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I've read been introducted to more rigorous definitions of the reals multiple times, but I'm still having some difficulties wrapping my head around the fact that

  1. the real numbers cannot be sorted and
  2. there aren't any gaps in the reals

For 1, I believe that real numbers cannot be fully ordered otherwise they could be put in a one-to-one correspondence with the rationals (since the rationals are dense in R). If they could be ordered, then we could take the rational number between each real and end up with a corresondence. My though would be that you could make a construction that goes something like this:

Construct the list $\mathbb{R}_2$ as follows. Then take some $r \in \mathbb{R} \wedge r \notin \mathbb{R_2}$ and insert $r$ in a location such that all items to the left of $r$ are less than $r$ and all items to the right are greater than $r$. $r$ will always lie between some elements since $r$ is not already in $\mathbb{R}_2$. It seems like you could continue this construction for at least countably infinite steps, but why would it fail for uncountably infinite number of steps? [Note: I haven't specified how to keep order. I think you could maybe make a tuple with (r,indice) or something similar. I'm not exactly sure how to construct something that keeps track of the order, but I feel there must be some nice way].

For 2, it seems like there should be numbers be gaps in the reals. Or more specifically I feel that the least upper bound of rational numbers doesn't always exist in the real numbers. Consider {1,1/2,1/3,...}. The infimum seems as though it should be greater than 0. In particular, why is that an infinitesimal can't be considered the infimum, since $\frac{1}{n} > \epsilon>0$?

I think the heart of my question is why this: why is that the surreal numbers are not counterexamples to certain properties of the real numbers? The surreal numbers seems to contain numbers that are between the gaps of reals and provide lower supremum and higher infimum. Furthermore, they are completely ordered, which would seem to contradict the idea that the real numbers can't be completely ordered.

In response to the comments, and me realizing I had not very well articulated my question, I'll add some more specifications on what I meant. I thought that if the reals could be put into an ordered chain that contained all reals numbers, like this: $r1<r2<r3<...<rn$, where every where number is included in this chain, then that would imply that the reals were only as large as the rationals. This is because the rationals are dense in the reals, so it seems like on could do this: $r1<q1<r2<q2<r3<q3<r4<...<qn-1<rn$. But this is clearly wrong, so it seems like the only way to avoid this is to say that its impossible to construction something like: $r1<r2<r3<...<rn$.

I also felt that this was wrong too though, because the surreal numbers are a way to make that chain of real numbers. Surreal tree You could take the surreal numbers at a specific day start at the far left, and then order each item as you move to the right. Or, it seems like you could step-by-step/recursively insert real numbers into a set to get the ordered chain of real numbers by inserting the real numbers so that to the left are numbers less and numbers to the right are greater.

So, I was hoping I could get help understanding what I'm doing wrong, and why none of these things prove that the rationals are the same size as the real.

Also, thank you so far to everyone whose commented, and I'm sorry the question wasn't worded very well

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    $\begingroup$ The reals are definitely totally ordered by their usual order. What they aren't is well ordered by that order. $\endgroup$
    – Ian
    Oct 26 '20 at 1:25
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    $\begingroup$ "Consider {1,1/2,1/3,...}. The infimum seems as though it should be greater than 0." Why? $\endgroup$ Oct 26 '20 at 1:38
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    $\begingroup$ @CalebBriggs Just because there are a lot of numbers left at each stage, doesn't mean that we don't throw them all away eventually. Think about natural numbers: for each $n\in\mathbb{N}$ there are infinitely many $k>n$, but there are no natural numbers bigger than every $n\in\mathbb{N}$. $\endgroup$ Oct 26 '20 at 1:46
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    $\begingroup$ The problem is that $\epsilon$ isn't a real number. There is no real number that works as a bound. $\endgroup$
    – user147556
    Oct 26 '20 at 1:46
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    $\begingroup$ What do you mean by "the surreals...are completely ordered...the real numbers can't be completely ordered"? This sounds backwards: the real numbers have an order that is Dedekind complete and the surreals don't. There is a least upper bound of $\{-1,-1/2,\ldots\}$ (negative reciprocals of naturals) in the reals (it's $0$), but there's no least upper bound in the surreals. $\endgroup$
    – Mark S.
    Oct 26 '20 at 1:57
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The real numbers $\mathbb{R}$ have no gaps—the technical way to say this is that $\mathbb{R}$ is a complete space. In other words, whenever you have a sequence of points $x_1,x_2,x_3,\ldots$ that ultimately get arbitrarily close together, then the sequence has a limit, and that limit point belongs to $\mathbb{R}$. For example, the sequence 3, 3.1, 3.14, 3.141, 3.14159, etc. has a limit of $\pi$.

Completeness is an internal property—strange as it sounds, you can spot "holes" in the space where limits fail to exist, just by examining these arbitraily-bunched-up ("Cauchy") sequences within the space. For example, the rational numbers $\mathbb{Q}$ are incomplete; you can tell because in $\mathbb{Q}$, the sequence 3, 3.1, 3.14, etc. gets arbitrarily close together, but doesn't converge to any rational number. There's a hole where a limit $\pi$ "ought" to be.

One way of defining the real numbers is as the completion of the rational numbers. It fills in the gaps by defining real numbers as (equivalence classes of) Cauchy sequences, so that a number like $\pi$ is axiomatically defined as the sequence 3,3.1,3.14,3.141, etc.

Of course, even though $\mathbb{R}$ is metrically complete, you can still add other elements to it. You can extend it to the plane $\mathbb{R}^2$, or the complex numbers $\mathbb{C}$, or the surreal numbers $\mathfrak{s}$. You can add new numbers in between, add infinitesimals and infinities, and do many other things besides. But $\mathbb{R}$ by itself is gap-free (complete) because every Cauchy sequence in the space converges to a point in the space;in terms of limits of sequences, it has no holes that need filling.


One way to define the limit $L$ of a sequence $x_1,x_2,\ldots$ is to say that every ball, no matter how small, around the limit point $L$ contains an entire tail of the sequence. By this definition, you can prove that $0$ is the limit of the sequence 1/2, 1/3, 1/4, ..., because every ball around 0 contains an entire tail of the sequence.

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  • $\begingroup$ When you'll be discussing non-archimedean things like the surreals, "get arbitrarily close together" is perhaps not the best gloss for something like "get closer together than an arbitrary positive rational" (and similarly for "every ball"). The tension between surreals being able to get even closer together than any two rationals/reals could despite the reals "having no gaps" in the dedekind completeness sense seems to be what the question is about. Using Cauchy completeness seems risky to me as it doesn't imply the field is archimedean, so an "$\varepsilon$" that confused the OP could exist. $\endgroup$
    – Mark S.
    Oct 26 '20 at 2:16
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    $\begingroup$ Oh, I think I understand much better now. The internal property makes sense. There are still "holes" as you said, but those holes never arise unless you add new elements into the set. In other words, as limits converge in R, but that doesn't mean that those limits are the closest number possible that exists to what a sequence converges to. Is that right? $\endgroup$ Oct 26 '20 at 2:17
  • $\begingroup$ +1. @CalebBriggs Yes, that's basically right. (Although that said this does indeed pose a limitation on the ways we can add new numbers - see my answer.) $\endgroup$ Oct 26 '20 at 2:18
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The first question is difficult to tackle: the idea you have is too vague to really have a specific point of failure.

I can say some things about the second question though. It's true that there are bigger number systems than the reals and which have lots of new elements lurking between old real numbers. However this does not mean that the reals don't have any holes, at least not in the sense that's meant when we say that. "Every nonempty bounded-below set has a greatest lower bound" is a property of a number system on its own: the fact that $0$ isn't the g.l.b. of $\{{1\over n}: n\in\mathbb{N}\}$ in the sense of the surreals doesn't contradict the fact that it is the g.l.b. of that set in the reals.

"The reals have no holes" has a very specific technical meaning. You can think of it as a statement about a particular kind of new number which we can't add to $\mathbb{R}$. Specifically, any finite surreal number (that is, not bigger than every real and not less than every real) is "stuck to" some real number, in the sense that it's infinitesimally close to exactly one real number. There's no place where you can put a new number "on its own," in a sense. Contrast this with $\mathbb{Q}$: when we add $\pi$ to $\mathbb{Q}$, there's no element of $\mathbb{Q}$ closest to $\pi$, so in a sense $\pi$ really is "off on its own."

(That said, if you're interested in the idea of a "maximally big number system" - which is not what "complete" means in the context of $\mathbb{R}$ - this article is a good source. Basically, the surreals do indeed have a special place of honor.)

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  • $\begingroup$ This is really helpful, and thanks for including a link, I'll definitely check that out. So if I understand, any number which attempts to extend the reals will end up not adding anything new, in the sense that all the new numbers will be arbitarily close to a number that was already in the reals? So, new properties won't really emerge, because these new numbers will act esentially the same as their real counterparts. For example, if $s$ is a surreal number and $r$ is the arbitarily close real number, then: $\alpha s$ is arbitarily close to $\alpha r$, $\beta+s$ is arbitarily close to $\beta+r$ $\endgroup$ Oct 26 '20 at 2:30
  • $\begingroup$ @Caleb, as I said to user326210, I wouldn't recommend using "arbitrarily close" in this sort of context. A big new property emerges when you build a bigger ordered field out of the reals is you get positive numbers smaller than every positive rational. Your conjecture basically requires $\alpha$ to be real. $\endgroup$
    – Mark S.
    Oct 26 '20 at 2:39
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    $\begingroup$ @CalebBriggs Well, there are some caveats. First of all, this is only for finite numbers. Infinite numbers of course are far away from every real. Also it's not true that you don't get any new properties: adding an infinitesimal for example gives you the property of ... having an infinitesimal. :P But it is true that they don't amount to a hole in the reals that we're thinking about. $\endgroup$ Oct 26 '20 at 2:41

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