Well, to be honest, I’ve never taken the time to learn much about infinitesimals. Still, I’m just thinking that if e is an infinitesimal, then e is strictly between 0 and 2[sup]n[/sup] for any positive integer n–e is “in the way”, so the sequence can’t possibly limit on zero.
Anyway, my earlier suspicion about this was right, there has been research done in this area:
(pdf file)
One of the nice things about the hyperreals is that statements that only involve real numbers have the same truth value when taken over the hyperreals. The statement limit(1/2[sup]n[/sup]), n -> infinity) = 0 is such a statement, and since it’s true for the reals, it’s true for the hyperreals as well.
Ask Dr. Math has a rather nice introductory article on the subject that you might like to read.
I still can’t see how it can be zero. Again, there’s an open neighborhood of zero, (-e,e), which the sequence never enters, so it can’t limit on zero. In your example it looks to me like your just using an alternate notation for real numbers without involving hyperreals at all. Wouldn’t the limit of 1/2[sup]n[/sup] be something like (1/2,1/4,1/8,…,1/2[sup]n[/sup],…)?
All that math looks great, but I don’t have a clue about it - i’m not being rude when I don’t comment on it 
I was thinking - in a situation with infinite possible outcomes (a situation that is only possible in an infinite universe) there is an infinite number of probability values between 0 and 1 for each outcome. If you take any possible outcome at all there is always 1 more with a higher probability so you can never accurately predict what might be the case. Like Xeno’s paradox, but without the paradox.
Make sense?
Read the article I linked to, paying close attention to the section on the transfer principle. That’s what matters here.
I’m not seeing anything there that relates to this; limits don’t work the same in reals vs. hyperreals:
http://www.cs.nyu.edu/pipermail/fom/2000-June/004119.html
The definition of convergence says that x[sub]n[/sub] converges to a if and only if for any open neighborhood U of a, there is an N such that x[sub]n[/sub] is in U for all n > N. I’ve already exhibited an open neighborhood (-e,e) of 0 such that 1/2[sup]n[/sup] is never in (-e,e). Where’s the flaw in my argument?
It looks like the definition of covergence doesn’t transfer to the hyperreals because it’s a statement involving subsets of R. As I said earlier, read the article I linked to with special attention to the section on the transfer principle.
The statement in question is thus:
For any [symbol]e[/symbol] > 0, there is an N such that whenever n > N, |1/2[sup]n[/sup]| < [symbol]e[/symbol].
The only objects in that statement are real numbers and functions in R [symbol]®[/symbol] R, and such a statement is well-formed.
I agree that the notion of convergence doesn’t transfer. However, it’s not an issue of whether the definition I gave transfers, since that is the definition of convergence in an arbitrary topological space, including the hyperreals. No transference required.
I’m not following you here, because now you seem to be claiming that it does transfer, after all. How can you claim that it transfers, when this statement is clearly true for the reals, and clearly false for the hyperreals when e > 0 is an infinitesimal?
Wait, I think I may see the problem. When you transfer the above statement to the hyperreals, you get:
For any hyperreal e > 0, there is a hyperreal N such that whenever n > N, |1/2[sup]n[/sup]| < e.
Such an N couldn’t possibly be an integer, for the reason I already mentioned, it would have to be an infinite (positive unlimited) hyperreal number (greater than any integer).
The problem with that, however, is then we’re no longer talking about the sequence 1/2[sup]n[/sup], but rather an extension of a sequence allowing n to take on transfinite values. So I still claim that if we specify that we’re talking about the sequence 1/2[sup]n[/sup] in the hyperreals, it can’t possibly converge to zero.
I’m not absolutely sure that’s the problem, but something is definitely wrong with saying the sequence 1/2[sup]n[/sup] converges to zero in the hyperreals.
Let x = 1/e. Let N = be the smaller integer such that 2[sup]N[/sup] > x.
Clearly 1/2[sup]N[/sup] > x < 1/x < 1/e.
So, you have a number 1/2[sup]N[/sup] inside the neighborhood (-e;e).
Cabbage:
That is correct. When working with sequences in the hyperreals, you need to consider the hyperintegers as well ( a hyperinteger is a hyperreal number y such that y = for some hyperreal number x).
An infinite sequence <a[sub]n[/sub]> is said to converge to a real number L if a[sub]H[/sub] is infinitely close to L whenever H is a positive infinite hyperinteger.
In the case under discussion, 2[sup]-H[/sup] is infinitessimal whenever H is positive infinite, and the sequence converges to 0.
Sérgio: Cabbage and ultrafilter are both quite familiar with the rules of standard analysis. The subject under discussion is the behaviour of hyperreal numbers.
Sérgio, that’s true in the reals, but not in the hyperreals. One of the properties of the hyperreals is that it has infinitesimals, and one of the properties of an infinitesimal e is that 0 < e < 1/2[sup]n[/sup] for any positive integer n. And, consequently, 1/e > 2[sup]n[/sup] for any positive integer n , so there is no such integer N as you claim.
You may also like to look at ultrafilter’s link from above:
Thanks, Jabba, your post snuck in there while I was making mine; I think that clears up the confusion.
OK, I went to the site. For what I read, zero is a hyperreal. And zero fits the requisites to be the limit of the sequence. Since the limit must be unique, no infinitesimal can be candidate.
Sérgio, the problem with that is that the hyperreals consist of infinitesimals. An infinitesimal is a numberthat is closer to zero than any real number; a defining property could be x is infinitesimal if and only if nx < 1 for all positive integers n.
In an earlier post I gave the definition of a convergent sequence in an arbitrary topological space: The definition of convergence says that x[sub]n[/sub] converges to a if and only if for any open neighborhood U of a, there is an N such that x[sub]n[/sub] is in U for all n > N.
We’ve been talking about the sequence 1/2[sup]n[/sup]. Now, if we allow n to range over all positive hyperintegers, then yes, the limit of this sequence is zero.
However, when I was talking about the sequence 1/2[sup]n[/sup], I was only talking about the range of values for when n is a positive integer, not hyperinteger. Now this sequence doesn’t limit on zero.
To be clear:
The sequence {1/2[sup]n[/sup]:n is a positive hyperinteger} converges to zero.
The sequence {1/2[sup]n[/sup]:n is a positive integer} doesn’t converge to zero.
I’m not terribly familiar with the hyperreals, but this much I’m sure of.
The reason the latter sequence doesn’t converge to zero is because if I take a positive infinitesimal e and consider the open neighborhood (-e,e) around zero, the latter sequence never even enters this open set (since e < 2[sup]n[/sup] for all positive integers n); this is a clear violation of the necessary condition for convergence, as I defined above.
I would expect that the latter sequence converges to the hyperreal infinitesimal (1/2, 1/4, 1/8, 1/16,…, 1/2[sup]n[/sup],…), but this may be where my lack of familiarity is a problem. Perhaps it doesn’t converge at all, but I feel more confident that my first statement is more likely to be correct.
I’m confused! For what I read in the site you suggested, the reals are a subset of the hyperreals. Since zero is a real, it is also a hyperreal.
In the field of the reals zero is the limit of the sequence, with N an ordinary integer. In the field of the hyperreals zero is the limit with N a hyperinteger.
Hyperreals consist of all the reals, plus the infinitesimals plus the infinite numbers. Am I wrong?
If you only allow integer indices on the sequence, it should fail to converge, for the reasons that Cabbage cited. However, if you allow hyperintegers (which are greater than any normal integer), it does converge to 0. Is that a little clearer?
You’ve all been getting ahead of my math/probability knowledge, so forgive me if this has been refered to earlier in a format I did not understand.
Imagining an infinite series of 1’s and 0’s each digit randomly chosen. Is there a difference between the probabilitites of the result being
000…, probability=0 but can happen, and
222…, probability=0 but cannot happen.
The first case has a positive probability for every finite list of zeros, whilst the second case has zero probability for every finite list of twos.
Strictly speaking, it doesn’t make sense to talk about the probability of the second case.
A probability function is a mapping P [symbol]Î[/symbol] 2[sup]S[/sup] [symbol]®[/symbol] [0, 1] for some set S which has some nice properties. P is extended to S by P(s) = P({s}), s [symbol]Î[/symbol] S. If E is not a subset of S, E is not in the domain of P, and so P(E) is undefined.
If you want to informally say it has probability 0, that’s fine, but not the mathematical way to do it.
Cabbage:
The number [symbol]e[/symbol] is a positive infinitesimal if it is positive and is less than positive real number. A negative infinitesimal is defined similarly. A number is infinitesimal is it is a positive infinitesimal, a negative infinitesimal or zero. It may seem strange to define zero as an infinitesimal, but there is a good reason. For example, we want the statement “the sum of two infinitesimals is infinitesimal” to be true, and it is not if we exclude zero.
Thanks for the clarification, Jabba. I was thinking of “infinitesimal” as meaning what you’ve defined as positive infinitesimal.