User garabed gulbenkian - MathOverflow

A question about large real closed fields

2013-05-18T18:54:42Z

A real closed field can be ordered in one and only one way, and is therefore provided with a unique order topology. Given any infinite cardinal number k, does there always exist a real closed field F (whose cardinal number is greater than k), such that no non-empty subset of F having a cardinal number not greater than k has a limit point in F? The criterion for a point of F to be a limit point of a subset of F is dtermined by the order topology of F. Since we are dealing here with arbitrarily large cardinal numbers, let us assume that we are working within the set theory ZFC.

A question about simple arcs in higher dimensional Euclidean spaces.

2013-05-06T17:48:10Z

Let E(n) be n-dimensional Euclidean space. It is known that there exist subsets of E(n) which are simple arcs and have positive n-dimensional Lebesgue measure when n=1 or 2. Does this continue to be true for arbitrarily large n? If not, what is the largest n for which it holds and is there a simple proof of this fact? Intuitively, I feel that there should be no upper bound, but cannot see how to prove it.

A question about "paradoxical" sentences in the language of ZF set theory.

2013-05-05T17:55:44Z

Let F(x) be a formula belonging to the language of first order ZF in which x is the one and only variable that occurs free and let N(x) be the negation of F(x). Are any examples known of an F(x) and an N(x) such that not just one but each one of the two sentences "The set (x:F(X)) exists" and "The set (x:N(x)) exists"-which can be expressed in the language of first order ZF-are inconsistent with ZF? An obvious candidate for F(x) is the formula which defines the Russell set but in that case N(x) defines the null set (because of the Axiom of Foundation in ZF) and there is no inconsistency.

A question about indecomposable continua.

2010-09-01T19:15:49Z

The term "continuum" is often used to mean a compact and connected metric space. But it is also used in a broader sense to mean any infinite, complete, separable and connected metric space-which is not necessarily compact. This is the sense in which we use it here. A "continuum" is called "indecomposable" if it is not the union of two of its proper infinite subsets, each of which is itself a "continuum". It is known that a compact "indecomposable continuum" has uncountably many proper infinite subsets that are themselves "continua". If C is a non-compact "indecomposable continuum" and S is the set of all its proper infinite subsets that are themselves "continua", what can be said about the cardinal number of S?

A question about Kunen's inconsistency theorem

2013-04-13T18:17:51Z

It would seem as though the sentence s(E) which expresses the existence of a non-trivial elementary embedding of the universe V into itself-and which can be formalized in the first order language of NBG-could also be formalized in the language of Quine's NF. These two set theories would seem to share the same first order language (since NBG does not really need separate variables and quantifiers for sets and for proper classes.) I am interested in this question because I understand that Kunen's proof of the inconsistency of NBG+s(E) depends upon the axiom of choice (which is taken as one of the axioms of NBG.) Now the axiom of choice does not hold in NF (or holds at most for Cantorian classes- and V is not a Cantorian class.) Is it possible that NF+s(E) might be consistent if NF is, and should this be the case, would the set theory NF+s(E) be of any interest in its own right? I realize there may be some very simple facts-which I am missing-that imply my questions obviously have negative answers and probably should be closed. But I'll take the chance and ask.

A question about connected subsets of metric spaces

2013-04-08T19:10:30Z

Let M be a metric space. Let T(M) be the topology of M (i.e. the collection of all open subsets of M) and let C(M) be the collection of all connected subsets of M. In my opinion one often has a much clearer intuitive picture of the topology of a metric space M if one knows C(M), than if one knows T(M). I would be interested in characterizing the largest collection S of metric spaces M for which T(M) is uniquely determined when C(M) is specified. To show that all Banach spaces belong to S, let the term "*continuum" denote any non-empty closed and connected (but not necessarily compact) subset of a metric space. We can define *continua in terms of connected sets by saying that a *continuum is either identical with the metric space M (if M is connected) or else it is a non-empty proper subset Q of M such that the union of Q and the singleton of p is not connected, whenever p is a point of M that does not belong to Q. Now if B is a Banach space whose dimension is not less than 2, then the complement of each bounded open ball of B is a *continuum. Hence the complements of the *continua of B are a base for the topology of B. If B has dimension 1 and if H is any unbounded proper subset of B that is open and connected, then the complement of H is a *continuum. Hence the complements of the *continua of B are a sub-base for the topology of B. This proves that the collection S contains all Banach spaces. But could the collection S actually include all complete, connected and locally connected metric spaces?

A question about closed curves

2013-03-28T19:56:45Z

Does three dimensional Euclidean space contain unbounded closed curves that do not cross themselves? It does not seem possible to find examples of such anmd yet it is not clear just what is standing in the way. More precisely, let n be any positive integer not less than 3 and let E(n) be n-dimensional Euclidean space. Does there exist a subset S of E(n) with the following properties that is not compact? (1)S is closed, connected and locally connected. (2) If any point of S is removed, the resulting space is still connected and is homeomorphic to a straight line. (3) If any distinct pair of points of S are removed, the resulting space is no longer connected and has two components. Finally can S be a non-compact subset of a separable and infinite dimensional Hilbert space?

A question about formalized theories that may be both consistent and w-consistent

2013-03-11T18:58:21Z

Let T be a first order set theory formalized in the language L(ZF) of ZF, which has "membership" and "=" as its only atomic predicates. For each positive integer n, let P(n) be the sentence which expresses that "there exists a set having exactly n elements". P(n) can be formalized in L(ZF) and is an axiom of T for each positive integer n. Note that L(ZF) does not need to contain any terms that are constants for this to be possible. Now let Q be the sentence stating that "there exists a finite set (in Tarski's sense of finite) which can be mapped onto every non-empty set". Q is also an axiom of T that can be formalized in L(ZF) and should be consistent with every finite collection of sentences of the form P(n). However with infinitely many axioms it would seem appropriate to call T at least a w-inconsistent theory. But could T still be consistent? The answer is not clear to me since it would depend upon what other axioms T has. T must have some other axioms (such as the axiom of pairing) in order to define the mapping of one set onto another as the sentence Q describes. We must be careful that these additional axioms are not inconsistent with Q. We should probably not want a power set axiom, for example. But, with a proper choice of the new axioms could we end up with a consistent T? If so, T would be an example of a consistent but w-inconsistent formalized theory whose language need contain no constants and no formulae of the form N(x) intended to be interpreted as "x is a positive integer".

A question about Universal sets.

2013-02-22T19:55:15Z

In several set theories, among which Quine's NF is one of the best known and most extensively investigated, the existence of a Universal set V can be proved. V is the set of all sets-indeed all "proper classes", including V itself, are elements of V. Although it might be considered "meaningless" to ask what the "cardinal number" of V is, there are some questions of this type which one could ask. As far as I know, the consistency of NF relative to ZF is still an open question. But have any of the well known "large cardinal axioms" (assuming that they can be expressed in NF) been proved to be inconsistent with NF? If so, this might throw some light on the question of how "large" V really is.

A question about definable non-empty sets containing no definable elements.

2013-02-11T15:01:11Z

Can anyone provide an example of a set S which is definable in ZFC and provable in ZFC to be denumerably infinite, while at the same time, no set definable in ZFC can be proved in ZFC to be an element of S? Such examples are easy to find if S is allowed to be uncountable-for instance S could be the set of all non-measurable sets of real numbers. In most of the examples where S is uncountable the axiom of choice is needed just to prove that S is non-empty and the cardinal number of S is greater than the cardinal number of R.

Does there exist a non-trivial Ultrafinitist set theory?

2012-09-23T19:17:57Z

Does there exist a set theory T-which has not yet been proved to be inconsistent-and in which one can prove the existence of (1) the empty set (2) sets that are singletons and (3) sets which have non-empty proper subsets. T has no axiom of infinity but-as with Quine's NF-one can prove in T the existence of a universal set (i.e a set of all sets). However-unlike Quine's NF-the universal set of T should be finite. One can think of T as being formalized in the classical first order predicate calculus, using the same language as ZF. My motive in seeking a set theory such as T is to find out whether there exist set theories that might be acceptable to an ultrafinitist (as conforming to the principles of that viewpoint), while still allowing a certain amount of arithmetic to be carried out in them.

A question about well ordered subsets of totally ordered countable sets

2013-01-15T18:37:34Z

Let us assume ZFC and let Q be the set of rational numbers ordered according to size. There is a well known theorem which implies that if S is any totally ordered countable set containing a subset ordinally similar to Q, then S contains well ordered subsets having arbitrarily large countable ordinal numbers. Is there a converse to this theorem which implies that if S has the second of these properties, then it has the first? I have been unable to find any mention of such a converse theorem or to come up with any obvious counter-examples. (I apologize if my question is not considered appropriate for "mathoverflow.net")

A question about a hierarchy of metric spaces arising from an operation defined by Hausdorff.

2013-01-02T19:16:10Z

Given any metric space M, Hausdorff defined a new metric space h(M) whose "points" are the non-empty closed and bounded subsets of M. The hierarchy emerges from the following iteration process. Let H(0,M)=M and for each non-negative integer n, let H(n+1,M)=h(H(n,M)). If M is (for example) a finite dimensional Euclidean space or Hilbert space, does this process ever reach a fixed point-in the sense that there exists a non-negative integer k for which the spaces H(k,M) and H(k+1,M) are homeomorphic (or even isometric)? If there is no fixed point in the case of some particular metric space M, is it possible to continue this iteration process into the transfinite?

A question about the Axiom of Choice and straight lines in the Euclidean plane.

2012-11-22T20:18:39Z

Let E be the Euclidean plane. Does there exist a collection C of subsets of E whose union is E and which are all straight lines such that (1) No two distinct straight lines belonging to C are parallel (2) Every subset of E which is a straight line not belonging to C is parallel to exactly one straight line belonging to C and (3) No three pairwise distinct straight lines belonging to C are concurrent? The problem is trivial, of course, without condition (3). All the straight lines belonging to C can intersect in the same point. If such a collection as C exists, could anybody give an example of one. Or is some version of the Axiom of Choice needed to prove its existence? Finally, while still requiring conditions (1) and (2) to be fulfilled, one could ask whether it is possible for the union of all the straight lines belonging to C can be a closed proper subset of E-rather than E itself. But this, I suspect, is impossible-even with the Axiom of Choice and even though we have not imposed condition (3).

A question about local connectedness

2010-08-23T20:12:21Z

Let C be a connected and completely metrizable subset of the Euclidean plane. Can C fail to be locally connected at each of its points?

A delicate measure-theoretic question about Jordan curves and arcs in the plane.

2012-11-13T19:05:49Z

Let E be the Euclidean plane and let M(X) be two-dimensional Lebesgue measure defined for each Borel subset X of E. Suppose that s is an arc in E and that e is a positive real number. Does there always exist a bounded connected open subset Z of E such that (!) s is a subset of Z (2) M(Z)-M(s) is not greater than e (3) Z is the interior of a Jordan curve? It is not hard to show that such a Z exists if only conditions (1) and (2) are required to be satisfied. But how does one show that Z can be the interior of a Jordan curve, or even be simply connected? Remember that s can have an infinity of wiggles and can have positive two-dimensional Lebesgue measure.

A question that arises in trying to make mathematically precise a well known informal statement about analytic functions

2012-11-09T20:13:16Z

It is often stated that a single-valued analytic function f(z) is uniquely and completely determined if (1) it is analytic at all points of a convergent sequence of points in the complex plane and at their limit point and (2) one is given the points of the sequence and the values of f(z) at each of these points.

Let z(1),z(2),...,z(n)... be a convergent sequence of complex numbers which are strictly decreasing in absolute value as n increases, and whose limit point is zero. Let f(z) be analytic at all the points of this sequence and at their limit point. Supposing that for each positive integer i one is given z(i) and f(z(i)). Does there then always exist a unique power series P(z) centered at zero such that (3) the radius of convergence R of P(z) is positive (or infinite) and (4) if k is any positive integer for which the absolute value of z(k) is less than R, P(z(k))=f(z(k))?

If such a unique power series exists, how do we obtain its coefficients from the data we are given? One can set up an equation for these unknown coefficients involving two infinite column vectors and an infinite Vandermonde matrix. The rows of the matrix are all of the form 1,z(j),z(j)^2,z(j)^3...where j is a positive integer. But I do not know what conditions are needed to insure that such matrices have a unique inverse.

Is this a "folk theorem" about analytic functions of a complex variable?

2012-10-24T15:56:37Z

In a comment on question 110345 I made a claim that might be incorrect. I claimed that if f(z) is a non-constant analytic function defined by a power series whose circle of convergence C has a positive radius, then f(z) cannot be bounded at all points in the interior of C. But is this really true? Or am I just imagining that I learned it somewhere. I could not come up with any simple counter-examples. It sounds like some weird generalization of Liouville's theorem.

A question about the comparability of large cardinals.

2012-10-22T20:03:40Z

Are there any examples of two large cardinal axioms $AX$ and $AY$, in the language of first order $ZFC$, which satisfy the following conditions.

Each of them defines a unique cardinal number - $C(AX)$ for $AX$ and $C(AY)$ for $AY$ - not like the axiom of measurable cardinals which defines a whole collection of cardinal numbers.
If $T$ denotes the theory $ZFC+AX+AY$, then $T$ has not yet been proved inconsistent.
$T$ proves that each of $C(AX)$ and $C(AY)$ is larger that the smallest strongly inaccessible cardinal number.
The sentences of $T$ stating that $C(AX) < C(AY)$ and that $C(AY) < C(AX)$ are each consistent with $T$, if $T$ is consistent.

A question concerning a well known "law" about triangles.

2012-10-18T17:42:02Z

Let a,b,c denote the lengths of the sides and let A,B,C denote the corresponding opposite angles of a triangle. In the Euclidean plane we have the law of sines. a/sin(A)=b/sin(B)=c/sin(C). A recent article pointed out that these 3 equal ratios are also all equal to the diameter of a circle whose circumference contains the 3 vertices of the triangle. There is a similar law of sines for spherical triangles. sin(a)/sin(A)=sin(b)/sin(B)=sin(c)/sin(C). Are these 3 equal ratios also all equal to some important geometric quantity associated with the spherical triangle? If so, what is this quantity and does the same sort of phenomenon (involving an analogous "law of sines") occur with triangles in some other metric spaces-such as hyperbolic space?

A question about formulating first order axioms for group theory.

2012-10-20T20:14:49Z

Let T be a first order theory whose set of axioms is one of the standard finite sets of axioms for an arbitrary group. Is it possible to have an axiomatizable first order extension of T which would characterize just those groups that are finite, simple and "sporadic"? In other words I am asking whether the properties of being finite, simple and "sporadic" can each be expressed by a recursively enumerable (or perhaps even finite) set of first order axioms? I realize, of course, that if we know the "Monster" to be the largest such group and know its cardinal number, then our task becomes considerably simpler. But I would consider this to be cheating and don't want to assume that we have such knowledge. I confess that I don't know much about group theory and have never seen a clear precise statement of what it means for a group to be "sporadic". Otherwise I would have refrained from asking these questions which I suspect will all sound stupid to a group theorist.

A question about first order theories having only finite models.

2012-10-16T20:31:41Z

Suppose all the models of a first order theory are finite, have the same cardinal number, and are isomorphic. Is the theory then necessarilly complete? Normally I would not ask such a"specialized" question on "mathoverflow.net", a question whose answer must be well known. However I have been unable to find an answer after quite a bit of searching through various books and articles that deal with finite models. I did find a converse-completeness implies categoricity for finite models. Nevertheless, I am almost sure my question has a "yes" answer though I don't quite see how to prove it.

Some questions about polyhedra in 3-dimensional Euclidean space, E3

2012-10-14T15:01:27Z

Let $e$ and $d$ be real numbers such that $0 < e < d$. Are there known functions $B(e,d)$ that are upper bounds (close to or even equal to least upper bounds) for the surface area of the boundaries of (not necessarily convex) polyhedra in $E^3$ which have a diameter not greater than $d$ and every distinct pair of whose vertices have a distance apart not less than $e$? It is easy to construct examples showing that if we keep $d$ fixed and allow $e$ to approach 0, then $B(e,d)$ approaches infinity. Questions like this arise in connection with some recent theories of physics in which space (and perhaps also time) is "quantized". There is a minimum length $e$. Furthermore, the maximum amount of information that can be contained in any bounded region of space is limited. This limit is proportional, not to the volume of the region, but to the surface area of its boundary. One final question: Are there any simple necessary and sufficient conditions on a finite set of points $S$ in $E^3$ for there to exist a (convex or non-convex) polyhedron whose set of vertices is identical to $S$?

A question about recursively enumerable sets of rational numbers

2012-10-13T18:23:47Z

Let (Q*,<) denote the ordered set in which the elements of Q* are just the positive rational numbers less than 1 and "<" is the ordering relation of the ordered field (of all rational numbers) Q. Let f be a fixed effective enumeration (without repetitions) of Q*-many examples of which are well known. It is also well known that if S is any denumerable ordered set, there exist many subsets X of Q* such that (X,<) is ordinally similar to S. Consequently there exist many sets N* of positive integers which have the property that (f(N*),<) is ordinally similar to S. I am interested in the case where S is well-ordered. MY question is this. If N* is any recursively enumerable set of positive integers such that (f(N*),<) is well-ordered, is its ordinal number always constructive (in the sense of Church and Kleene)? Or if larger non-constuctive ordinal numbers are attainable by means of this procedure, how large can they get? Various theorems about recursive ordinal numbers make it seem likely that one would not be able to reach beyond these numbers in this way. But one aspect of the situation renders it difficult to work out a straightforward solution to this problem. If one thinks of recursively enumerable sets such as N* as being generated by Turing machines, the problem of determining whether the resulting ordered set (f(N*),<) is well-ordered appears to be as computationally unsolvable as the the general halting problem for Turing machines-if not more so.

Could this peculiar set theory be of any interest even though it is trivial?

2012-10-11T20:47:31Z

.......Let T denote the first order theory of dense linear order with no infimum and no supremum. The only atomic formulae in the language of T are "$x \lt y$" and "$x=y$" (where "x" and "y" are variables). This theory T is axiomatizable (it has about 6 or 7 axioms). T has no finite models and is categorical in the smallest infinite cardinal. Therefore, by a theorem of Vaught, T is consistent, complete and decidable. One would hardly call T a trivial theory, since many of its theorems are used to prove results in more general theories of ordered sets. .......The first order set theory T* is obtained from T by substituting the symbol for set membership for the symbol "$\lt$" wherever "$\lt$" occurs (in all the formulae of T). The axioms of T* are the transformed axioms of T after the substitution. The language of T* is the language of first order ZF. Then (although there may be a gross error here) I claim that T* shares with T the properties of being consistent, complete and decidable. Every sentence of first order ZF can be proved or disproved in T* (including all the sentences that are undecidable). If anybody wants to know what T* has to say about the Continuum Hypothesis, they can even work it out for themselves-because T* is decidable. Nevertheless, T* seems like the most trivial set theory that it is possible to have. What can one do with it? Not even the weakest sub-theory of arithmetic appears to be interpretable in it. What can one say about it? The set-theoretic universe of T* is infinite, yet not one of the sets that it contains is specifically definable within it. It violates the axiom of foundation in the worst possible way. It negates almost all axioms of comprehension that (in other set theories) are used to prove the existence of sets satisfying a variety of conditions. Last, but not least T* thumbs its nose at Godel"s incompleteness theorem. Can it be be that all trivial set theories are not created equal?

A question about Quine's set theory NF.

2012-09-30T18:42:15Z

This question might not really be considered appropriate for mathoverflow.net but I'll risk asking it and apologize in advance if I have commited a booboo. It is often said that in NF one can prove the existence of infinite sets without the help of any special axiom of infinity. Now Tarski's definition of an infinite set (which includes more sets than Dedekind's definition when the axiom of choice is not available) states that a set X is infinite just in case there exists a non-empty set T of subsets of X such that if we are given any element u of T there is an element v of T which is a proper subset of u. In order to prove that even the universal set V is infinite one would need to exhibit a specific non-empty sub-cllection V* of V which is certified to be a set in NF and is such that for every element y of V* there is an element z of V* which is a proper subset of y. I have never seen a specific example of such a set V* and cannot think of how to define one. Note that most infinite sub-collections of V are not sets in NF because of the stratification requirements. Is there (an example of) such a V* and if not how can one really say that NF proves the existence of infinite sets?

A question about J.H. Conway's SURREAL NUMBERS

2011-09-29T20:17:35Z

My quesion is: What set theory are the mathematicians who are developing the theory of these numbers working in-or are they, in fact, working outside any of the standard set theories?. Each surreal number is a mapping of an ordinal number into the pair (+,-) so that the collection S of all these numbers is a proper class. Moreover S is a real closed (ordered) field containing sub-collections which are ordinally similar to the class of ordinal numbers and to the set of real numbers (in their usual order). Since S is densely ordered but not order-complete, there exists an order-complete ordered collection C (constructed from the Dedekind cuts of S), which contains a dense sub-collection that is ordinally similar to S. Now the elements of C are proper classes and if we are going to have theorems about sub-collections of C (such as closed intervals), then the underlying set theory (if any) must be one that allows some proper classes to be elements of collections.

A question about ordinal definable real numbers

2010-03-09T14:50:01Z

If ZFC (Zermelo-Fraenkel set theory with the Axiom of Choice) is consistent, does it remain consistent when the following statement is added to it as a new axiom?

"There exists a denumerably infinite and ordinal definable set of real numbers, not all of whose elements are ordinal definable"

If the answer to the above question is negative, then it must be provable in ZFC that every denumerably infinite and ordinal definable set of real numbers is hereditarily ordinal definable. This is because every real number can be regarded as a set of finite ordinal numbers and every finite ordinal number is ordinal definable. Garabed Gulbenkian

A question about local connectedness in metric spaces

2012-01-19T20:08:42Z

Must every compact and connected metric space be locally connected at at least one of its points?

A question about how far projective geometry can be extended

2012-01-07T21:13:11Z

Is there a natural way to extend or embed a separable and infinite-dimensional real Hilbert space H into a "projective space" (i.e. a space which is "infinite-dimensional" and satisfies some standard version of the axioms of projective geometry.)? In other words, is there an "infinite-dimensional" analogue of the situation in which finite-dimensional Euclidean spaces can be considered to be sub-spaces of projective spaces of the same dimension? If such an "infinite-dimensional" projective space can exist would it's topology necessarily make it non-metrizable?

Comment by Garabed Gulbenkian

2013-05-20T18:57:05Z

In my question Y was the Euclidean plane and I could obtain its "non-standard" analogue Y* by using a two-dimensional "Euclidean" distance function. This satisfies all the axioms for a metric and is defined for all the elements of each real closed field.

Comment by Garabed Gulbenkian

2013-05-20T18:38:08Z

Joel, in reference to that question: Given any metric space Y (in the classical sense) and a set R* of standard and non-standard real numbers (which I assume is some real closed field like F in my question) how does one define the non-standard analogue Y* of Y? If d,d* are the respective distance funcions of Y,Y*-how does one define d* when d is given? Many standard metrics involve functions which are exponential or inverse trigonometric and which could cause difficulties when one tried to extend them to larger real closed fields.

Comment by Garabed Gulbenkian

2013-05-19T15:07:40Z

My question about limit points arises in trying to work out whether- and if so, how-it might be possible to define some sort of "completion" for F^2 that still allows the "metric" that we have defined to exist.

Comment by Garabed Gulbenkian

2013-05-19T14:56:33Z

Many thanks for this helpful information. If (x(1),y(1)) and (x(2),y(2)) are elements of F^2, then the formula: ((x(2)-x(1))^2)+((y(2)-y(1))^2)^(1/2) satisfies all the axioms for a metric on the "space" F^2, although the the resulting "distance" is a non-negative element of F and not necessarily a non-negative real number. Of course if F is sufficiently large, then F^2 is not metrizable (in the classical sense) and although it is dense in itself, it is not "complete".

Comment by Garabed Gulbenkian

2013-05-11T15:06:24Z

Your arguments and examples convince me that if my set theory T is to be a sub-theory of ZF, neither Foundation nor Replacement can be an axiom of T. Otherwise there will be too many sets X such that the statement "X does not exist" will already be a theorem of T-and my question will be too easy to answer. In any case no pair of sets that are complements of one another can ever both exist or else the axiom of Union will allow the Universal set to exist and Russell's paradox to be derived (from the axiom of Separation).

Comment by Garabed Gulbenkian

2013-05-09T18:38:51Z

Are you saying that if I delete Replacement as well from the list of axioms of my set theory, I will still have the same trouble with my question? Maybe I need a much weaker set theory.

Comment by Garabed Gulbenkian

2013-05-08T14:25:16Z

@ Asaf and Goldstern: I see now what is wrong with my question. I failed to realize that the Foundation axiom of ZF already prohibits -as you point out-the existence of sets containing elements whose rank is arbitrarily high. This makes my question (in its present form) absurdly easy to answer. I must modify it by stipulating that the axioms of my set theory are just those of ZF other than Foundation. Sorry about that.

Comment by Garabed Gulbenkian

2013-05-07T19:42:50Z

Thanks alot for your answer. I have searched extensively to find it in the literature but with no success.

Comment by Garabed Gulbenkian

2013-05-06T18:46:48Z

Perhaps I did not formulate my question clearly enough. I am working within ZF in which there is no distinction between sets and classes-all sets are classes and all classes are sets. And, I am not looking for a pair of sentences which contradict each other. I am looking for a pair, such that each sentence-by itself- is inconsistent with ZF. The sentences I have in mind are the sort that lead to well known paradoxes such as Curry's or "the paradox of the set of all grounded sets". Such sentences are usually called Axioms of Comprehension.

Comment by Garabed Gulbenkian

2013-04-17T20:09:50Z

Thanks. I will try to digest all this.

Comment by Garabed Gulbenkian

2013-04-16T15:30:11Z

Is GBC the same as NBG and, if so, does the "C" stand for Cohen?

Comment by Garabed Gulbenkian

2013-04-15T19:12:34Z

Anyway, if I want to bring Quine's NF into this picture (as well as the axiom of choice) it seems as if I should really be talking about "second order NF". Although such a theory must certainly exist, I never heard of anybody taking any interest in it.

Comment by Garabed Gulbenkian

2013-04-15T19:01:26Z

Thanks for clearing up my mistaken notion about the languages in which Kunen's theorem could be formalized. I thought it could be done in NBG but did not realize that second order NBG would be required if the axiom of choice was going to be needed in the proof. When set theorists prove theorems in a second order set theory, they must use some axiomatizable sub-theory of that theory since the logical axioms of second order classical logic are not recursively enumerable.

Comment by Garabed Gulbenkian

2013-03-29T19:35:20Z

Thanks for both of your illuminating responses. Anton, that one-line demonstration is very nice! I wonder whether the no answer would still hold if I changed condition (2) to state that S minus any one of its points was still connected but did not stipulate that it should be homeomorphic to a straight line. Condition (1), of course, implies that S is arc-wise connected. It seems also that bringing Hilbert space into the picture doesn't change the situation in any way.

Comment by Garabed Gulbenkian

2013-03-13T18:51:02Z

Thanks, Joel, for a very complete answer. Your axiom Q+ accomplishes everything that axiom Q does, while avoiding the problem of some additional axioms being possibly needed to define mappings. Nothing new would seem to be needed to express that a set is finite in the sense of Tarski. So, if the only axioms of T are axiom Q+ and an infinite collection of sentences of the form P(n), T may be one of the simplest possible examples of a formalized theory that is both consistent and w-inconsistent.