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Let $E$ denote a particular (arbitrarily chosen, but fixed) method of representing ordinals as real numbers. Let $\alpha$ denote any large countable ordinal. Suppose that we want to refer to a particular real number $r$ that encodes the corresponding copy of $\alpha$ (assuming that $0 < r < 1$).

For example, consider the following “definition”:

Let $\alpha=\omega_\omega^{\text{CK}}$. The number $r$ is equal to the minimal real number that encodes (according to the algorithm of $E$) any copy of $\alpha$ and is greater than the number $v = \pi - 3$ (that is, there does not exist another real number $w$ that encodes any copy of $\alpha$ by the algorithm of $E$ and satisfies the condition $v < w < r$ ).

Is this “definition” a mathematically correct definition of some particular (unique) real number $r$? If no, then what are possible options to refer to unique real numbers that encode copies of $\alpha$?

One option is using the notion of definability by formulas in a chosen formal language, so that some $i$-th formula will somehow correspond to the definition of $r$ that encodes a particular copy of $\alpha$. But are there any other options?

lyrically wicked
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2 Answers2

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No, the "definition" you've given is not satisfactory. Regardless of what you mean by "encodes," you would need to justify the existence of a minimal real encoding each ordinal. SSequence's answer explains why this should strike you as implausible rather than plausible.

In fact, there is a precise sense in which what you want cannot be done. That is:

There is no good way to assign to each countable ordinal a specific real "coding" that ordinal.

There's just no way around this.

Of course, ZFC proves that there is a way to assign a real to each countable ordinal, that is, that there is an injection from $\omega_1$ to $\mathbb{R}$. The point is that the existence of such an injection doesn't say anything about the existence of such an injection which is in any way nice!

Specifically, at the most abstract level you're asking for a "reasonably definable" injection from $\omega_1$ to $\mathbb{R}$. It turns out that such a thing probably doesn't exist (the "probably" being a response to the inherent weaselliness of the phrase "reasonably definable"). This is a consequence of results in descriptive set theory.

  • First of all, the mere existence of an injection $\omega_1\rightarrow\mathbb{R}$ at all is not provable from ZF, so to do what you want you'll need to use the axiom of choice in a fundamental way. This tends to rule out the hope of anything so produced being "reasonably definable." And in particular, the axiom of determinacy - which tends to make sets of reals behave incredibly well - proves that no such function exists. "Reasonably definable" objects tend to be compatible with AD, so this provides a further point of evidence.

  • Now even within ZFC, this cannot be done in a Borel way. Precisely, if $f:\omega_1\rightarrow\mathbb{R}$ is an injection, then we cannot have $ran(f)$ and $\{\langle f(\alpha),f(\beta)\rangle: \alpha<\beta<\omega_1\}$ both be Borel. In fact, if the continuum hypothesis fails, there isn't even a Borel set of cardinality $\omega_1$ since the Borel sets have the perfect set property!

  • In fact, under additional set-theoretic hypotheses, this can be pushed further - e.g. strong but still fairly reasonable large cardinal hypotheses imply that there is no projective well-ordering of a set of reals of ordertype $\omega_1$, and even no injection from $\omega_1$ to $\mathbb{R}$ which is definable from reals and ordinals.

Noah Schweber
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  • I remember reading that we cannot precisely describe large countable ordinals, but we can define them in a suitable formal system (for example, by a formula based on ZFC). Does this mean that we can define such ordinals in some abstract form, but we cannot define any particular real number that encodes a copy of such ordinal? – lyrically wicked Sep 29 '18 at 17:12
  • Well it's going to depend on what you mean exactly by "define," but basically yes: it is in general easier to describe an ordinal than to describe any particular real "encoding" it. More generally, the following phenomenon is surprisingly common: a "simply definable" collection with no "simply definable" element. E.g. (i) the set of non-ordinal-definable reals is ordinal definable, (ii) the set ${0^#}$ is $\Pi^1_2$ while the real $0^#$ is $\Delta^1_3$, (iii) beyond the reals see also here. – Noah Schweber Sep 29 '18 at 17:19
  • I am starting to experience some difficulty: if such real numbers are impossible to refer to, then I don't understand how a function in this question can be well-defined when the underlying ordinal does not correspond to a particular well-ordering. But if we have a particular well-ordering, then there exists a way to define a real number that encodes this well-ordering, and referring to such number is what I want... – lyrically wicked Sep 29 '18 at 17:36
  • @lyricallywicked I'm not sure what tension you're seeing. The whole point of that answer was that while specific reals coding $\alpha$ exist, there's no obvious way to pick out one of them as particularly nicely definable. You seem to be conflating the statement "there is a (= at least one) real coding $\alpha$" with "there is a 'specifically definable' real coding $\alpha$." – Noah Schweber Sep 29 '18 at 17:41
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    Basically: we have (i) for any countable ordinal $\alpha$, there is a (in fact, many) real encoding a copy of $\alpha$, and (ii) to each such real we can get an "$\alpha$-hierarchy" as per my other answer. However, (iii) at no point have we claimed that there is, to an arbitrary countable ordinal $\alpha$, a "best" (or "obvious" or "most natural" or ...) real coding a copy of $\alpha$. Point (iii) should be seen as reinforcing the issues re: copies/notations that I've pointed out in answers to your other questions: that we hit unavoidable non-uniqueness quite quickly. – Noah Schweber Sep 29 '18 at 17:47
  • Regarding (i) in the previous comment, I want to clarify: assuming that the method of coding well-orders as reals is fixed, are there finitely many or infinitely many reals encoding a copy of $\alpha$? – lyrically wicked Sep 29 '18 at 18:23
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    @lyricallywicked Infinitely many of course, indeed continuum many, for any reasonable notion of coding I can think of. E.g. there are continuum-many binary relations $R$ on $\omega$ such that $(\omega;R)\cong\alpha$ whenever $\alpha$ is an infinite ordinal. – Noah Schweber Sep 29 '18 at 18:24
  • Thank you, that was very important clarification. I am also guessing that there does not exist a particular well-order $R0$ such that $f_{R0}(x)$ dominates all other $f_{Ri}(x)$, where $Ri$ corresponds to a copy of $\alpha$. Because if such $R0$ existed, it would be possible to identify it as unique... – lyrically wicked Sep 29 '18 at 18:32
  • @lyricallywicked Yup, we can always do better than any given one. It's really as annoying as it could possibly be! – Noah Schweber Sep 29 '18 at 18:36
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First of all let us observe that a function $r:\mathbb{N} \rightarrow \{0,1\}$ can be easily used to easily "store" all the information that a well-order relation with order-type $\alpha$ will have. For example, let $less:\mathbb{N}^2 \rightarrow \{0,1\}$ denote the well-order relation (for a specific well-order of $\mathbb{N}$ with order-type $\alpha$). Now just define $r(x)=less(first(x),second(x))$, where $first:\mathbb{N} \rightarrow \mathbb{N}$ and $second:\mathbb{N} \rightarrow \mathbb{N}$ are your usual inversion/extraction functions corresponding to a pairing function (https://en.wikipedia.org/wiki/Pairing_function).

Regarding the sort of definition you put in quotes, I am sure this can't work. If, instead of $r:\mathbb{N} \rightarrow \{0,1\}$, one works with a slightly different function, it is very easy to see. So first I will try to explain that example, and then a specific example for $r$. Define a function $I:\mathbb{N} \rightarrow \mathbb{N}$ as: $$I(x)=\left |\{n\,|\,(n<x) \,\,\land \,\,less(n,x)=1 \}\right |$$ Note that the bar sign means the number of elements in the given "finite" set.

Now suppose we are given two different functions $I_1(x)$ and $I_2(x)$ that are formed from different well-order relations (as described above) but with same order-type $\alpha$. Now while comparing the values of two different functions $I_1(x)$ and $I_2(x)$ let $a$ denote the smallest value for which $I_1(a) \neq I_2(a)$. Now we write $I_1 < I_2$ for two such functions iff $I_1(a) < I_2(a)$.

Now somewhat analogous to what you quoted, we could try to define the "smallest" function $I$ that describes the well-order relation for $\alpha$. But a closer examination will show that such a function $I$ is not well-defined. By contradiction, suppose there was such a function $I_{min}$. If $I_{min}$ indeed gives the description of some well-order relation for $\alpha$ then there must be some smallest value $a$ for which $I_{min}(a)\neq 0$. It is easy to see that one can describe a new function $I$ such that $I<I_{min}$ by "setting" it to have the property that $I(a)=0$.

While this isn't "exactly" as you described in quotes, the situation described above is quite similar. And illustrates that generally setting up such a simple property will not work to give a unique function (that "stores" all the information of well-order relation). As for the more specific description you gave in quotes, I think for $\omega \cdot 2$ and bigger values one will start to see that there is no such minimum real number (I might add a specific example later on).

SSequence
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  • I should clarify a few things. Firstly, can we be absolutely sure that it is impossible to obtain a method of representing well-orders as real numbers that will allow the given “definition” to work? Secondly, regarding the note “setting up such a simple property will not work to give a unique function” — but is it possible to define some more complex property that will allow to identify an unique real number that corresponds to a particular copy of $\alpha$? – lyrically wicked Sep 29 '18 at 13:25
  • I am not an expert on this at all, but this problem is "essentially" equivalent to giving (fundamental) sequences all the way up till $<\omega_1$. This thread may probably be of some relevance (specifically some parts in the answers): https://math.stackexchange.com/questions/2883221. – SSequence Sep 29 '18 at 13:36
  • @lyricallywicked Regarding my personal (and probably somewhat naive) understanding, once you have AC, then there also must exist a well-order of $\mathbb{R}$ and that, I "think", could then be used to give what you are asking for. However, a well-order of $\mathbb{R}$ in itself is not a simple matter. Some threads related to that which I found using a quick search:(i) https://math.stackexchange.com/questions/6501 (ii) https://math.stackexchange.com/questions/209033 – SSequence Sep 29 '18 at 14:11