Is the analysis as taught in universities in fact the analysis of definable numbers? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T17:53:43Z http://mathoverflow.net/feeds/question/44102 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/44102/is-the-analysis-as-taught-in-universities-in-fact-the-analysis-of-definable-numbe Is the analysis as taught in universities in fact the analysis of definable numbers? Anixx 2010-10-29T10:47:27Z 2012-05-07T16:41:08Z <p>Ten years ago when I studied in the university I had no idea about <a href="http://en.wikipedia.org/wiki/Definable_number" rel="nofollow">definable numbers</a>, but I came to this concept myself. My thoughts were as follows:</p> <ul> <li>All numbers are divided into two classes: those which can be unambiguously defined by a limited set of their properties (definable) and such that for any limited set of their properties there is at least one other number which also satisfies all these properties (undefinable).</li> <li>It is evident that since the number of properties is countable, the set of definable numbers is countable. So the set of undefinable numbers forms a continuum.</li> <li>It is impossible to give an example of an undefinable number and one researcher cannot communicate an undefinable number to the other. Whatever number of properties he communicates there is always another number which satisfies all these properties so the researchers cannot be confident whether they are speaking about the same number.</li> <li>However there are probability based algorithms which give an undefinable number in a limit, for example, by throwing dice and writing consecutive numbers after the decimal point.</li> </ul> <p>But the main question that bothered me was that the analysis course we received heavily relied on constructs such as 'let's $a$ to be a number that...", "for each $s$ in interval..." etc. These seemed to heavily exploit the properties of definable numbers and as such one can expect the theorems of analysis to be correct only on the set of definable numbers. Even the definitions of arithmetic operations over reals assumed the numbers are definable. Unfortunately one cannot take an undefinable number to bring a counter-example just because there is no example of undefinable number, but still how to know that all those theorems of analysis are true for the whole continuum and not just for a countable subset?</p> http://mathoverflow.net/questions/44102/is-the-analysis-as-taught-in-universities-in-fact-the-analysis-of-definable-numbe/44118#44118 Answer by arsmath for Is the analysis as taught in universities in fact the analysis of definable numbers? arsmath 2010-10-29T12:27:34Z 2010-10-29T12:27:34Z <p>"Definable numbers" are numbers that are definable in terms of first-order logic over set theory. There are perfectly intelligible numbers that cannot be defined in your sense. For example, suppose you have a sequence of definable numbers, $a_n$ that is bounded by a constant $C$. Then $b = \sup a_n$ is a number that is unique and has an unambiguous meaning, but $b$ is not necessarily definable. Each $a_n$ is given by a formula $\phi_n(x)$, but if the formulas are sufficiently different then there is no way to write down a single formula $\phi$ for $b$. The Wikipedia page you link to talks about this issue in terms of Cantor diagonalization.</p> http://mathoverflow.net/questions/44102/is-the-analysis-as-taught-in-universities-in-fact-the-analysis-of-definable-numbe/44129#44129 Answer by Joel David Hamkins for Is the analysis as taught in universities in fact the analysis of definable numbers? Joel David Hamkins 2010-10-29T13:15:25Z 2011-07-25T11:03:18Z <p>The concept of <em>definable</em> real number, although seemingly easy to reason with at first, is actually laden with subtle metamathematical dangers to which both your question and the Wikipedia article to which you link fall prey. In particular, the Wikipedia article contains a number of fundamental errors and false claims about this concept.</p> <p>The naive treatment of definability goes something like this: In many cases we can uniquely specify a real number, such as $e$ or $\pi$, by providing an exact description of that number, by providing a property that is satisfied by that number and only that number. More generally, we can uniquely specify a real number $r$ or other set-theoretic object by providing a description $\varphi$, in the formal language of set theory, say, such that $r$ is the only object satisfying $\varphi(r)$.</p> <p>The naive account continues by saying that since there are only countably many such descriptions $\varphi$, but uncountably many reals, there must be reals that we cannot describe or define.</p> <p>But this line of reasoning is flawed in a number of ways and ultimately incorrect. The basic problem is that the naive definition of definable number does not actually succeed as a definition. One can see the kind of problem that arises by considering ordinals, instead of reals. That is, let us suppose we have defined the concept of definable ordinal; following the same line of argument, we would seem to be led to the conclusion that there are only countably many definable ordinals, and that therefore some ordinals are not definable and thus there should be a least ordinal $\alpha$ that is not definable. But if the concept of definable ordinal were a valid set-theoretic concept, then this would constitute a definition of $\alpha$, making a contradiction. In short, the collection of definable ordinals either must exhaust all the ordinals, or else not itself be definable.</p> <p>The point is that the concept of definability is a second-order concept, that only makes sense from an outside-the-universe perspective. <a href="http://en.wikipedia.org/wiki/Tarski%27s_undefinability_theorem" rel="nofollow">Tarski's theorem on the non-definability of truth</a> shows that there is no first-order definition that allows us a uniform treatment of saying that a particular particular formula $\varphi$ is true at a point $r$ and only at $r$. Thus, just knowing that there are only countably many formulas does not actually provide us with the function that maps a definition $\varphi$ to the object that it defines. Lacking such an enumeration of the definable objects, we cannot perform the diagonalization necessary to produce the non-definable object.</p> <p>This way of thinking can be made completely rigorous in the following observations:</p> <ul> <li><p>If ZFC is consistent, then there is a model of ZFC in which every real number and indeed every set-theoretic object is definable. This is true in the minimal transitive model of set theory, by observing that the collection of definable objects in that model is closed under the definable Skolem functions of $L$, and hence by Condensation collapses back to the same model, showing that in fact every object there was definable.</p></li> <li><p>More generally, if $M$ is any model of ZFC+V=HOD, then the set $N$ of parameter-free definable objects of $M$ is an elementary substructure of $M$, since it is closed under the definable Skolem functions provided by the axiom V=HOD, and thus every object in $N$ is definable.</p></li> </ul> <p>These models of set theory are <em>pointwise definable</em>, meaning that every object in them is definable in them by a formula. In particular, it is consistent with the axioms of set theory that EVERY real number is definable, and indeed, every set of reals, every topological space, every set-theoretic object at all is definable in these models.</p> <ul> <li>The pointwise definable models of set theory are exactly the prime models of the models of ZFC+V=HOD, and they all arise exactly in the manner I described above, as the collection of definable elements in a model of V=HOD.</li> </ul> <p>In recent work (soon to be submitted for publication), Jonas Reitz, David Linetsky and I have proved the following theorem:</p> <p><b>Theorem.</b> Every countable model of ZFC and indeed of GBC has a forcing extension in which every set and class is definable without parameters.</p> <p>In these pointwise definable models, every object is uniquely specified as the unique object satisfying a certain property. Although this is true, the models also believe that the reals are uncountable and so on, since they satisfy ZFC and this theory proves that. The models are simply not able to assemble the definability function that maps each definition to the object it defines.</p> <p>And therefore neither are you able to do this in general. The claims made in both in your question and the Wikipedia page on the existence of non-definable numbers and objects, are simply unwarranted. For all you know, our set-theoretic universe is pointwise definable, and every object is uniquely specified by a property.</p> <hr> <p><b>Update.</b> Since this question was recently bumped to the main page by an edit to the main question, I am taking this opportunity to add a link to my very recent paper <a href="http://de.arxiv.org/abs/1105.4597" rel="nofollow">"Pointwise Definable Models of Set Theory", J. D. Hamkins, D. Linetsky, J. Reitz</a>, which explains some of these definability issues more fully. The paper contains a generally accessible introduction, before the more technical material begins.</p>