User andreas blass - MathOverflow

Answer by Andreas Blass for Differentiable manifolds by Serge Lang question

2013-05-18T19:42:34Z

I'm not sure what Lang had in mind with "unnecessary double dualization", but here's an example that occurred to me in ancient times when I was trying to understand differential geometry better. Some (many? most?) people define tangent vectors to a manifold to be certain derivations on the smooth functions, and they define the cotangent space to be the dual of the tangent space. So a cotangent vector is a function taking as arguments tangent vectors, which are themselves functions taking as arguments smooth functions. In this sense, a cotangent vector is a doubly-dualized function. But one can avoid these dualizations by defining a cotangent vector at a point $p$ to be an element of $m/(m^2)$ where $m$ is the maximal ideal in the ring of germs of smooth functions at $p$. In other words, $m$ consists of the smooth germs that vanish at $p$ and $m^2$ consists of those that vanish to second order, so the quotient is "first-order data about a germ at $p$, omitting the value (zero-order data) at $p$." That picture captures pretty well my intuition of what a cotangent vector should be. (I think that the $m/(m^2)$ definition is used more in algebraic geometry than in differential geometry.)

Answer by Andreas Blass for Germs at infinity of sequence of integers

2013-05-16T12:14:57Z

This abelian group, which can also be described as the quotient of the direct product $\prod_{\mathbb N}\mathbb Z$ by the direct sum $\sum_{\mathbb N}\mathbb Z$, is isomorphic to the direct sum of the following two pieces. The first piece is a torsion-free, divisible abelian group (so you can view it as a vector space over $\mathbb Q$) of rank (i.e., dimension over $\mathbb Q$) equal to the cardinality of the continuum. The second piece is the direct product, over all primes $p$, of the direct product groups $\prod_{\mathbb N}\mathbb Z_p$, where $\mathbb Z_p$ is the additive group of $p$-adic integers. I believe this result is due to Balcerzyk.

Edit: I believe the Balcerzyk reference is "On factor groups of some subgroups of a complete direct sum of infinite cyclic groups" [Bull. Acad. Polon. Sci., Sér. Sci. Math., Astron., Phys. 7 (1959) 141-142.

Answer by Andreas Blass for Notation of a pregallery

2013-05-03T13:02:20Z

It seems to me that the $K$'s and $S$'s should be the same size. They play the same role as the edges and vertices in a path in a graph. The reason they are written on different levels is just to make the alternating structure easier to see. In particular, the K's are not exponents, not are the S's subscripts.

In LaTeX, I'd code this as a two-row matrix in which half of the entries are blank.

Second Edit: I just got the following email confirming what I had written above:

Dear Professor Blass,

I just saw your comment on Math Overflow on a question regarding the thesis of Harm van der Lek. Van der Lek wrote his thesis under my guidance and I can affirm that your answer is essentially correct. I am not a registered user of Math Overflow (and I have no way to find out who Callister is or what his email address is) and so perhaps you can forward to him his message (or add it as a comment).

With kind regards, Eduard Looijenga

Answer by Andreas Blass for The Kunen inconsistency and definable classes

2013-05-03T13:16:47Z

I won't try to say what set theorists generally do, but I usually handle the problem as follows. Most of the time, I work in ZFC and I use "class" to mean a class definable with set parameters. This is adequate most of the time --- for example, when I want to talk about $V$, $L$, $L[U]$, $L(\mathbb R)$, the elementary embeddings arising from measures, extenders, etc. In situations where it isn't adequate, for example in saying how Kunen's theorem goes beyond Suzuki's, I would work in ZFC with the assumption that there is an inaccessible cardinal $\kappa$, and I would (temporarily) use "set" to mean an element of $V_\kappa$ and use "class" to mean a subset of $V_\kappa$. (As long as I don't need anything of even higher rank than classes, this is pretty much equivalent to working in Morse-Kelley set-class theory. But, once I'm working in a world that goes beyond what I'm calling sets, I figure I might as well continue the cumulative hierarchy naturally rather than stopping after just one layer of non-sets.)

Answer by Andreas Blass for Approximating a function via definable functions II

2013-04-22T11:58:52Z

Let $M$ be the structure consisting of a countable universe with the following structure. There is a binary function $E$ such that, given any finitely many distinct elements $d_1,\dots,d_k\in M$ and any (not necessarily distinct) $a_1,\dots,a_k\in M$ there is some $q\in M$ with $E(q,d_i)=a_i$ for all $i=1,\dots,k$. (I'll abuse notation by letting $E$ serve as both a function symbol and its interpretation in any model, rather than writing $E^M$ here and $E^N$ below.) Let $N$ be a highly saturated elementary extension of $M$. What I need from saturation is that every function $F:M\to M$ is of the form $E(q,-)$ for some $q\in N$. Thus, all functions $F:M\to M$ are the restrictions to $M$ of functions parametrically definable in $N$. Furthermore, every such $F$ agrees on any finite subset of $M$ with $E(q,-)$ for some parameter $q\in M$. Since there are uncountably many such $F$'s and only countably many can be definable from parameters in $M$, we have a negative answer to your question. (You also wanted a topology on $M$ with a definable base, but this seems to be just a remnant of your earlier question, since the topology plays no real role here. If you really want a topology. take the discrete topology with the base consisting of singletons.)

Answer by Andreas Blass for Existence of limit measure

2013-04-20T18:48:19Z

Why isn't the following a (locally compact) counterexample? Let $X$ be the set of natural numbers, with the metric where the distance between every two distinct points is 1. So the topology is discrete, and the only balls are the singletons and the whole space. Let $\mathcal C$ consists of the finite sets and the whole space. Let $\mu_n$ be the probability measure concentrated at the point $n$. Then for each finite set $A$ we have $\lim_n\mu_n(A)=0$, while for the whole space $X$ we have $\lim_n\mu_n(X)=1$. These limits are not the values of any countably additive measure $\mu_\infty$.

Answer by Andreas Blass for Approximating a function via definable functions

2013-04-20T13:34:46Z

With your clarification (in a comment) that $U$ should be from the definable basis, the answer to your question seems to be negative. Notice first that the discrete topology on any model has a definable (with parameters, as you wrote in the question) basis, consisting of the singletons. Now you can uniformly define a family of functions $(f_a)_{a\in M}$ to be the family of constant functions, i.e., $f_a$ is constant with value $a$. Then any $F:M\to M$ whatsoever will agree locally with this family, i.e., $F$ is constant on each singleton. So you can't conclude anything about definability of $F$ (unless your model $M$ is such that all functions on it are definable, which can only happen when either $M$ is finite or its language is bigger than the model itself).

Answer by Andreas Blass for existence of equivalence checking algorithm

2013-04-17T12:34:20Z

[Edited in light of Joel Hamkins's correction] The nonexistence of an equivalence-checking algorithm follows from Rice's Theorem, whose intuitive content is that no property of the function computed by a program is decidable as a predicate of the program (except, of course, the "always true" and "always false" properties).

Answer by Andreas Blass for Isometry on a Hamming cube

2013-04-09T12:40:46Z

This might not be research-level, but here's an answer anyway. I'll take the cube to be $\{0,1\}^n$ . Let an isometry $\phi$ be given. Adding a constant vector, if necessary, we can assume that $\phi(\vec0)=\vec 0$. Then the unit vectors (the vectors with exactly one component equal to 1), being the vectors at distance 1 from $\vec 0$, are mapped to each other, in a one-to-one way. That gives us a permutation $\pi$ of the coordinate positions. Finally, to show that $\phi$ acts on all vectors $\vec v$ by permuting components according to $\pi$, proceed by induction on the number of non-zero components of $\vec v$, taking into account which vectors with fewer 1's are at distance 1 from $\vec v$.

Answer by Andreas Blass for Is there infinite generated reflexive module?

2013-04-02T17:43:48Z

Yes, over the ring of integers. The free abelian group on countably many generators clearly has, as its dual, the direct product of countably many copies of $\mathbb Z$. The dual of the latter is, by a theorem of Specker (1950) [maybe already in a paper of Baer, 1937] again the free abelian group we started with, and the canonical embedding is an isomorphism.

Answer by Andreas Blass for Grzegorczyk-hierarchy, growth-rate and functions with finite image

2013-03-25T14:35:09Z

I'm not an expert on subrecursive hierarchies, so the following idea comes with no warranty, but it looks reasonable to me. Once $i$ is not absurdly small ($i\geq 3$ should suffice), the class $\mathcal E_{i+1}$ should contain a binary function $u$ that is universal for $\mathcal E_{i}$ functions in the sense that, for every unary $\mathcal E_{i}$ function $f$, there is an index $e$ such that $f(n)=u(e,n)$ for all $n$. If that's right, then you can diagonalize, defining $g(n)=0$ when $u(n,n)>0$ and $g(n)=1$ when $u(n,n)=0$, to get a 2-valued function $g$ that is in $\mathcal E_{i+1}$ but not in $\mathcal E_{i}$.

Answer by Andreas Blass for What is the way to after finding Cohen-Macaulay semigroup as ring of monomial?

2013-03-25T14:17:47Z

The question is nearly incomprehensible, but I'll answer what I think might have been intended by the first and last numbered items. The middle one is probably asking for an expository article on Cohen-Macaulayness, which I'm not qualified to write (and which wouldn't be appropriate for MO anyway).

For the first question: The lowest tier (not counting the 0 vector) consists of the generators, which are chosen arbitrarily (except that the vectors on the coordinate axes are always present.)

For the third question (labeled 2): There is no end; the semigroups continue arbitrarily far to the right.

Answer by Andreas Blass for Surreal numbers and large cardinals

2013-03-23T15:01:52Z

I'm not aware of references that use universes in the study of surreal numbers. The reason --- for the non-existence of such references or for my non-awareness of them if they do exist --- is that there seems to be very little new to be said here. If $U$ is a Grothendieck universe, then there is a model of NBG (and in fact of the stronger Morse-Kelley theory) that has the elements of $U$ as its sets and the subsets of $U$ as its classes. Thus, whatever has been done in NBG (or MK) can be automatically translated to the setting of a Grothendieck universe.

Your idea of looking at large cardinals, for example measurable ones, in the setting of surreal numbers certainly makes sense. All the ordinal numbers, including in particular the cardinal numbers, are among the surreal numbers, so any large cardinals that exist can be regarded as surreal numbers. The key issue here will be whether something can be said about large cardinals in the surreal context that isn't just a direct, routine translation of what can be said in the usual set-theoretic context. I'm not aware of any such results, but there may well be some that have escaped my notice.

Answer by Andreas Blass for How to prove a quadratic equation has at most two roots in first order theory of field

2013-03-20T12:53:59Z

According to the OP's comment, he wants "just the important steps", and I think wccanard has provided those, for a standard proof, in the first comment to the question. Here are the "important steps" for a slightly different and perhaps quicker approach. Suppose you had three distinct solutions $x,y,z$ for the given equation. Write down the equations for $x$ and $y$, and subtract to get $x^2-y^2+a(x-y)=0$ and therefore, since $x$ and $y$ are distinct, $x+y=-a$. Similarly, $x+z=-a$. But the last two equations contradict the assumption that $y\neq z$.

Answer by Andreas Blass for Standard model of ZFC

2013-03-17T00:32:21Z

Joel has explained that the existence of standard models and a negative answer to the question follow from large cardinal axioms. He also mentioned, in the context of philosophical justification for large cardinal axioms, "the reflection paradigm, the view that truths in the whole set-theoretic universe are increasingly found in proper initial segments of it". Let me add that the existence of standard models follows from (ZFC plus) mere acceptance (in an appropriate sense) of the very notion of "truths in the whole set-theoretic universe". More precisely, suppose we add to the language of ZFC an additional predicate symbol Sat, intended to denote satisfaction, in the whole universe, of formulas of the original language of ZFC. Suppose we add to the ZFC axioms the clauses that specify Sat by induction on formulas. And suppose we also extend the axiom scheme of replacement by allowing formulas of the enlarged language. Then, by formalizing a downward Löwenheim-Skolem argument, we can prove that ZFC has standard models, in fact ones of the form $V_\kappa$. (We actually get more, namely $V_\kappa$'s that are elementary submodels of the universe, with respect to the original language of ZFC.) In other words, for the limited sort of "truths in the whole set-theoretic universe" needed in proving the existence of standard models of ZFC, the reflection principle mentioned by Joel becomes provable, once one incorporates the relevant notion of truth into the theory.

Answer by Andreas Blass for Is choice needed to establish the existence of idempotent ultrafilters?

2012-04-20T00:32:38Z

Yes, it's still weaker. To build a model of ZF in which choice fails but $\beta\mathbb N^+$ has idempotents, start with a model of ZFC (which will, of course, have idempotent ultrafilters in $\beta\mathbb N^+$). Add a lot of Cohen-generic subsets of some regular cardinal $\kappa$ well above the cardinal of the continuum; forcing conditions are partial functions of size $<\kappa$. No new reals are added, and your idempotent ultrafilters from the ground model are still idempotent ultrafilters (and AC still holds). Now pass to the symmetric submodel given by the group of automorphisms of your forcing that permutes the names of the added Cohen subsets, with the filter determined by supports of size $<\kappa$. That model violates choice, because you can't well-order the power set of $\kappa$. But the ground model's reals and ultrafilters haven't been touched, so you still have the same idempotent ultrafilters that you had to start with.

Answer by Andreas Blass for Periodic point-free maps and free ultrafilters.

2013-03-11T13:01:34Z

Although the question has already been thoroughly answered, it might be worthwhile to point out that the result can be separated into the combinatorial "meat", which doesn't involve ultrafilters, and a small corollary where the meat is fed to the ultrafilter problem. The meat is the following theorem, which is, if I remember correctly, explicit in the early references; it is essentially proved (though not stated) in Joel Hamkins's answer here. Given any set $X$ and any function $f:X\to X$, there is a partition of $X$ into four disjoint (possibly empty) sets $X=A_0\sqcup A_1\sqcup A_2\sqcup A_3$ such that $A_0$ is the set of fixed points of $f$ and each of the other three $A_i$'s is disjoint from its image under $f$. Once one has this result, one immediately sees that any ultrafilter on $X$ must contain one of the four $A_i$'s and if that $A_i$ isn't $A_0$ then the ultrafilter can't be $f$-invariant as in the question.

Answer by Andreas Blass for A yes no question concerning induced group

2013-03-09T18:48:26Z

The "right" way to make $G$ act on $\Omega$ is $\hat g(\omega)(x)=\omega(g^{-1}(x))$. Not only does this give you a left action (rather than a right action), but it fits nicely with the "set of ordered pairs" view of functions. If you regard $\omega$ as set of ordered pairs $(x,\omega(x))$, then the action of any $g\in G$ amounts to applying $g$ to the only place where it makes sense, namely the first components of the ordered pairs.

Answer by Andreas Blass for Reflection principles

2013-03-07T14:33:28Z

I suppose the "paradox" you're asking about is the passage marked with >> at the link you gave, but with "$\omega$-model" in place of "model" and with "has an $\omega$-model" in place of "is consistent". But then there is no longer any justification for the statement (on lines 9 & 10) that there's a proof in ZFC of the negation of con(ZFC) (which now becomes the negation of "ZFC has an $\omega$-model"). What you have is rather that this negation holds in all $\omega$-models of ZFC, but that doesn't immediately translate into a syntactic fact about existence of a proof, which you could then translate into English.

I conjecture that, if you write down carefully just what the "paradox" is supposed to be, it will disappear.

Answer by Andreas Blass for On the notion of partial semigroup

2013-03-05T13:43:56Z

The "right" definition of "partial semigroup" probably depends on the use one wants to make of these structures. Vitaly Bergelson, Neil Hindman, and I needed partial semigroups in our paper "Partition Theorems for Spaces of Variable Words", and we used the definition that says if either of $(x*y)*z$ and $x*(y*z)$ is defined then so is the other and they are equal. That seems quite a natural definition, and it worked well for our purposes. I don't know (though I may have known when working on the paper) whether some other definition would have worked as well. [The paper is in Proc. London Math Soc. (3) 68 (1994) pp. 449-476, and a version of it is on my web site at http://www.math.lsa.umich.edu/~ablass/bbh.pdf .]

Answer by Andreas Blass for Why do we choose the standard total order on the integers?

2013-02-27T21:21:17Z

The standard order is (up to isomorphism) the only total order on $\mathbb Z$ that makes it an ordered group under addition. So I'd expect it to be useful in situations where addition plays a role; these are probably most (though certainly not all, as Ryan Budney pointed out in a comment to the question) of the situations that arise in practice

Answer by Andreas Blass for A question of Allan on infinite divisibility

2013-02-26T14:02:37Z

What's to stop you from just freely building a counterexample? That is, let the ring be generated by elements $x,a,b_1,b_2,\dots,b_n,\dots$ subject to (only) the relations $a=x^nb_n$ for all positive integers $n$. Then $a$ is in $I(x)$. Unless I'm making a stupid mistake, the only solutions $q$ of $a=xq$ are finite linear combinations (with integer coefficients adding to 1) of the elements $x^{n-1}b_n$, and none of those are in $I(x)$.

Answer by Andreas Blass for What can the degrees of constructibility be?

2013-02-22T02:06:18Z

In addition to the work on what can occur in the constructibility degrees, there's a theorem of Bob Lubarsky about what cannot occur. If the lattice of constructibility degrees is countable and has a top element, then it must be a complete lattice. [Lattices of c-degrees, Ann. Pure Appl. Logic 36 (1987) 115-118]

Answer by Andreas Blass for On duality on finite projective planes

2013-02-11T18:54:05Z

I'd expect that, in the duality principle that you quoted from "most (if not all) projective geometry texts", the symbol $\mathfrak P$ refers to the theory of projective planes, not to an arbitrary particular projective plane. One reason for this expectations is that theories, not planes, are the sort of entity that can "have" theorems (planes can satisfy statements, including theorems). Another reason is the observation you noted in your question; it's possible for a projective plane to satisfy a statement but not the dual statement. A final reason in favor of my expected interpretation of $\mathfrak P$ is that it makes the principle of duality true.

Answer by Andreas Blass for Building a scale by diagonalization and transfinite induction

2013-02-04T19:24:41Z

The inequality relation between functions $\omega\to\omega$ that is written $f<g$ in the question must mean that $f$ is eventually (not everywhere) below $g$, i.e., that $f(n)<g(n)$ holds for all sufficiently large $n$. (If it meant "for all $n$, then the induction would indeed break down at the first limit ordinal.) Now to construct an upper bound for countably many functions $f_\alpha$, first re-enumerate them in an $\omega$-sequence, say as $g_n$, and then use the function $k\mapsto 1+\max\{g_n(k):n<k\}$ .

By the way, the usual notation for the "eventually $<$ " relation is not $<$ but $<^*$ .

Answer by Andreas Blass for Is the empty graph a tree?

2013-02-02T14:32:16Z

I don't consider the empty graph to be a tree, or a connected graph, because I prefer the following definition of connectedness: A graph $G$ is connected if, whenever it is the disjoint union of a family of graphs, then one of the graphs in that family is $G$ itself. The empty set does not satisfy this, because it is the disjoint union of the empty family.

A category-theoretic version would define connectedness to mean that, whenever $G$ is expressed as a coproduct, one of the coproduct injections must be an isomorphism. That's less elegant than the "Hom preserves coproducts" definition in Todd Trimble's answer, but I think it's closer to common, non-category-theoretic intuition.

The same style of definition deals (in my opinion correctly) with the question whether 1 is prime. Define a positive integer $p$ to be prime iff, whenever it is expressed as a product, one of the factors is $p$ itself. This definition makes 1 not prime, because it is the product of the empty family.

Answer by Andreas Blass for Models of ZFA corresponding exactly with a particular class of groups

2013-01-23T16:16:21Z

The answer to the first question is negative, because the permutation model generated by $G$ and $\mathcal F$ doesn't really "see" the group $G$. Consider the following "shrinking" operation. Replace $G$ by a subgroup $H$ that belongs to $\mathcal F$ and replace $\mathcal F$ by its restriction to $H$ (i.e., $\{K\in\mathcal F:K\subseteq H\}$ ). Then the permutation model doesn't change, but whether $\mathcal F$ satisfies the hypotheses of Theorem 1 can change. Specifically, if $\mathcal F$ is a Ramsey filter, as witnessed by a subgroup $H$, then using this $H$ in the shrinking construction produces the same permutation model but now with a group and filter satisfying the hypotheses of Theorem 1. So no property of the permutation model can exactly match the hypotheses of Theorem 1. (You could think of "Ramsey filter" as the hypothesis of Theorem 1 made invariant under (inverse) shrinking so that it can be a property of the permutation model.)

The same observations give a positive answer to your second question, with $\phi$ being BPIT itself. The desired $G'$ and $\mathcal F'$ are obtained by suitable shrinking as above.

Answer by Andreas Blass for A question on hereditary Lindelof number

2013-01-19T15:02:54Z

In one direction, suppose $\kappa$ is a cardinal and $X$ has a subspace $Y$ with an open cover $\mathcal U$ that has no subcover of size $<\kappa$. Define in parallel a $\kappa$-sequence of points $y_i\in Y$ and a $\kappa$-sequence of sets $U_i\in\mathcal U$ by the following induction of length $\kappa$, in which one new $y_i$ and one new $U_i$ are chosen at each stage. At any stage, there are fewer than $\kappa$ $U_i$'s chosen at previous stages, so there is a point in $Y$ not covered by those $U_i$'s; choose one and make it the next $y_j$ in your sequence. Then choose a set in $\mathcal U$ that contains this $y_j$ and make it the next $U_j$. After $\kappa$ steps, the chosen $y_i$'s form a set $S$ of size $\kappa$, and it is right separated because any initial segment, say up to but not including $y_i$, is the intersection of $S$ with $\bigcup_{j<i}U_j$ .

For the other direction, suppose $X$ has a right-separated subset $S$ of size $\kappa$. We may assume that the length of the well-ordering witnessing right-separation is $\kappa$, because if it isn't we can just delete some elements from the end and retain only the first $\kappa$ points of $S$. The proper initial segments of $S$ form an open cover of $S$. If $\kappa$ is a regular cardinal, then this cover of $S$ has no subcover of size $<\kappa$, so we're done. If $\kappa$ is singular then there is a subcover of size cf$(\kappa)$, so we have to work a little harder. The previous argument can be applied to every regular cardinal $\lambda<\kappa$, in particular to every successor cardinal $<\kappa$. It gives a subset of $X$ with an open cover of size $\lambda$ with no subcover of smaller cardinality. So the hereditary Lindelöf number of $X$ is at least $\lambda$. Since these $\lambda$'s are cofinal in $\kappa$, we're again done.

Answer by Andreas Blass for Can one always Decide whether a Systems of Nonlinear Equations with Bilinear terms is Feasible?

2013-01-17T17:42:18Z

Over the reals, this sort of question is decidable, because it's in the first-order theory of the real field, which is decidable by an old theorem of Tarski. Over the rationals, this is as hard as the general question of solvability of Diophantine equations. The point is that, even if a Diophantine equation involves products of more than two variables, it's equivalent to a system in which additional variables are introduced for sub-products, and these can be defined by quadratic equations. And quadratic expressions that aren't bilinear, like $x^2$, can be replaced by $xy$ where $y$ is fresh variable and the equation $x=y$ is added to the system.

Answer by Andreas Blass for A question about well ordered subsets of totally ordered countable sets

2013-01-15T19:02:10Z

If a linear order doesn't contain a copy of $\mathbb Q$, then, by a theorem of Hausdorff, it can be obtained by a transfinite sequence of steps, starting with singletons, and at each step forming well-ordered or reverse-well-ordered sums of previously constructed orderings. If the final result is to be a countable set, then the transfinite sequence will be only countably long, and each of the well-ordered or reverse-well-ordered index sets used along the way will also be countable. I believe this will allow a proof, by induction along the transfinite sequence, that at no stage does it become possible to embed arbitrarily large countable ordinals. Unfortunately, I don't have time to work out the details right now. I'll come back to it later if no one else does it first.

Comment by Andreas Blass

2013-05-21T16:23:01Z

Is there a clear and reasonable criterion for when two constructions are different?

Comment by Andreas Blass

2013-05-21T16:04:11Z

Actually, the first version of choiceless polynomial time was proposed by Shelah and was (not surprisingly) fairly complicated. The "official" definition resulted from my and Yuri's effort to understand (and write up) Saharon's original proposal. I confess that I've forgotten all details about the original version, but I believe that Saharon continued to use it, for example in reference [15] of the paper that Asaf linked to. Note that the linked paper began as Yuri's and my effort to simplify and explain [15].

Comment by Andreas Blass

2013-05-20T17:36:21Z

Now I can add "DAG graph" to my pet peeve list that includes such repetitions as "ICBM missiles", "AC current", "ATM machines", and "PIN numbers" (the last of which are occasionally called "personal PINs").

Comment by Andreas Blass

2013-05-19T19:11:07Z

If one works in ZFA, the version of ZF that allows atoms, then the basic Fraenkel model and the second Fraenkel model (as defined in Jech's book "The Axiom of Choice") show the consistency of "there is a set with no linear order", while Mostowski's linearly ordered model shows that linear orderings of all sets don't yield the axiom of choice. So these ZFA consistency results go back to 1922 and 1937 respectively. The former transfers automatically to ZF by the Jech-Sochor metatheorem. I haven't checked whether perhaps the latter transfers also by Pincus's stronger metatheorems.

Comment by Andreas Blass

2013-05-19T18:45:53Z

I'd go a bit further than "there are very similar proofs ... using independent sets and independent partitions." I think of the existence of continuum many independent partitions of $\mathbb N$ and the separability of a product of continuum many separable spaces as being essentially the same theorem. Each is deducible easily from the other. Being more combinatorial than topological, I tend to view the former (and its generalizations to higher cardinals) as the main point, and to view separability as a nice way to make it look topological for those whose tastes differ from mine.

Comment by Andreas Blass

2013-05-19T18:40:24Z

Concerning "the Rudin-Keisler ordering measures the size of an ultrafilter": Did you mean "the size of an ultrapower"?

Comment by Andreas Blass

2013-05-18T19:12:46Z

Any $\kappa^+$-saturated real-closed field has the property you asked for. Since you allow the field to be significantly larger than $\kappa^+$ if necessary, there's no difficulty producing such fields.

Comment by Andreas Blass

2013-05-17T17:46:35Z

"Random numbers" makes sense only in the context of a specified probability distribution.

Comment by Andreas Blass

2013-05-16T20:50:54Z

The $M$ at the end of my previous comment should have been $nM_n$, i.e., the sum of the $X_i$'s.

Comment by Andreas Blass

2013-05-16T20:27:45Z

The sentence beginning "I get that" seems confused. $M_n$ is the sum, not the convolution, of the $X_i$'s, divided by $n$. Convolution is what gets done to the distribution functions of the $X_i$'s in order to produce the distribution function of $M$.

Comment by Andreas Blass

2013-05-16T20:21:43Z

@Wlodzimierz: No harder than Abel, which I misspelled in my first comment.

Comment by Andreas Blass

2013-05-16T12:40:45Z

In an edit correcting the misspelling of "Balcerzyk" in my original answer, I added a reference that includes a new misspelling of the same name. I'll edit again, but if I produce yet another error, I'll just give up.

Comment by Andreas Blass

2013-05-16T12:22:43Z

A possibly more accessible reference is volume 1 of Fuchs's book "Infinite Ablian Groups", page 177.

Comment by Andreas Blass

2013-05-15T20:00:08Z

"Using elemeentary mathematics" does not mean "easily". Elementary proofs are often much harder than proofs of the same theorem using more sophisticated machinery.

Comment by Andreas Blass

2013-05-12T19:27:13Z

If I correctly understand the OP's four most recent comments, they are 3 to 1 in favor of convexity. Because of that and because the non-convex case has been given easy answers, I suggest we assume "convex" is intended.