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[EDIT] Is zero the limit to the following sequence?

$1/2, 1/4, 1/8, 1/(2^4),.....$

Why it shouldn't be $1/\aleph_0$? Since $\aleph_0$ is the limit to sequence:

$2,4,8,2^4,..,2^n,...$

$1/ \aleph_0$ can be understood as an infinitesimal, i.e. larger than zero yet smaller than any finite number.

The context is that of including transfinite numbers, thereby violating the Arachimedean property.

$\aleph_0$ is a limit in the sense of being a minimal upper bound, i.e. if we view the numbers in the second sequence as finite cardinal numbers, then $\aleph_0$ would be a the least (with respect to cardinal smaller than) cardinal number that is cardinally greater than all of them.

$1/ \aleph_0$ can be understood as an infinitesimal, i.e. larger than zero yet smaller than any finite number.

Here $1/ \aleph_0$ would be a limit in the sense of being a maximal lower bound on the first sequence. Since whatever algebra we want to define division in for such milieus, then this must contain a reciprocal for every number, so if there is a number bigger than $1/\aleph_0$ and yet is a lower bound on the first sequence, then the reciprocal of it would be smaller than $\aleph_0$ and yet an upper bound on the second sequence, which is implausible!

Of course this necessitate defining a certain algebra to handle division in such milieus. But for now I'm merely concentrating upon the naive aspect of this idea, which I think is a plausible one.

Zuhair
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    The conventional notion of the number line and of limits does not have room for infinitesimals. Look for the hyperreals or the surreals. However, even in the theories that do have room for infinitesimals, $1/\aleph_0$ is not the name they give to them. – Arthur Apr 21 '19 at 20:14
  • $\aleph_0$ is the name given by Cantor to the first infinite cardinality larger than any finite cardinality. Its natural to assume that $1/\aleph_0$ would be a kind of infinitesimal, even if that's not what infinitesimals are named. – Zuhair Apr 21 '19 at 20:16
  • There are other kinds of infinities than cardinalities. Some of them make more sense to invert than $\aleph_0$. The surreals have $1/\omega$, for instance. And I'm not very familiar with limits in the surreals (it even if they make sense), but I think that that's what the limit of $1/2,1/4,1/8\ldots$ is there. – Arthur Apr 21 '19 at 20:19
  • @Arthur, yes correct, there are the transfinite ordinals, still in the standard set theory $\aleph_0$ is itself the first infinite ordinal $\omega_0$, so that won't make much of a difference here. – Zuhair Apr 21 '19 at 20:22

2 Answers2

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The explicit questions

"Is zero the limit to the following sequence?"

Yes. Unless you clarify that you're doing something very nonstandard, you must be using one of the compatible default meanings of limit everyone would expect, with a context of rationals or real numbers, and the limit is indeed $0$.

Why it shouldn't be [something other than $0$]?

There is good reason for this: in the context of the reals or the rationals, $0$ is the unique number $\ell$ for which the differences $1/2-\ell, 1/4-\ell,\ldots$ become arbitrarily small (smaller than any positive rational/real). The reals/rationals don't have any number greater than all of the $2^n$s (like "$\aleph_0$") to let you build a counterexample by taking the reciprocal.

In a sense, the real numbers are the maximal ordered field without "infinite"/"unbounded" elements greater than all of the integers (this is the archimedean property).

$\aleph_0$ is the limit to sequence $2,4,8,\dots$

This is incorrect. Under the usual definition of limit, we would usually say this diverges (since it has no real number limit) to $\infty$ (see wikipedia on this), and the $\infty$ symbol doesn't really have a connection to any cardinality. In some contexts (e.g. topology), you could think of $\infty$ as a point in a special "real line with endpoints" called the extended reals. But even then, there wouldn't be a good reason to associate $\infty$ with any other cardinal.

Additionally, the claim is not easy to fix even with a charitable reading. Even if you wanted to let some sort of infinite number be some sort of limit of the sequence $2,4,8,\dots$, it would be reasonably be a number where you could add $1$ and get a new number. Using cardinal notation like $\aleph_0$ strongly implies you want something where $\aleph_0+1=\aleph_0$ (see cardinal addition), which isn't likely to work as a nice infinite number you would use here.

In a comment, you mentioned the fact that in standard set theory we take $\aleph_0$ to be $\omega_0$ (a.k.a. $\omega$), and claimed that this doesn't make a difference. But when you're trying to explain a non-standard idea like this, even if there's technically no difference in denotation, the difference in connotation (in what it implies about the expected properties of addition) is significant.

Is there something close to the intuition?

Setup

As Arthur suggested in a comment, in non-archimedean ordered fields and similar (like the surreal numbers), there is some hope since there are infinite and infinitesimal elements.

Indeed, in the surreal numbers, there is a copy of the ordinals, except that the arithmetic operations aren't the usual ones, they're the so-called natural sum and product. Notably, $\omega+1$ works out the same either way.

To talk about examples like the one in the question, we don't need all of the surreals, just a chunk that has numbers like $\omega$ in it.

"Limits" for the surreals

In several works of Norman L. Alling on the Surreals, including the paper Conway's Field of Surreal Numbers, Alling has defined a sense of "limit" different than the usual one, for which certain sequences (and similar - the indexing set could be any limit ordinal) have unique surreal limits.

Very briefly, if you have a sequence that is "very Cauchy" (my phrasing) in the sense that the difference between $|a_k-a_j|$ is infinitesimal compared to $|a_j-a_i|$ whenever $i<j<k$, then there are many "pseudo-limits", but there is a unique pseudo-limit that is the simplest (in the sense of simplicity/birthday of the surreals), which Alling calls "the limit".

This definition, while interesting, does not apply to $1/2,1/4,1/8,\ldots$ as that sequence is merely Cauchy in the usual sense, not "very Cauchy".

"Limits" for the surreals - take 2

We can take still take inspiration from Alling's work, though. We can just take the work to find pseudo-limits and hope that it'll still make some sense. (In doing so, we replace his original notion of pseudo-limits with one that uses absolute value in place of a valuation of the location in a hierarchy of infinitesimals.)

For example, $1/3,1/9,1/27,\ldots$ would have to have something like pseudo-limits between lower bounds $1/9-|1/3-1/9|=-1/9, 1/27-|1/9-1/27|=-1/27,\ldots$ and upper bounds $1/9+|1/3-1/9|=1/3, 1/27+|1/9-1/27|=1/9,\ldots$. There are many surreal numbers between those negative and positive bounds, but the simplest one is $0$.

But if we take the example from the question, $1/2,1/4,1/8,\ldots$ would have to have something like pseudo-limits between lower bounds $1/4-|1/2-1/4|=0, 1/8-|1/4-1/8|=0,\ldots$ and upper bounds $1/4+|1/2-1/4|=1/2, 1/8+|1/4-1/8|=1/4,\ldots$. There are many surreal numbers between $0$ and the $1/2^n$s, but the simplest one is $1/\omega$.

We still may not say that "the limit of $1/2,1/4,1/8,\ldots$ is [anything but $0$]". However, we could say something like "If you extend the definitions of Alling for surreal limits to certain cauchy sequences where something reminiscent of pseudo-limits still happen to exist, then the 'limit' of $1/3,1/9,1/27,\ldots$ is $0$, and the 'limit' of $1/2,1/4,1/8,\ldots$ is $1/\omega$."

A note about the reciprocals

Note that even with my implied re-definition of Alling's definition of surreal limits, sequences like $2,4,8,\ldots$ still don't have limits because they're not anything like Cauchy. Taking inspiration from the monotone convergence theorem, since it's increasing, we might want to say that its "limit" should be the simplest (in the sense of the surreals) upper bound.

In that sense, $2,4,8,\ldots$ and $3,9,27,\ldots$ both have "limit" $\omega$. This is unfortunate for the intuition in the question, since the sequences of reciprocals have (for a different meaning of "limit") two different limits: $0$ and $1/\omega$.

As an aside, since $1/2,1/4,\ldots$ and $1/3,1/9,\ldots$ are decreasing, we might want to examine the simplest lower bound for each sequence. Fortunately or unfortunately, the answer is $0$ for both.

Community
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Mark S.
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  • I don't know much about the surreals, but naively speaking, if we define reciprocals of infinite cardinals, and allow the transfinite cardinals of Cantor, then the limit to sequences $1/2^n$ and $1/3^n$ should be the same, namely the rational transfinite number $1/\aleph_0$. There is no infinite rational number greater than it that is smaller than all items in those sequences? all the other infinite rationals like $1/2^{\aleph_0}$, $1/2^{2^{\aleph_0}}$,... are strictly smaller than $1/\aleph_0$. – Zuhair Apr 22 '19 at 05:33
  • Thanks a lot for that informative answer. But I just have a simple comment? When we are speaking of infinite sequences like for 2,4,8,.. we say it diverges because there is no "real number limit" for it. However if we extend the system with Cantor's transfinite numbers, then it appears to me that $\aleph_0$ being a cardinal limit to finite cardinals, seems to be in conformity to what "limit" is ought to mean, which is the minimal upper bound, i.e. the cardinally smallest number greater in cardinality than all naturals. – Zuhair Apr 22 '19 at 10:28
  • You said: "Even if you wanted to let some sort of infinite number be some sort of limit of the sequence 2,4,8,…, it would be reasonably be a number where you could add 1 and get a new number". My question why it is reasonable for the limit to be incremental by additions of 1? To me it appears that the most important property is that the limit must be smaller than some number and we can define the $<$ relation, now we have $\aleph_0 < 2^{\aleph_0} < 2 ^ {2^{\aleph_0}},...$ where $<$ is cardinal smaller than; that is besides of course the limit being $\geq$ the numbers it limits. – Zuhair Apr 22 '19 at 10:32
  • @Zuhair (re: your first two comments) If there is a transfinite number $H$ greater than all of the integers, then if I can do some arithmetic with it compatible with the order, then $1/(H-1)$ and $2/H=1/H+1/H$ are greater lower bounds than $1/H$ for $1/2^n$ than $1/H$. And $H-1$ and $H/2$ are lower upper bounds for $2^n$ than $H$. To force $1/2^n$ to have a greatest lower bound you have to either say we're not allowed to add two copies of $1/H$ nor multiply $1/H$ by $2$, or say that we can but it doesn't play nicely with the order (e.g. $1/H<1/8$ would no longer imply $2/H<2/8$). – Mark S. Apr 22 '19 at 12:12
  • @Zuhair "why...additions of 1?" I was making an assumption that in a context where you had division (e.g. $1/4$ and "$1/\aleph_0$"), you probably had addition. If you meant to only have multiplication and division, but $2$ is around, then again, $H/2$ would be a lower upper bound for $2,4,8,\ldots$ unless division does not interact nicely with the order (e.g. so $8<H$ would no longer imply $8/4<H/2$). If you don't allow nice operations on your transfinite numbers, then $\aleph_0$ could be the least upper bound, but then $1/\aleph_0$ doesn't mean "multiplicative inverse" anymore, so why bother? – Mark S. Apr 22 '19 at 12:17
  • I like to stick to $\aleph_0$ as the limit to the finite numbers. Now to address your comment. $\aleph_0 - 1 = \aleph_0$, so $1/(\aleph_0 -1) = 1/\aleph_0$. Similarly we do have $\aleph_0/2 = \aleph_0$. Also we have $2/\aleph_0 = 1/\aleph_0$, [see my prior comments]. So there is no number interevening between $\aleph_0$ and the known finite rationals, nor there is any number intervening between the finite rationals and $1/\aleph_0$, which establish them as bounds over the finite rationals. Also $8 < \aleph_0/1$ does imply $8/4 < \aleph_0/2$ – Zuhair Apr 22 '19 at 12:30
  • to address the second of your comments, of course I'm thinking of a system in which there is addition, subtraction, multiplication and division, and all numbers in it can be considered as "rationals" whether finite or infinite. And I think we can extend the system to close over exponentiation and rooting as well. I still fail to see why one needs additions by 1 to be incremental over limits? this is certainly not required in the definition of a limit in the first place. – Zuhair Apr 22 '19 at 12:35
  • @Zuhair I think the short answer is that operations like the ones you describe are very nonstandard, since the basic defining properties of subtraction and division are discarded. Your subtraction does not properly undo addition, etc. So why don't limits work the way you want? Because no one lets arithmetic work that way. – Mark S. Apr 22 '19 at 12:40
  • I think if one takes Cantor very seriously, then we do have a cardinality that clearly abides by the definition of a limit as the minimal upper bound. And if one extend the system to have reciprocals of those with careful re-definition of subtraction, division and exponentiation, then one can NAIVELY get a system which includes something like infinitesimals, and infinities, and that is closed totally over all six known operators. Actually I think I can even get it to close over division by zero, but this is another story. – Zuhair Apr 22 '19 at 12:45
  • but yet I agree, the results would be non-standard, in the sense of how it compares to the finite arithmetic, much as Cantor's arithmetic of the transfinite numbers is also non-standard in that sense. However, I do believe that the naive approach I'm speaking about can result into what would be considered one day as the STANDARD approach to those infinite numbers, much as Cantor's transfinite numbers is the STANDARD approach to the infinite numbers. In nutshell, what I'm trying to do here, is to extend Cantor's paradise in a naive manner to the infinitesimal realm. – Zuhair Apr 22 '19 at 12:47
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You can defined number system where such bizarre limits exist, but in the real numbers it is not the case. By definition of convergence in the real numbers $(1/2)^n\to 0.$

Melody
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  • Yes zero is the "real" number limit to $(1/2)^n$, of course. Clearly $1/{\aleph_0}$ is not a "real" number. My question is beyond the "real" limit. I'm asking about the true limit of that sequence, if we are to maintain that transfinite numbers exist. – Zuhair Apr 21 '19 at 20:19
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    There is no such thing as a "true limit". "Limit" is just a word, and it means whatever we define it to mean. – Eric Wofsey Apr 21 '19 at 20:20
  • @EricWofsey, agreed! what I meant is that if we are to hold that transfinite numbers exist, then it would be natural to try to define inverts of them, and to see those as a kind of infinitesimals, and by then those can be defined as the limits to those sequences. – Zuhair Apr 21 '19 at 20:24
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    @Zuhair inverts of them with respect to what operation? Cardinal multiplication and even ordinal multiplication aren't nice enough to admit such a thing. That said, "natural multiplication" on ordinals is nice enough, and this is compatible with the operations used in the surreal numbers. – Mark S. Apr 22 '19 at 00:37
  • @MarkS. with respect to cardinal multiplication. I thought the problem with this was that it can lead to "indeterminacy" like for example with $\aleph_0/\aleph_0$ having any finite number or $\aleph_0$ being a multiplicative inverse, but this could be fixed by taking the maximal as the value of division! So division itself is to be re-defined as the maximal reciprocal of multiplication, so $\aleph_0/\aleph_0 = \aleph_0$. Would that make them "nice enough"? – Zuhair Apr 22 '19 at 05:10
  • naive thinking about the matter make one thing that the finite rationals lie between $\aleph_0/1$ and $1/\aleph_0$, so we have $\aleph_0/ k = \aleph_0/1$ if $ 1/\aleph_0 \leq k \leq \aleph_0$, and for the reciprocals we have $k/\aleph_0 = 1/\aleph_0$ for $ 1/\aleph_0 \leq k < \aleph_0$ – Zuhair Apr 22 '19 at 06:29