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This question is motivated by a comment of Robert on the question Can any Real number be typed in a computer? :

Can you "think of" an undefinable number? – Robert Israel

I would like to reformulate this (ironic) question to:

Can you give an (concrete) example for an undefinable number?

My first guess would be "No!" because when you give an example of a number you need a definition for it which contradicts that the number shall be undefinable. But we are talking about mathematics and I learned that the first guess is not always true ;-) So I can imagine that finally the answer is "Yes" to the above question (maybe it is possible to give an undefinable number in first order logic by using higher order logic).

Note: I mean "undefinable" in the sense "undefinable in first order logic".

Community
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Stephan Kulla
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    You might find more information if you drop "undefinable" and use "non-computable", and I refer you to https://en.wikipedia.org/wiki/Chaitin's_constant and related topics. This is just another example of uncountable infinity brushing up against intuition, and it can be resolved if you're willing to read some of the excellent resources out there. But computability is a gold mine for this question. – Eric Tressler Nov 08 '15 at 23:05
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    Note that the notion of definability makes sense only relatively to a structure. (or when the thing whose definability you are questionning is actually definable) – nombre Nov 08 '15 at 23:41
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    First order logic over what structure? Over $\langle\Bbb R,<\rangle$? Every number is undefinable over that structure. Over $\langle\Bbb R,+,\cdot,0,1,<\rangle$? You can show that the definable numbers are exactly the algebraic reals. Over $\langle\Bbb R,\pi\rangle$? You can show that anything other than $\pi$ is undefinable. Or maybe first-order definable in the universe of set theory, in which case you are running into a quagmire from which there is little escape. – Asaf Karagila Nov 09 '15 at 10:44
  • @AsafKaragila: Why it is quagmire if I would use the structure of ZFC? – Stephan Kulla Nov 09 '15 at 18:12
  • Let's start with the fact that it is consistent that every real number is definable. – Asaf Karagila Nov 09 '15 at 18:13
  • @AsafKaragila: To understand your answer: You mean that assuming ZFC each real number can be defined via a statement in first order logic with the language of ZFC (which only consists of the relation "$\in$" when I remember well)? Is this right? – Stephan Kulla Nov 09 '15 at 18:17
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    Almost. It is consistent with ZFC that every real number (and in fact every set) is definable with a first-order definition in the language of set theory which includes a single binary relation $\in$. But it is also consistent that there are non-definable real numbers as well. – Asaf Karagila Nov 09 '15 at 18:22
  • @AsafKaragila: Okay, now I understand. Thanks! ;-) When you like, you can copy your comments to an answer so that I can upvote it (your comments helped me to understand the issue) – Stephan Kulla Nov 09 '15 at 18:47
  • Isn't it contradictory to describe a number that is not definable? A bit like the smallest uninteresting number? –  Aug 09 '24 at 07:42

3 Answers3

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You say you mean "undefinable" in the sense "undefinable in first order logic," but this is still ambiguous: what is the language (i.e., the collection of non-logical symbols) you are using?

For instance:

  • ${1\over 2}$ is undefinable in $(\mathbb{R}; +)$.

  • $e$ is undefinable in $(\mathbb{R}; +, \times)$.

  • $\pi$ is (almost certainly :P) undefinable in $(\mathbb{R}; +, \times, exp)$.

And so forth.

There are even some examples which may seem to strain the limits of possibility:

  • Fix a countable language $\Sigma$, and enumerate the $\Sigma$-formulas as $\varphi_i$. Let $r$ be the real whose $n$th bit (in binary) is $0$ iff $\varphi_n$ holds of some real whose $n$th binary bit is $1$. Then I have "defined" $r$, in a language larger than $\Sigma$, but still somehow "$\Sigma$-ish."

  • Much more weirdly: Suppose there is a definable well-ordering of the reals - this is a consequence of e.g. $V=L$. Then there is a least (according to this well-ordering) real which is not definable in the language of set theory$^*$! The reason this isn't a contradiction is that "definable" isn't definable. :P Nonetheless, in some sense I have "given" you a real . . .

And, of course, there is no real which is "absolutely" undefinable: we can always expand the language to include a constant for that particular real!

(A natural question at this point is to ask whether there is a sense in which some reals are "more definable" than others. Depending how you ask this, you wind up in computability theory, descriptive set theory, model theory, . . . )

$^*$I'm assuming that we're working with the true reals, or at least a model which is well-founded and whose $\mathbb{R}$ is truly uncountable; beware of weird countable models, e.g. the pointwise definable ones (http://arxiv.org/abs/1105.4597)!

Noah Schweber
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  • $V=L$ promises you an ordering of the reals where every non-empty definable collection of reals has a first element. It doesn't guarantee that an undefinable collection (such as the collection of undefinable reals) has a first element. [We may find ourselves in a model with non-standard integers, for example -- then there are even undefinable subcollections of $\mathbb N$ that have no least element]. (Yes, the footnote, if interpreted expansively enough, may take care of that, though I'm not completely sure). – hmakholm left over Monica Nov 08 '15 at 23:21
  • @HenningMakholm This is why I had the footnote! If we're working with the true reals, then obviously this isn't a problem. (The second half of my footnote is incomplete, though, and should read ". . . which has the true $\omega$ and whose $\mathbb{R}$ is truly uncountable." I've fixed it.) – Noah Schweber Nov 08 '15 at 23:24
  • @ruakh Derp, yes, of course. :P – Noah Schweber Nov 08 '15 at 23:38
  • I'm on a bit shaky ground here, but I think one could stil have a model in which $\omega$ is standard, but $\omega_1$ has non-standard ordinals in it. If so it would still pass the revised footnote condition. (But I can't actually prove such a model would exist off the cuff; perhaps I'm wrong). – hmakholm left over Monica Nov 08 '15 at 23:39
  • @HenningMakholm you're absolutely right, I'm tired - I meant to say "well-founded model." Fixed. – Noah Schweber Nov 08 '15 at 23:40
  • In $(\mathbb{R}; +)$, are there any definable numbers besides 0? – Nate Eldredge Nov 08 '15 at 23:41
  • @NateEldredge No, there are not. Much more is true: the automorphism group acts transitively on the nonzero reals. – Noah Schweber Nov 08 '15 at 23:41
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    @Nate: No, because there's an automorphism that fixes only $0$ (namely multiplication by any real not in ${0,1}$). – hmakholm left over Monica Nov 08 '15 at 23:41
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    What do you mean by "true" reals or "truly" uncountable? – user2357112 Nov 08 '15 at 23:49
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    By true I mean, well, true - i.e., not a countable model, but the actual reals. Similarly with "truly uncountable." I.e., I'm assuming the existence of a "background universe" $V$ of set theory, and I mean the reals in $V$ or uncountable in $V$. If you dislike talk of the "true" universe of set theory, that's perfectly reasonable - I do too, much of the time - but it's very convenient language to describe how, the "nicer" a model of ZFC+V=L is, the weirder a phenomenon occurs. – Noah Schweber Nov 08 '15 at 23:53
  • Out of curiosity, why the down-vote? – Noah Schweber Nov 09 '15 at 00:38
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    Am I right in guessing that both $\pi$ and $e$ are definable in $(\Bbb R;+,\times,<)$? (Since, with $<$, we can define limits.) – Akiva Weinberger Nov 09 '15 at 01:30
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    @AkivaWeinberger No, they're not - "definable" here means "definable by a first-order formula." The issue, basically, is that to describe a sequence for $\pi$, you need an infinitely long formula. Now on the one hand it's not really that bad - $\pi, e$, and basically every real you've ever heard of are definable by computable infinitary first-order formulas - but it's still not first-order. (Also, side note: the ordering is already definable once we have $+$ and $\times$, since $x>y\iff \exists z(z\times z=x-y)$.) – Noah Schweber Nov 09 '15 at 02:23
  • @NoahSchweber I think you mean x >= y. – Doug McClean Nov 09 '15 at 15:21
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    @DougMcClean Quite right, but the point stands. :) – Noah Schweber Nov 09 '15 at 17:30
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If the choice of language is the type of language we use in everyday life, Ian Stewart's How to Cut a Cake gives the example of Richard's paradox:

"In the English language, some sentences define positive integers and others do not. For example 'The year of the Declaration of Independence' defines the number 1776, whereas 'The historical significance of the Declaration of Independence' does not define a number. So what about this sentence: 'The smallest number that cannot be defined by a sentence in the English language containing fewer than twenty words.' Observe that whatever this number may be, we have just defined it using a sentence in the English language containing only nineteen words. Oops."

Of course, this can also be extended to Spanish, French, German, etc., instead of just English.

404UserNotFound
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As in non-recursive? There are lots. At a higher level of complexity, index the sentences of arithmetic in one of the usual ways. Let $u$ be the number whose $n$-th decimal digit is $1$ if $n$ is not the index of a sentence, $2$ if $n$ is the index of a sentence true in the natural numbers, and $3$ if $n$ is the index of a sentence false in the natural numbers. Then $u$ is an undefinable real, by Tarski's result on the undefinability of truth of arithmetical sentences.

André Nicolas
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    But you can't prove that $u$ exists. – nombre Nov 08 '15 at 23:32
  • The definition is of a standard mathematical kind. One can take the point of view that the only objects that exist are the ones for which we can provide a constructive proof, but that is very much a minority point of view. – André Nicolas Nov 08 '15 at 23:37
  • @nombre Why not? $T$ isn't definable, but it exists, and then all that's needed is that the three sets partition $\mathbb{N}$ (which is true by definition). – Steven Stadnicki Nov 08 '15 at 23:40
  • Well, how would you prove that $u$ (or $T$) exists? – nombre Nov 09 '15 at 00:02
  • @StevenStadnicki The question isn't whether it exists, but whether you can prove it exists. I think. – Akiva Weinberger Nov 09 '15 at 01:32
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    @nombre Of course it depends what your axioms are, but e.g. ZFC is more than enough to prove the existence of $u$, since in ZFC we can define truth or falsity of first-order sentences in (set-sized) structures. – Noah Schweber Nov 09 '15 at 02:25
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    Yes, for any model $\mathcal{M} = (M,\in_{\mathcal{M}})$ of ZF, one can define the sets $T_{\mathcal{M}}$ and $F_{\mathcal{M}}$ of Gödel numbers of arithmetical sentences true and false in $\mathbb{N}{\mathcal{M}}$. However these sets need not satisfy $\forall n \in T{\mathcal{M}}, n \in_{\mathcal{M}}\mathbb{N}{\mathcal{M}}$ so it doesn't make sense to say that the decimal expansion of a real number of $\mathcal{M}$ has the value $2{\mathcal{M}}$ at $n \in T_{\mathcal{M}}$. – nombre Nov 09 '15 at 09:43
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    (One can also define the set $T$ of Gödel numbers of sentences true in $\mathbb{N}$ in ZFC but if you define $u$ using this definition of $T$, then $u$ is tautologicaly definable.) – nombre Nov 09 '15 at 11:20