau$ as well.

Answer by Todd Eisworth for A result of Shelah about the nonstationary ideal

2011-10-03T14:31:10Z

Try also Chapter XVI of "Proper and Improper Forcing" (entitled "Large ideals on $\aleph_1$ from smaller cardinals"). It's hard to tell exactly what's in there, but he does say in the chapter he will "keep old promises from 84-85 mentioned in [Sh:253]", where [Sh:253] is the paper Michael mentions, and he does claim to be replacing certain hypotheses used earlier by the assumption "$\lambda$ is a Woodin cardinal".

Answer by Todd Eisworth for A question about J.H. Conway's SURREAL NUMBERS

2011-09-29T21:17:56Z

Philip Ehrlich at Ohio University has written extensively on Conway's surreal numbers, and somewhere in his work he has the details for formalizing the theory of surreal numbers in NBG set theory. This should qualify as a "standard" set theory despite its use of classes as it's a conservative extension of ZFC, as mentioned above. His forthcoming paper for the Bulletin of Symbolic logic gives his paper

Absolutely saturated models. Fund. Math. 133 (1989), no. 1, 39–46.

as a reference for this formalization.

Best,

Todd

Answer by Todd Eisworth for Some Pcf Theory

2011-09-28T17:41:42Z

Hi. I know this is over a year late, but I the proof you're looking at matches that given on page 61-62 of "Cardinal Arithmetic". The argument can be finished along the following lines:

(1) First, we may as well assume $|\mathfrak{a}|^+<\min(\mathfrak{a})$, as we can derive the result you want if we get it in this more restricted situation.

(2) You know that $\prod\mathfrak{a}$ has true cofinality $\lambda$ modulo the ideal $J_{<\lambda}[\mathfrak{a}]$ by way of Lemma 1.8 on page 9 of Cardinal Arithmetic. Let $\langle f_\alpha:\alpha<\lambda\rangle$ witness this. Without loss of generality, this sequence of functions is an element of $N_0$, and each $f_\alpha\in N_0$ as well. (Here we are using $|N_0|=\lambda$)

(3) Given one of your $g_i$, we know there is an $\alpha<\lambda$ with $g_i < f_\alpha$ modulo $J_{<\lambda}[\mathfrak{a}]$ .
By the usual sorts of argument, there is a single $\alpha<\lambda$ such that $g_i<f_\alpha$ modulo $J_{<\lambda}[\mathfrak{a}]$ for all $i<|\mathfrak{a}|^+$ .

(4) Now look at the collection $\langle\mathfrak{c}_i:i<|\mathfrak{a}|^+\rangle$ , where $\mathfrak{c}_i$ is defined to be the set of $\theta\in\mathfrak{a}$ for which $f_\alpha(\theta)\leq g_i(\theta)$ . You've arranged that $g_i<g_j$ (everywhere) whenever $i<j$ , so the sequence $\langle \mathfrak{c}_i:i<|\mathfrak{a}|^+\rangle$ is an increasing sequence of subsets of $\mathfrak{a}$ . Such a sequence of length $|\mathfrak{a}|^+$ must be eventually constant, say with value $\mathfrak{c}$ .

(5) This set $\mathfrak{c}$ has three important properties:

(5a) $\mathfrak{c}$ is in $J_{<\lambda}[\mathfrak{a}]$ (as all $\mathfrak{c}_i$ are in this ideal),

(5b) $g_i(\theta)<f_\alpha(\theta)$ whenever $i(*)\leq i<|\mathfrak{a}|^+$ and $\theta\in \mathfrak{a}\setminus\mathfrak{c}$, and

(5c) $|\mathfrak{c}|\in N_{i(*)+1}$ as it is definable from $g_{i(*)}$ and $f_\alpha$.

(6) we want to now apply our induction hypothesis to $\mathfrak{c}$ in the model $N_{i(*)+1}$ . This model has cardinality greater than maxpcf $\mathfrak{c}$, and so $N_{i(*)+1}$ is going to contain every member of a dominating family for $\prod\mathfrak{c}$.

(7) Consider now the function $g_{i(*)+1}$. It is not an element of $N_{i(*)+1}$; in fact, it is not even bounded by an element of $N_{i(*)+1}$. However, there is a function $g$ in $N_{i(*)+1}\cap\prod\mathfrak{c}$ dominating $g_{i(*)+1}$ on $\mathfrak{c}$. And the function $f_\alpha$ dominates $g_{i(*)+1}$ on the set $\mathfrak{a}\setminus\mathfrak{c}$. By pasting $g$ and $f_\alpha$ together, we get a function in $N_{i(*)+1}$ which dominates $g_{i(*)+1}$, and this is a contradiction.

Best,

Todd

PS: This is only the 2nd time I've attempted a post here; I'm not sure if the LaTeX looks right but I'll try to edit things so it looks OK.

PPS: Well, that took longer than I thought to edit, but maybe it looks OK now.

Answer by Todd Eisworth for Exact consistency-strength of "all projective sets are Ramsey"

2011-09-28T15:27:36Z

Hi David,

This is my first foray onto MathOverflow as well, so this answer is an experiment to see if I can get things to work, rather than an attempt to convey a lot of serious information.

As Andres said, the problem is still open as far as I know. I worked on this with Shelah a bit in the late 90s and we generated many things that led nowhere. Some related material:

1) Roslanowski and Shelah investigated "sweetness" and "sourness" (properties of ccc posets motivated by the constructions in Shelah's "Can you take Solovay's Inaccessible Away") in a series of quite technical papers early in the 2000s. This was partially motivated by the problem of getting all nicely definable sets to be Ramsey without using an inaccessible.

2) CH + "every set of real in L(R) is Ramsey with respect to every Ramsey ultrafilter" is equiconsistent with the existence of Mahlo cardinals. Mathias got the consistency result assuming a Mahlo, and the other direction is in a paper of mine from 1999 or so. The trick I used didn't seem to shed any light on whether or not the inaccessible is needed when we drop the reference to ultrafilters.

Best,

Todd

Comment by Todd Eisworth

2013-06-05T02:18:07Z

"But I haven't managed to prove what Shelah omitted" should be printed on t-shirts sold at ASL meetings because we've all been there...

Comment by Todd Eisworth

2013-05-31T13:33:00Z

Oh, that's an old induction argument using the "inaccessible + tree property" characterization of weakly compact cardinals. Kanamori's "Higher Infinite" and Jech's "Set Theory" have it as an exercise, but Hajnal and Hamburger's "Set Theory" has it written out on page 220 if you've got access to that.

Comment by Todd Eisworth

2013-05-30T04:29:56Z

Not quite, Andreas! Saharon was quite involved with the choice of "666" for that paper...he thought it a fine joke! (He was working on this paper when I first started working for him in Jerusalem.)

Comment by Todd Eisworth

2013-05-15T14:46:31Z

I wish I knew this result while Arhangelskii was still around our department. I could've answered lots of his questions!

Comment by Todd Eisworth

2013-05-10T15:19:15Z

It won't be proper if $S$ is co-stationary (as the stationarity of the complement of $S$ is destroyed), but it will preserve $\omega_1$ because you get generics for "enough" elementary submodels.

Comment by Todd Eisworth

2013-05-09T20:59:51Z

Answered this quickly as I'm leaving work. If on the way home I realize I've said something silly, I'll edit!

Comment by Todd Eisworth

2013-04-16T18:38:05Z

This error can be found in Gamow's famous book "One, Two, Three...Infinity", and even the Oxford English Dictionary contains a quote in their definition of "aleph null": "There is no infinite number between aleph-null (the number of positive integers) and aleph-one (the number of real numbers)." (apparently pulled from a Scientific American article)

Comment by Todd Eisworth

2013-03-14T14:51:13Z

Maybe it was Balogh. I know Fremlin and Nyikos were working on the problem as well, but that's all well before my time!

Comment by Todd Eisworth

2013-03-14T02:55:02Z

It's been a while since I've thought about these things, but I do have a couple of comments: In the work of mine you mention above, normality is a bit of a red herring as it is "perfect" and "regular" that allow us to build the notion of forcing needed for the proof to go through. It is also consistent with ZFC that every first countable, countably compact regular space is either compact or contains a homeomorphic copy of $\omega_1$. (I think that's originally due to Fremlin and Nyikos, but Peter Nyikos and I had a paper in the 2000s showing this consistent with CH as well.)

Comment by Todd Eisworth

2013-02-26T15:53:30Z

Luke, if you're having trouble getting people to answer your questions on stackexchange, then you might try the "ask a topologist" forum: at.yorku.ca/cgi-bin/bbqa Those guys are usually pretty quick to respond and they can provide helpful feedback. And again, the answer to your question is quite easy, as every infinite set in A has a point of accumulation (by countable compactness of X) in A (by your assumption on A).

Comment by Todd Eisworth

2013-02-26T04:01:03Z

That is, you make sure $V_{i+1}\subseteq\overline{V}_{i+1}\subseteq V_i$, and then $\cap V_i = \cap \overline{V}_i$ is closed and a $G_\omega$ set.

Comment by Todd Eisworth

2013-02-26T03:56:54Z

The nesting guarantees that the intersection of the V_i is the same as the intersection of their closures, hence closed.

Comment by Todd Eisworth

2013-01-11T15:32:56Z

I had a series of papers in set-theoretic topology where sets of this form were critical to analyzing a notion of forcing, and I never came across a standard name...

Comment by Todd Eisworth

2012-12-09T18:14:01Z

Is the notion of "realcompact" what you are looking for?

Comment by Todd Eisworth

2012-11-06T20:52:41Z

No idea about the case where the regular cardinal is successor of singular though!