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From the axiom of choice we get that every set can be ordered in a way that will make it a well ordered set, including $\Bbb R$. However, since the ordinal of such a well-ordered set of $\Bbb R$ will be at least $\omega_1$, and since as far as I understood from what I read and the questioned I researched over the website, it is only proveable that $\Bbb R$ is well ordered with the axiom of choice, it's supposedly impossible to 'show' one, as in, not to show that one exists, but to point to a specific one.

After the Course's test that thought bothered me quite a bit. I've been thinking about a specific ordering of $\Bbb R$, that seems to me as if it 'covers' all of R indeed, and is well-ordered as well. Obviously it is wrong, due to what I wrote earlier, but I can't figure out why. Any explanations to what's the 'problem' with it, or what's wrong with it will be highly appreciated. Does the problem come from the non-uniqueness of decimal representation?

The ordering I thought of:

Let $r\in\Bbb R$, then every such r can be defined as:

$r=k.n_1n_2n_3n_4...n_n...$ where $k,n_i\in\Bbb N$ and $0\le n_i\le9$.

For example, 3.14$\in\Bbb R$ would have k=3, $n_1=1, n_2=4,$ and for every $i \gt 2, n_i=0$.

let <' be an order on $\Bbb R$, so that =

{0,1,2,3,4….

-1,-2,-3,-4…

0.1, 1.1, 2.1 ,3.1…

-0.1, -1.1, -2.1, -3.1…

0.2, 1.2, 2.2, 3.2…

-0.2, -1.2, -2.2…

….

-0.9, -1.9, -2.9….

0.01, 1.01, 2.01…

…

-0.09, -1.09…

0.11, 1.11…

-0.11, -1.11…

0.12, 1.12…

-0.12, -1.12…

…

-0.19, -1.19…

0.21, 1.21…

…

-0.99, 1.99…

0.001, 1.001,…

……………….}

Let me try to explain each line. The first two lines are of $\Bbb Z$ where the first one is the positives (an 0), ordered by <, and the second one is the negatives, ordered by >. Every two lines are so (regarding the positives and negatives). Next line is the 'tenth's', as in for every $i \gt 1, n_i=0$, and we go with $n_1=0, n_1=2$ and so forth. Then we reach the hundreth, as in for every $i\gt 2$ $n_i=0$, and we go around with $n_1=0$ and 'run' with the $n_2$'s, then again with $n_1=1$ and so forth. Then the thousandth's, and so forth where we set for every $n_i$'s which's i is lesser or greater than the i on the $n_i$ which we 'run' on is 0, and then we go backwards and increase it by one every time, keeping the greaters on 0 until we reach '$999...$'.

That's the general idea, hope I made it clear enough. Eitherway, it's easy to see the order is well-ordering, and every line's ordinal is $\omega$. Question is - what did I do wrong, as this is obviously wrong according to what I stated above, if I got it right at least. I hope I didn't write anything too stupid.

Edit 1: Following Mauro ALLEGRANZA's comment, it seems the issue with the above ordering that it only covers finite number of strings. So we have to deal with r's with infinite number of strings, i.e. infinite number of strings that doesn't go to $n_i=0$ from $i \gt n_0$. If we take said number $r_w$ with smallest countable infinite strings, the set of such strings can be mapped to and have the ordinal of $\omega$. Question is whether I can follow it up from here - say that for every $n_i$>$n_w$ we set it to zero, as well as every $n_i$<$n_w$, and then we go backwards adding 'ones', or will it be impossible to say that since well, it doesn't make all too much sense. Any thoughts on the line of thought suggested in here? If it is possible to state so, then it's possible to go transfinite with this. Does something stops me from doing so, is it wrong because again, I limit it to certain place?

Martin Sleziak
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Studentmath
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  • "... every such $r$ can be defined as ...". With a finite set of symbols (the figures) you can only write down a countable set of finite strings; so you cannot cover uncountable many real numbers. See your $3.14$: it is the rational $314/100$. – Mauro ALLEGRANZA Feb 13 '14 at 11:24
  • Even though I did not limit them to be finite, as in, ni, where i goes to infinity? I am pretty sure I saw somewhere such a definition for real number exists. Edit: here http://en.wikipedia.org/wiki/Decimal_representation – Studentmath Feb 13 '14 at 11:27
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    I understand you - but if you stop the decimal representation after a finite amount of figures, however long may be the representation, you will cover only rationals; if you will not "stop it", how you handle infinite long strings of symbols ? – Mauro ALLEGRANZA Feb 13 '14 at 11:58
  • I get it now. Need to think that part over. – Studentmath Feb 13 '14 at 12:17
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    This is not what you were asking about, but a confusion that should be cleared up anyway: it is consistent that there is a definable well-ordering of the real numbers (so we can not only point it out, we can describe it precisely, at least in some cases). There is a formula which defines a well-ordering of the constructible universe $L$, so assuming $V=L$, any set (including the set of real numbers) has a definable wellordering, though it does not follow from ZFC alone. – tomasz Feb 13 '14 at 12:40
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    @tomasz : see the post about A way to well-order real line : we know that w-o exists (we all agree on it), but the issue is how to "describe" it (this - I think - is the aim of the OP's question). What about "However we cannot describe, as I did with [...] above, any way to well order the real numbers" ? – Mauro ALLEGRANZA Feb 13 '14 at 13:32
  • That's precisely what I aimed with my question, Mauro. – Studentmath Feb 13 '14 at 13:43
  • @Mauro: it can be argued that you cannot explicitly define a well-order of the real numbers in $\sf ZFC$ using forcing (that is, there is no formula which $\sf ZFC$ proves well-orders all the real numbers). Suppose there was, add two Cohen reals and by permuting them you can easily prove that the formula cannot decide which of the Cohen reals come first in the order. – Asaf Karagila Feb 13 '14 at 14:32
  • @Asaf - thanks for your patience and please, excuse me if I'm noisy, but (see also tomasz's comment), Wiki has : "for example, it follows from $ZFC+V=L$ that a particular [emphasis mine]formula well-orders the reals, or indeed any set". It may be wrong... but if not, "a particular formula" sound different from a non-costructive proof of existence. So, what I'm missing ? – Mauro ALLEGRANZA Feb 13 '14 at 14:41
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    @Mauro: There is a formula which uniformly well-orders every model of $V=L$. But not every model if a model of $V=L$. In fact $L$ is very very rigid, in the sense that it has no "inner models" and that every forcing extension violates $V=L$ immediately. In $L$ there is a definable and nice well-order. In some other models too. But there is no one unique formula which $\sf ZFC$ proves always to well-order all the real numbers. – Asaf Karagila Feb 13 '14 at 14:49
  • Thanks Asaf for clarfying that, and thanks Mauro for the further input! That truly solved my question. – Studentmath Feb 13 '14 at 15:16
  • @Asaf Skolem showed that the AC implies the existence of the universal selection operator $\sigma(a)$ which chooses an element of any nonempty set $a$. This operator (while not constructible in any way) trivially wellorders any set. – Spencer Jun 16 '17 at 02:22
  • @Spencer: That is demonstrably false, since that implies global choice which is not even provable from ZFC. But putting this aside, proving existence of something and writing an explicit definition for it are two incredibly distinct things in modern mathematics, so I just don't see the point of your comment here... – Asaf Karagila Jun 16 '17 at 02:25
  • @Asaf It is demonstrably true, at least in Von Neumann-Gödel-Bernays set theory (with class formalism). In his book Axiomatic Set theory, Bernays reduces his choice of AC equivalents (every set of ordered pairs contains a subset with the same domain, that is also a function) to a global choice operator. From there it's just iteration. My point is that you already have an explicit well-ordering of every set as soon as you let AC in the door. – Spencer Jun 16 '17 at 02:45
  • @Spencer: Yes, many formulations of NBG include global choice, not just set choice. But it seems that you don't understand what explicit means. It means a formula which has two free variables, and provably defines a well ordering of the reals, or the power set of $\omega$ if you want to be concrete. And I can prove to you there is no such formula, simply by adding a Cohen real to the universe. – Asaf Karagila Jun 16 '17 at 05:17
  • @Asaf I agree that you can't construct a formula like that without AC. But with AC you don't need one. You've already conceded the ground. – Spencer Jun 16 '17 at 11:02
  • @Spencer: It might be a good idea for you to first learn what "explicit" means in this context, before arguing with people whose dayjob is this subject. – Asaf Karagila Jun 16 '17 at 11:16

1 Answers1

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Some things to point out.

  1. It suffices to well-order the interval $[0,1)$, since it has the same cardinality as $\Bbb R$.

  2. $\omega_1$ is the least uncountable ordinal, but it is consistent that $|\Bbb R|=\omega_2$, or other values.

  3. As Mauro notes, this covers only finite decimal expressions. Which in fact is a countable set. Your plan for the infinite decimal expressions is not clearly defined, and I'm not sure what you're trying to define. But there is no number with "smallest infinite expansion", if you divide it by $10$, you will get a smaller number with infinite expansion.

    Moreover once you start "going backwards" you will have to go infinitely many steps down, which will ultimately end as not a well-order.

Community
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Asaf Karagila
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  • Interesting regarding point 1, and I get it regarding point 3. I thought of trying to define each next line with w, then w+1, the sets of strings having these ordinals.. but I see it's not possible, or at least can't see how it will be useful as once I get there, going backwards will end it as being not a well-order. Regarding point two - obviously, I meant more so it has at the very least the ordinal of $w_1$. Cheers for the clear up and thoughts from all! – Studentmath Feb 13 '14 at 13:44
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    @tomasz & Asaf - Asaf, please, see my comment above regarding your answer about a previous post (I'm loosing myself in this self-referencing loop ...) : we know they exists (w-e); is it possible to "describe", i.e.to define them in some way ? Thanks ... and be patient with my sloppiness about "to describe" and "to define". – Mauro ALLEGRANZA Feb 13 '14 at 14:00