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Are there real numbers that cannot be uniquely expressed with a finite number of symbols? Is this the same thing as an uncomputable number?

I can show that if there is one such number then there are infinitely many, and the set of all these numbers is not compact if it is nonempty, but I don't know if any such numbers exist.

Suggestions for tags are welcome.

Edit: I want to add that by "symbols" I mean numbers, mathematical symbols, or English language descriptions (or any language really). So for π, you could write it as "the ratio of a circle's circumference to its diameter" and that would count.

Zachary F
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  • Relevant soapboxing: http://karagila.org/2015/name-that-number/ – Asaf Karagila Mar 13 '17 at 14:33
  • Also http://math.stackexchange.com/questions/1125213/definable-real-numbers and http://math.stackexchange.com/questions/961558/undefinable-real-numbers are relevant here. Which may or may not already answer your question in detail. – Asaf Karagila Mar 13 '17 at 14:34
  • I think that all the mentioning of "language" in this post refer to mathematical language. Defining something means giving a definition in a mathematical language. But which one? The language of orders? Of fields? Of ordered fields? Of exponential fields? Of set theory itself? Etc. etc. – Asaf Karagila Mar 13 '17 at 14:53
  • @AsafKaragila I was thinking any description whatsoever – Zachary F Mar 13 '17 at 15:06
  • @AsafKaragila essentially anything that can be written down. – Zachary F Mar 13 '17 at 15:08
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    I guess that would mean "using the whole set theoretic universe to our disposal". But this could also mean that the language you allow for this is uncountable, e.g. one that has a constant symbol for each and every real number, and therefore all the reals are definable. And don't even get me started about asking what is the logic in which we are allowed to write the formulas. I'm sorry if this sounds annoying or pedantic, but this is a very naive question that has an answer once formalized properly, but the formalization is usually very not-naive and technical. – Asaf Karagila Mar 13 '17 at 15:08
  • @AsafKaragila that's a good point, but that answer definitely falls outside the realm of "acceptable answers" under the rules I want to use. I'll try to think of a better phrasing. – Zachary F Mar 13 '17 at 15:16
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    I can only recommend that you look at the links I gave in the first couple of comments. – Asaf Karagila Mar 13 '17 at 15:18

3 Answers3

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Internally, most of them cannot be so expressed, assuming you have a finite alphabet. The set of all finite strings from a finite alphabet is countable, but the reals are uncountable. Consequently, every function from the set of strings to the set reals is missing uncountably many reals.

Externally, it is possible (but not necessary) that every real number can be uniquely expressed in the same metalanguage you use to describe the set theory you're using. Here is a related question asking whether, not just each individual real number, but every set can be defined.

hmakholm left over Monica
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  • Thanks for the explanation! Could every member in a countable infinite set of number (for example $\mathbb N$) be represented by an alphabet of a limited length (for example, 10 pages)? – High GPA Oct 01 '17 at 01:05
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Yes.

Just note that there are uncountably many real numbers, while there are only countably many finite strings of symbols (unless you are willing to use uncountably many symbols, but then I guess the question is moot, as you might as well have one symbol for each real), and most of them don't even describe a real number.

The numbers you are asking about are called definable numbers. They do include computable numbers, but there are definable numbers which are not computable.

tomasz
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    What about models of ZFC which are pointwise definable? – Asaf Karagila Mar 13 '17 at 14:30
  • @AsafKaragila To be fair, this answer is probably talking about the "actual" real numbers, and "obviously" the actual universe isn't pointwise definable. :P (I love pointwise definable models, so weird!) – Noah Schweber Mar 13 '17 at 14:44
  • @Noah: But one can always code all the reals into a GCH pattern without adding new reals. So in a slightly larger universe, all the reals are definable! – Asaf Karagila Mar 13 '17 at 14:45
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    @AsafKaragila Well, that's not quite accurate - in a slightly larger universe, all reals are definable from ordinal parameters (you have to find the right part of the continuum pattern), and that's not really surprising since there are plenty of those. – Noah Schweber Mar 13 '17 at 14:47
  • @AsafKaragila: Well, those models clearly have no idea (well, no correct idea) what a finite string of symbols is (that, or "countable"), and hence of no interest when the question is about actual real numbers and actual finite strings of symbols. – tomasz Mar 13 '17 at 14:48
  • @Noah: I'm pretty sure that you can code them in intervals of length $\omega$ above $\omega_1$. Then you can just say that there is some limit ordinal $\alpha$ such that $x$ is the GCH pattern between $\aleph_\alpha$ and $\aleph_{\alpha+\omega}$ (more or less). – Asaf Karagila Mar 13 '17 at 14:49
  • @tomasz That's only partly right - these models can have a correct understanding of the "actual" set of finite strings from a countable alphabet. They will, however, think that certain sets are uncountable which (externally) are actually countable. However, this happens in a much broader way too by the Lowenheim-Skolem theorem. – Noah Schweber Mar 13 '17 at 14:49
  • @AsafKaragila That doesn't let you define an individual real in a parameter-free way - multiple reals will satisfy the definition "I am coded into the continuum pattern above some limit $\alpha$" . . . – Noah Schweber Mar 13 '17 at 14:50
  • @NoahSchweber: Hence my parenthetical remark. – tomasz Mar 13 '17 at 14:50
  • @tomasz Yes indeed - I just wanted to clear things up around that. – Noah Schweber Mar 13 '17 at 14:50
  • @Noah: Okay, sure, I concede to that remark, then. But nobody said "no ordinal parameters!" :-) Also, since the language wasn't specified, maybe it includes all the reals as constants, eh? :P – Asaf Karagila Mar 13 '17 at 14:51
  • @AsafKaragila: That in turn is addressed by the parenthetical remark in my answer. ;) – tomasz Mar 13 '17 at 14:52
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I'd appreciate it if a set-theorist chimes in to shred this answer but here is a half-serious attempt:

Apply a well-ordering to the real numbers.

Define $$E = \{x \in \mathbb R : \text{$x$ cannot be uniquely expressed with a finite number of symbols}\}.$$

Assume $E \not= \emptyset$. Then $E$ has a least element $e$. Thus $e$ cannot be uniquely expressed with a finite number of symbols, yet we can unambiguously state

$e$ is the least element of $\mathbb R$ (unique, under the given well-ordering) that cannot be uniquely expressed with a finite number of symbols

which uniquely expresses $e$ with a finite number of symbols. This is a contradiction which leads to $E = \emptyset$.

p.s. My own guess is that the argument fails because the well-ordering itself is not expressible with finitely many symbols.

Umberto P.
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    The argument fails because you don't tell us what is the language in which we are allowed to use to express these real numbers. If this is something like the language of ordered fields with some basic arithmetical augmentations (e.g. roots and exponents) then well-ordering is not at all definable. If this is the full-power of the language of set theory... well, then it is consistent that a well-ordering is in fact expressible. – Asaf Karagila Mar 13 '17 at 14:47
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    This is kind of an infinitary variant of Berry's paradox, which founders on the fact that you need to specify what "expressed" means, in a way such that this meaning can itself be "expressed". The problem of being sure which well-ordering you're talking about is just a cherry on top. – hmakholm left over Monica Mar 13 '17 at 16:11