Garabed Gulbenkian
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Registered User
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May 20 |
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A question about large real closed fields 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. |
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May 20 |
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A question about large real closed fields 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. |
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May 19 |
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A question about large real closed fields 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. |
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May 19 |
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A question about large real closed fields 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". |
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May 18 |
asked | A question about large real closed fields |
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May 11 |
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A question about “paradoxical” sentences in the language of ZF set theory. 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). |
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May 9 |
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A question about “paradoxical” sentences in the language of ZF set theory. 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. |
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May 8 |
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A question about “paradoxical” sentences in the language of ZF set theory. @ 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. |
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May 7 |
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A question about simple arcs in higher dimensional Euclidean spaces. Thanks alot for your answer. I have searched extensively to find it in the literature but with no success. |
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May 6 |
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A question about “paradoxical” sentences in the language of ZF set theory. 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. |
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May 6 |
asked | A question about simple arcs in higher dimensional Euclidean spaces. |
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May 5 |
asked | A question about “paradoxical” sentences in the language of ZF set theory. |
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Apr 17 |
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A question about Kunen’s inconsistency theorem Thanks. I will try to digest all this. |
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Apr 16 |
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A question about Kunen’s inconsistency theorem Is GBC the same as NBG and, if so, does the "C" stand for Cohen? |
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Apr 15 |
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A question about Kunen’s inconsistency theorem 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. |
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Apr 15 |
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A question about Kunen’s inconsistency theorem 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. |
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Apr 13 |
asked | A question about Kunen’s inconsistency theorem |
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Apr 8 |
asked | A question about connected subsets of metric spaces |
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Mar 30 |
awarded | ● Nice Question |
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Mar 29 |
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A question about closed curves 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. |
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Mar 28 |
asked | A question about closed curves |
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Mar 13 |
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A question about formalized theories that may be both consistent and w-consistent 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. |
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Mar 11 |
asked | A question about formalized theories that may be both consistent and w-consistent |
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Mar 6 |
awarded | ● Yearling |
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Feb 24 |
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A question about Universal sets. Thanks, all of you, for the abundance of information you have given me. Regarding the Axiom of Choice, is its inconsistency with NF still provable when we restrict it to Cantorian sets only? If not, then we could add this restricted axiom to NF and ask my question about a theory that could be called "NFC". I certainly hope that Holmes has answered what is one of my favorite open questions. Is that conference going to be held in Cambridge, Mass: or Cambridge, England? |
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Feb 22 |
asked | A question about Universal sets. |
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Feb 16 |
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A question about definable non-empty sets containing no definable elements. It is still not clear to me what the answer to my question No.17608 is. Some statements and comments of Ali Enayat seem to indicate that it should be YES. But if so, there ought somewhere to be a published proof or even a mention of the result in some book about set theory that covers the subject of ordinal definability. Likewise, if the question is still open. |
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Feb 13 |
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A question about definable non-empty sets containing no definable elements. Joel, that is a very interesting set that you came up with in your update. It would seem impossible to even be able to pin down its cardinal number. |
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Feb 13 |
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A question about definable non-empty sets containing no definable elements. I really asked my question to try and obtain a better understanding of why it seemed so much harder (or even impossible) to find "examples" among denumerably infinite sets than among sets having a larger cardinal number. Is there something important about the smallest infinite cardinal number and/or about the axiom of choice that is at the bottom of this phenomenon? |
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Feb 13 |
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A question about definable non-empty sets containing no definable elements. I was surprised that my question (or rather my example involving non-measurable sets) led to such an extended analysis of how one should construe "definability". There are two viewpoints here. One viewpoint will accept such expressions as "T implies P(x) and not-T implies Q(x)" to be a legitimate defining formula, while the other will not unless the underlying theory (in this case ZFC) can decide the sentence T. |
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Feb 12 |
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A question about definable non-empty sets containing no definable elements. The above result of Feferman seems to fit in better with our intuitions about S-that we cannot really "produce" any well-ordering of the continuum-in spite of what Joel has done. A different but perhaps more difficult version of my question might result if I changed the phrase "no set definable in ZFC can be proved in ZFC to be an element of S" so as to read "it is consistent with ZFC that there exists no ordinal definable element of S" |
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Feb 12 |
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A question about definable non-empty sets containing no definable elements. Let S be the set of all well-orderings of the continuum (perhaps a better example than the set of non-measurable subsets of R). Let P(x),Q(x) be defining formulae of ZFC and T be a sentence of ZFC. Then "T implies P(x) and not-T implies Q(x)" is also a defining formula of ZFC. BY properly choosing P(x),Q(x) and T-as Joel has shown-one can define an element of every set that can be proved in ZFC to exist and be non-empty-including S. However S. Feferman has proved that "there exists no ordinal definable well-ordering of the continuum" is consistent with ZFC. |
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Feb 11 |
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A question about definable non-empty sets containing no definable elements. With regard to Emil's comment, perhaps I should weaken my phrase "no set definable in ZFC can be proved in ZFC to be an element of S" so that it reads "no set definable in ZFC has yet been proved in ZFC to be an element of S" |
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Feb 11 |
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A question about definable non-empty sets containing no definable elements. I should emphasize, Joel, that I am referring to a formalized theory whose axioms are just those of ZFC. As you point out, if we add (as new axioms) sentences in the language of ZFC which are consistent with ZFC (like V=L or V=HOD that impose a particular well-ordering on the whole universe), then some sets which are not definable in ZFC may become definable. But they become definable in a new formalized theory having additinal axioms-not in ZFC. |
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Feb 11 |
asked | A question about definable non-empty sets containing no definable elements. |
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Jan 24 |
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A question about well ordered subsets of totally ordered countable sets Sorry. This is all wrong. I thought I had a definition for a countable ordering being "scattered" which could be expressed in a first order language, but I found a counter-example which shows that my definition does not work. It seems as if being "scattered" is just as much of a second order concept as being "well ordered". I was hoping to use my mistaken idea to come up with an easily understandable definition of the so-called "proof theoretical" ordinal numbers. How embarrassing. |
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Jan 21 |
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A question about well ordered subsets of totally ordered countable sets If X is any total and scattered ordering of the set N of all non-negative integers, let Su(X) denote the smallest ordinal number that is not the ordinal number of a well-ordered subset of X. Suppose that T is a formalized mathematical theory. Consider all the formulae of T containing two free variables which can be proved in T to represent a total scattered ordering X of the set N. If W(T) is the least upper bound of Su(X) for all such X, I wonder if W(T) is the proof theoretical ordinal number of T. |
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Jan 21 |
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A question about well ordered subsets of totally ordered countable sets I could not get access to a computer until now to reply to your responses. Many thanks to all of you for explaining so clearly how my question can be answered. Ali, I knew about scattered sets from topology but did not realize that the concept could be applied to arbitrary countable total orderings with no mention of topology. Furthermore, the proposition that an ordering of the set of non-negative integers is scattered can be expressed by a sentence of first order arithmetic-whereas well-ordering is a second order concept which cannot be expressed in this language. |
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Jan 15 |
asked | A question about well ordered subsets of totally ordered countable sets |
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Jan 4 |
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A question about a hierarchy of metric spaces arising from an operation defined by Hausdorff. It is true that for each non-negative integer n, the collection of all singletons of "points" of H(n,M) is a subspace of H(n+1,M) that is isometric to H(n,M). But if one attempts to use this fact to extend the hierarchy into the transfinite, and if w denotes the smallest infinite ordinal number, exactly how does one define the collection of points of the metric space H(w,M)-and exacly how does one define the metric for this space. Unfortunately, the set of relevant notions of category theory such as "direct limits" that I know anything about, has measure zero. |
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Jan 3 |
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A question about a hierarchy of metric spaces arising from an operation defined by Hausdorff. These comments are very interesting and I am trying to understand how they affect the possible answers to my questions. If M is a finite-dimensional Euclidean space, are you saying that a maximal set of points in the closed unit ball of H(n,M) such that no pair of these points are at a distance apart less than some preassigned positive number e, has a cardinal number that grows with n? And how does one cannonically embed H(n,M) into H(n+1,M)? |
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Jan 2 |
asked | A question about a hierarchy of metric spaces arising from an operation defined by Hausdorff. |
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Nov 28 |
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A question about the Axiom of Choice and straight lines in the Euclidean plane. Very nice! I thought my second question could not have a "yes" answer. But it seems that you have proven me wrong-and without either using the Axiom of Choice or giving up condition (3). |
