Does 0.999~ Equal 1?

[Hurkyl]

In binary, 0.1111~ = 1 This is the exact analogue of the problem in decimal. Now find me a binary number between 0.111~ and 1

Why, 0.111...01, of course.

(Please nobody construe this as being a serious reply!)

 

[Hurkyl]

Quick question, if you had line .999~ cm long vs a line 1 cm long would they be exactly the same or would the .999~ be missing the point at 1

If you had a line .999~ cm long, then something is seriously wrong, because lines extend infinitely in both directions.

If you had a line segment 0.999~ cm long, then it would have both its endpoints because, by definition, line segments contain both of their endpoints.

 

[russ_watters]

It seems to defy logic sense. Why? Because of limits. I thought the whole point is of a limit THAT YOU APPROACH SOMETHING but never actually reach it.

That isn't how limits work. A limit is an exact number - what tends to infinity isn't the limit (actually, I'm not sure if it has a name, but it's the "x" below...).

Ie:

lim 1/(1-x) = 1
x-> 0 (sorry - hate latex)

The equals sign is for real - that limit really is exactly equal to 1. The "x" is approaching something (infinity), but the limit itself is exactly equal to 1.


An interesting link, using polygons inscribed in a circle: http://www.coolmath.com/limit1.htm

-As n gets larger, the n-gon gets closer to being the circle.
-As n approaches infinity, the n-gon approaches the circle.
-The limit of the n-gon, as n goes to infinity, is the circle!

"n" approaches infinity, but the limit is exactly equal to the circle.

And that's the entire point of limits - to find exact values for things that are otherwise difficult to pin down (such as the slope of a curve at a single point).

I actually enjoyed calculus-I (yeah, I'm a freak), because when I got my arms around this concept (didn't just plug and chug, but actually understood the point of limits), calculus-I became a piece of cake. It also helped that I took it in conjunction with Newtonian physics and they are so related that both help you learn the other.

 

[sameandnot]

do you know that "there are twice as many numbers as numbers"?

 

[Mindscrape]

lim 1/(1-x) = 1
x-> infinity (sorry - hate latex)

Don't you mean zero?

 

[Sir_Deenicus]

do you know that "there are twice as many numbers as numbers"?

Word play.

---

I think the so called Weisserstras Rigorous Epsilon-Delta formulation of limits is every bit as artificial as the naive concept. Simply one level of abstraction high enough to delude one's self into acceptance.
---
nate808, on the ruler you would have the point. The best explanation I can think of is that 0.9... is 1 because we are clever enough to jump out of the system or loop of repetetive non stop addition to see what is actually going on with adroit manipulation of intrinsics.

 

[Sir_Deenicus]

I hope I didnt make too many mistakes. For the hyperreals , 0.9... <> 1. First I clarify and set up. If A is an *R statement then R --> *R (if true in R then R* for A and *R statement) , the converse, *R --> R is false. Thus simply stating that something is true in R makes it always true in *R is not enough. There is only a gurantee that if A is a properly defined statement then there will be a behaviour in *R that "functions" as expected. But it need not be the only one. An internal statement in *R is not necessarily true in all subsets of *R, especially depending on your language strength. Also, *N is an internal subset of *R but not N. For example, x + y = 3 is only true in *R when we have (x,y) = (1-e, 2 + e) otherwise, x + y <> 3 even though st(x + y)=3. The behaviour of infinitesimals and finites hint that there exists gaps in any cuts we make in *R (if we began with sets containing only finite hyperreals for example).

Note though, that A an R statement --> R <--> *R. But then *R is not Dedekind Complete, it is not possible to define cuts such that there will always be a number in a gap. It not possible to extend the completeness axiom over to *R. The proof for 0.999... = 1 depedends on an ability to define cuts at its heart and Dedekind completeness. I will not go into that but will argue intuitively.

There exists a hyperreal number e that is infinitely near to any finite number r. I use ~= to mean infinitely approximately (infiniproixmitely) close to. Since there exists e infinitely near to all r then there exists r ~= r - e ~= r + e. The real, r = st(r), the standard part of the hypernumber r. Suppose we have two finite hyperreals r ~= 0.99...9 and s ~=1 then you cannot prove that it converges or whatever because there exists hyperreals r,s,t members of some set S where r < t < s with t not in the interval. Which is undefineable anyway. There is no intreval between the two numbers since such a thing is undefined. You can still however, use hyperreals to study convergance and limits and all that lovely stuff since we would actually be operating on reals but using hyperreals in place of infinities, infinitesimals and stuff. Just as rigorous but much more efficient than epsilon-delta. limits.

st(0.9...) = st(s) = 1 ~= 1 - e < 1+e ~= 1 ~= 1 + e.

This hyperreals vs limits remind me of the old fight between quaternions and vectors and grassman's algebras. limits map to quaternions and hypers to Grassman.

 

[matt grime]

No idea what that proves. I know what you think it proves, but I'm afraid that 1/n still tends to zero in the hyperreals. Indeed the transfer principle you cite proves it. Arguing intuitively doesn't prove a single thing.

You do not construct hyper reals by taking limits of 'decimals' or real numbers, 0.9... still represents a real number, the limit of the partial sums, that limit is taken in the subset of the reals since the limits are purely objects lying in the reals, hence they are still equal in any *extension* of R.

 

[russ_watters]

Don't you mean zero?

Dang, me too... Fixed now, thanks.

 

[Sir_Deenicus]

No idea what that proves. I know what you think it proves, but I'm afraid that 1/n still tends to zero in the hyperreals. Indeed the transfer principle you cite proves it.

Perhaps we are talking about 2 different things? 0.9... does not tend to 1 on the hyperreal line because the hyperreal line contains infinitesimals!

The picture argument. On the hyperreal line, There exist numbers between every number that disallow the step required for 0.9... to converge at 1. We cannot make the inductive jump. 0.9...9, 0.9...9 + e, 0.9...9 + 2e and so on.

The formal attacks. Consider the first proof given http://www.cs.uwaterloo.ca/~alopez-o/math-faq/node41.html#SECTION00531000000000000000".
The proof fails since it is possible to choose e > 0 but less than or equal to all *R. Then e cannot equal 10 ^ (-1/d). Indeed the proof ends up saying something meaningless like infinitely approximately less than zero.

Again in http://en.wikipedia.org/wiki/Proof_that_0.999..._equals_1" it states:

If an upper bound less than 1 exists, it can be written as 1−x for some positive rational x. To bound 9⁄10, which is 1⁄10 less than 1, x can be at most 1⁄10. Continuing in this fashion through each decimal place in turn, induction shows that x must be less than 1⁄10^n for every positive integer n. But the rationals have the Archimedean property (they contain no infinitesimals), so it must be the case that x = 0. Therefore U(S) = U({1}), and 0.9999… = 1.

Since there exist non zero infinitesimals we can only say that the series gets aproximately close to 1 since we are approximately close to 0 (there exists 0 < e ~= 0). But in fact the whole endeavoure is meaningless since the cuts that created the entire decimals should not have been possible! Since we have e, the gaps created should have been empty. There is no concept of inifinite precision, just a line riddled with gaps as we go through all c in our original set. In fact, the proof is not possible since a meaningful upper bound cannot be found in *R due to the existence of infinitesimals.

Suppose we rewrite the decimal number as a sequence, then because there is no completeness [axiom] in *R, it cannot be shown to converge or diverge etc. I end like so:

The archimdean property is equivelant to the statement s = {1/n} converges to 0. Any system which contains infinitesimals is non-archimedean. The hyperreals contain infinitesimal and thus s = {1/n} does not neccessarily converge to 0. This fact, coupled with the lack of gurantee of convergance on monotonic sequences and a lack of least upper bound property all point that the series 9 * sum(n, infinity,1/10^n) over *R, whose properties are accounted for in their totality cannot be shown to converge at 1.

 

[matt grime]

So, what you're saying is that the statement

"1/n does not tend to zero"

is provably true in the hyperreals, which, implies that it is true in real analysis and moreover provably in real analysis, hmm, that seems more than a little problematic. perhaps you need to check what convergence means in the hyperreals (which is more than the statement that x_n-x lies between -e and e, something you point out is not meaningful in *R, I believe. something tends to zero if the limit is both infinitesimal and real, and thus zero since thaht is the unique real infinitesimal, so we are saying 1/n tends to zero because it is *real* at all points and the limits is an infinitesimal, hence 0, convergence is a different beast here.)*R is an extension of the real number system to a large set, results that are statable about elements of *R and that are true in R are true in *R, that is the transfer principle. The statement 1/n converges to zero is statable in *R, restricted to R is true so it is true in *R.

 

[Hurkyl]

I think the so called Weisserstras Rigorous Epsilon-Delta formulation of limits is every bit as artificial as the naive concept. Simply one level of abstraction high enough to delude one's self into acceptance.

How can accepting any axiom or definition be a delusion?

I can't even figure out what makes you think the definition is abstract anyways.

(The Weisserstras formulation is superior to the naive concept, of course, simply because it is a formulation -- I can't mathematically prove anything about the naive concept)

nate808, on the ruler you would have the point. The best explanation I can think of is that 0.9... is 1 because we are clever enough to jump out of the system or loop of repetetive non stop addition to see what is actually going on with adroit manipulation of intrinsics.

There are several problems with this:
(1) One's "cleverness" (or lack thereof) has nothing to do with what is actually deducible.
(2) Even if I grant you the liberty of calling an infinite sum "non-stop addition", there are plenty of formulations of the decimal numbers that never make reference to such a thing.
(3) In many formulations of the decimal numbers, 0.9~ = 1 (and similar equations) are part of the definition, and not some "adroit manipulation of intrinsics".
(4) Your whole response seems to imply that you think 0.9~ = 1 is a "bad" thing -- but you have to remember that the point of the decimal numbers is not so people can have fun pushing numerals around: it's so that they can be a complete ordered field. (i.e. a model of the real numbers) There are plenty of demonstrations why allowing 0.9~ and 1 to be different would mean this goal was not achieved. (Such as the famous 1.0~/3.0~*3.0~=0.9~ calculation)

I hope I didnt make too many mistakes. For the hyperreals , 0.9... <> 1. First I clarify and set up. If A is an *R statement then R --> *R (if true in R then R* for A and *R statement) , the converse, *R --> R is false.

Wrong. (Possibly due to sloppiness)

If P is an (internal) statement about the reals, and *P is the corresponding statement about the hyperreals, then P is true if and only if *P is true.

Thus simply stating that something is true in R makes it always true in *R is not enough.

Of course it is enough -- this is the direction that allows me, given the truth of the statement

0.9~ = 1

to conclude the truth of the statement

*0.9~ = *1

There is only a gurantee that if A is a properly defined statement then there will be a behaviour in *R that "functions" as expected. ... The behaviour of infinitesimals and finites hint that there exists gaps in any cuts we make in *R (if we began with sets containing only finite hyperreals for example).

I find this all confusing; I'm not sure what it's supposed to say. (e.g. x+y=3 is true in the reals if and only if there is an e such that x = 1-e and y = 2+e)

Note though, that A an R statement --> R <--> *R. But then *R is not Dedekind Complete, it is not possible to define cuts such that there will always be a number in a gap. Note though, that A an R statement --> R <--> *R. But then *R is not Dedekind Complete, it is not possible to define cuts such that there will always be a number in a gap. It not possible to extend the completeness axiom over to *R.

Yes, you can. Dedekind completeness is a true internal statement about the reals. Therefore it is true for the hyperreals.

You're confusing the internal and external statements of Dedekind completeness. It is true (by the transfer principle) that every internal Dedekind cut is generated by an element of *R, and therefore *R is internally Dedekind complete.

The proof for 0.999... = 1 depedends on an ability to define cuts at its heart and Dedekind completeness.

No, the classic proof is a basic epsilon-delta proof that
$$\sum_{n = 1}^{+\infty} 9 \cdot 10^{-n} = 1$$
and this is valid, for example, in the rational numbers, which are also not Dedekind complete.



I'll tell you what is true, and where you went wrong:

Suppose I take the mapping f : N -> R given by:

$$f(n) := \sum_{k = 1}^{n + 1} 9 \cdot 10^{-k}$$

Of course, this is simply the sequence of partial sums of 0.999~.

Now, suppose I also take the injection i : R -> *R.

It is true that the sequence (i o f) : N -> *R does not have a limit.

However, this has absolutely nothing to do with *0.9~.

Your mistake is that you've only transferred part of the whole setup. In the hyperreals, the value of *0.9~ is given by the limit of the function

*f : *N -> *R

which is the *-transfer of the previous function f.

If you don't want to use the transfer principle, it is straightforward to show directly that the limit of *f is 1. (The proof is identical to the real case -- you can get a closed form for the sums of hyperfinitely many terms, and then show the closed form goes to 1)



To sum it up, externally speaking, *0.9~ has more digits than 0.9~.

The archimdean property is equivelant to the statement s = {1/n} converges to 0. Any system which contains infinitesimals is non-archimedean.

But the hyperreals are hyperarchmedian! The sequence s = {1/n} does converge to zero when you take the hypernaturals as your index set.

 

[Sir_Deenicus]

(1) One's "cleverness" (or lack thereof) has nothing to do with what is actually deducible.

But it is what allows one to deduce something outside the system. See Godel, Escher Bach an Eternal Golden braid for my meaning.

(2) Even if I grant you the liberty of calling an infinite sum "non-stop addition", there are plenty of formulations of the decimal numbers that never make reference to such a thing.

Again, just an example.

Yes, you can. Dedekind completeness is a true internal statement about the reals. Therefore it is true for the hyperreals.

No it Isnt! I showed how there can exist gaps in cuts. Check any text on non standard analysis any you will see that there is no completeness axiom, and nor are they dedekind complete or archimedean. I do not think I could misudnerstand that. I think you use the transfer principle too liberally.

No, the classic proof is a basic epsilon-delta proof that
$$\sum_{n = 1}^{+\infty} 9 \cdot 10^{-n} = 1$$
and this is valid, for example, in the rational numbers, which are also not Dedekind complete.

I may be wrong but I think you implicitly just created a real number as the sum you have can be taken as a cut made in the rationals.

But the hyperreals are hyperarchmedian! The sequence s = {1/n} does converge to zero when you take the hypernaturals as your index set.

Oh wow! I didnt know that. I just learned a cool new word, thanks. EDIT: A search on google, I could not find any cases where hyperreals were linked to hyperarchimidean properties. Only multi values algebras (which gets the hyperacrhimedean property when all its elements are archimidean) whose meaning I am unsure of due to my lack of experience with them. But what of my sequence argument? Represent 0.9... as a sequence, in *R you cannot show convergance.

 

[Sir_Deenicus]

So, what you're saying is that the statement

"1/n does not tend to zero"

is provably true in the hyperreals, which, implies that it is true in real analysis and moreover provably in real analysis, hmm, that seems more than a little problematic. perhaps you need to check what convergence means in the hyperreals

*R is an extension of the real number system to a large set, results that are statable about elements of *R and that are true in R are true in *R, that is the transfer principle. The statement 1/n converges to zero is statable in *R, restricted to R is true so it is true in *R.

No, something true in *R does not need to exist in R, for example st() has no R counterpart. This by the way is what I was trying to say earlier Hurkyl, when you said I was wrong.

But what of my attack of the 2 proofs? They use properties (are not statements) only applicable to R an not *R . For example, an ability to create an upperbound set with a maximal element or the assumption of 0 being the only hyperreal etc. ?

 

[matt grime]

None of what you've written had indicated that 0.9... is not 1 in the hyperreals, and you've not rejected any of the arguments put forward that show that they are the same, you've only said that they are not necessarily the same for merely the same reason, and there is no arguning about that, and i agree entirely. convergence in *R is not the same as saying things about epsilon; you've not gotten round the argument that not only do 0.9... and 1 differ by a infinitesimal (which we both agree on) but that the difference is also *real* and hence since the only real infinitesimal is 0 then they are the same thing. Now, since the whole point of infinitesimals in the hyperreals is to provide another way of doing the analysis that agrees on R then how can you remotely believe the phrase 0.999... is 1 to be false in *R and true in R?

 

[Hurkyl]

I'm not going to try and quote to what I'm responding -- I don't think it would be helpful.


I confess that I made up the word "hyperarchmedian" -- I'm going along with the convention of prefixing the *-transfer of various concepts with the word "hyper". (e.g. hyperreal, hypernatural, hyperfinite)


One statement of Dedekind completeness is given by:

$$\forall S, T \in P(\mathbb{R}): (S \neq \emptyset \wedge T \neq \emptyset \wedge S \cup T = \mathbb{R} \wedge S \cap T = \emptyset \wedge \forall s \in S: \forall t \in T: s < t) \implies \exists r \in \mathbb{R}: \forall s \in S: \forall t \in T: s \leq r \leq t$$

This is a true statement in standard analysis. Therefore, by the transfer principle, the *-transfer of this statement must also be true. In other words:

$$\forall S, T \in P({}^\star \mathbb{R}): (S \neq \emptyset \wedge T \neq \emptyset \wedge S \cup T = {}^\star \mathbb{R} \wedge S \cap T = \emptyset \wedge \forall s \in S: \forall t \in T: s < t) \implies \exists r \in {}^\star \mathbb{R}: \forall s \in S: \forall t \in T: s \leq r \leq t$$

So, internally speaking, the hyperreals are Dedekind complete.

The catch is that P here is the internal power-set function. P(R) contains only the internal subsets of R. Of course, for the standard model, every subset of R is internal.

But in the nonstandard model, P(*R) is "missing" some subsets. For example $\mathbb{N} \notin P({}^\star \mathbb{R})$, because N isn't an internal subset of *R.

(Incidentally, this whole issue about the internal power-set operation was one of my biggest stumbling blocks in understanding this stuff!)

(I actually have a gripe about the usual treatment of second-order logic that relates to this context -- but that's for another thread!)


Incidentally, this fact is useful to prove certain sets are not internal. For example, I can prove that the set of finite hyperreals is not an internal set, because I can use it to construct a Dedekind cut around one of the gaps!



When I'm working in Q, I can still use all the normal definitions of things like continuity, infinite sums, et cetera. Of course, in Q, things like:

$$\sum_{n = 0}^{+\infty} \frac{1}{n!}$$

do not converge, despite the fact the sequence of partial sums is Cauchy.

But, it can still be shown (using the exact same proof as one would use in the reals) that the sum

$$\sum_{n = 1}^{+\infty} 9 \cdot 10^{-n}$$

converges to 1.



A sequence, incidentally is nothing more than a function written with different notation. The domain of this function is called the "index" set.

In general, if we do not say otherwise, we assume that all infinite sequences are indexed by N. (and finite sequences indexed by something of the form {a, a+1, a+2, ..., b-1, b}, usually with a=0 or a=1)

But in the nonstandard case, it becomes more appropriate to index sequences by *N, and not by N. In particular, this means that when we speak about a hyperdecimal, its digits should be indexed by *N and not by N.

You are correct in observing that the N-indexed sequence {1 / (n+1)} does not converge in the hyperreals. However, the *N-indexed sequence {1 / (n+1)} does converge to zero, which can be shown either through the transfer principle, or directly with an epsilon-N argument.



I would also like to point out that st() does not necessarily exist in a nonstandard model of R! It is another example of something external -- a function we can define set-theoretically, but have absolutely no reason to think it is a part of the model we're studying.


(Actually, models of R typically have no functions whatsoever -- what I mean to say is "model of real analysis", but that's too wordy. )

 

[Colonelteddy]

I apologize if this has come up before, i didn't see it, or it may have been in a different form, but here's the extent of my understanding of it:
let's say 0.99999999999999... = x
so then 10x = 9.999999999...
therefore 10x - x = 9.000...
so 9x=9
-> x = 1
Makes sense to me.

 

[Sir_Deenicus]

None of what you've written had indicated that 0.9... is not 1 in the hyperreals, and you've not rejected any of the arguments put forward that show that they are the same, you've only said that they are not necessarily the same for merely the same reason, and there is no arguning about that, and i agree entirely. convergence in *R is not the same as saying things about epsilon; you've not gotten round the argument that not only do 0.9... and 1 differ by a infinitesimal (which we both agree on) but that the difference is also *real* and hence since the only real infinitesimal is 0 then they are the same thing. Now, since the whole point of infinitesimals in the hyperreals is to provide another way of doing the analysis that agrees on R then how can you remotely believe the phrase 0.999... is 1 to be false in *R and true in R?

First I'd like to apologize for being so dense. I always assumed that when you said that the difference is real, you meant real, exists instead of Real, the numbering system. The statement would have to be true since both numbers are in R since and R is subset... Yes. Thank you for your patience. But another contention remains.

That of *0.9... equaling 1. But then, since the hyperreals is not complete, a cut in *R would not allow one one to create a decimal concept would it? 9/10^n + 9/10^(n + 1) + 9 /10^(n + 2) + ... where all the numbers are not necessarily reals but finite hyperreals, the series would not converge since there would be infinitesimals in the way [of precision]. Thus the question precludes itself. Why, if r finite, one could never be certain that each of the number chosen would be st(r) since in *R a approimately equal to b is possible without a = b as in R and there exists many r's for which r = st(r) [true].

 

[matt grime]

Just because some object (*R) is such that it doesn't have property P (all cuts being well defined) then that just means that the property P fails at *some* point, not at all points.

And in anycase you must apply the relevant notions in each space, that is to say if dedekind cuts aren't the correct notion to think about then you must ignore them and think about the correct notions of completion and convergence, whatever they may be in *R.

 

[Sir_Deenicus]

Hurkyl, I am still not sure I agree with your use of the transfer principle. If for no other reason that in Keisler's book, he states that a completeness statement is not a real statement and thus not transferable over to *R. This creates a clash of confusion for me and I am more willing to agree with the text. Of course as you may have noticed, I do not like to blindly accept things so will give my reasons as well. :D

First I will note that I pointed out in several posts previous that "An internal statement in *R is not necessarily true in *all* subsets of *R, especially depending on your language strength. Also, *N is an internal subset of *R but not N." To further that, to make it such that true in all R becomes possible then in *R, true becomes only applicable to internal subsets. Which is a major hitch.

Also I argue that what you gave can be considered to be more a statement of the properties of R than a real statement of R thus not valid by the transfer principle which itself varies dependant on the form of the treatment [axiomatic, formal languages etc.], so I remain hesitant to accept your *N argument.

 

[Sir_Deenicus]

Just because some object (*R) is such that it doesn't have property P (all cuts being well defined) then that just means that the property P fails at *some* point, not at all points.

And in anycase you must apply the relevant notions in each space, that is to say if dedekind cuts aren't the correct notion to think about then you must ignore them and think about the correct notions of completion and convergence, whatever they may be in *R.

Okay. I will keep this in mind as I learn more NA. Thank you very much for your help and time, I truly do appreciate it.

 

[Sir_Deenicus]

Hurkyl, on further thought, I will accept that the sequence converges in *N not because the transfer principle allows that *R is complete but because a N statement True in N <--> True in *N. Otherwise as Matt clarified for me NA would not be so useful. But then again, the converse is not true. That is an *N statement True in *N does not imply true in N. (Which I still feel is correct though you still say I am wrong, maybe I continue to be unclear?) Truth of that is based on language and the allowance of quantifications across all R is not always good. Furthermore, true in *N does not mean true in all *R. For example one can have true in *N but not in N both subsets of *R if the language allows extension through P(R).

So I remain skeptical of convergance across all *R. But I must learn more before I come to further conclusions.

 

[Hurkyl]

First, on the transfer principle

------------------------------------------------------
The whole point of the transfer principle is that a statement P is true in the standard model if and only if *P is true in the nonstandard model.

You can actually prove the "if and only if" if you merely assume the truth of Q implies the truth of *Q as follows:

Suppose *Q is true.
Assume Q is false, then:
~Q is true.
*(~Q) is true.
~(*Q) is true.
*Q is false.
This a contradiction, therefore Q is true.
And similarly for the case where *Q is false.


TMK, for the purposes of NSA, real analysis is cast into the following language:

The language of bounded first-order logic, with:
(1) Constant symbols for every real number and every set that could ever possibly arise in a real-analysis argument.
(2) The internal power set operator P.
(3) The internal membership relation $\epsilon$.

"Bounded" means that quantifiers are bound. You can say things like $\forall x \in T: x = x$, but not things like $\forall x: x = x$.

"Every real number and set that could possibly arise" means the elements of $\mathbb{R}$, or of $\mathcal{P}(\mathbb{R})$, or of $\mathcal{P}(\mathbb{R} \cup \mathcal{P}(\mathbb{R}))$, et cetera.


The "standard model" is simply the following map:
(1) Each constant symbol is mapped to the real number or set it denotes.
(2) P is mapped to the set-theoretic power set operator $\mathcal{P}$
(3) $\epsilon$ is mapped to the set-theoretic membership operator $\in$


Then, the complete first-order theory of real analysis simply consists of every true statement of this language. (If you prefer, it simply takes every true statement as an axiom)

So, Q is true in the standard model if and only if Q is in this theory.


Every consistent infinite first-order theory has a nonstandard model. (Which means that every statement of the theory must be true in the nonstandard model)


Because our theory of interest is complete, we know that for any statement Q, either Q or ~Q is in the language. It is then easy to show that Q is true in the nonstandard model if and only if Q is in this theory.


And thus, a statement in this language is true in the standard model if and only if it is true in the nonstandard model.

Happily, we can find a nonstandard model where:
(1) $\mathbb{R} \subset {}^\star \mathbb{R}$
(2) $\epsilon$ is mapped to $\in$
(3) $\forall T: P(T) \subseteq \mathcal{P}(T)$

The important thing to note, here, is that the internal power-set doesn't contain every subset -- it only contains the internal subsets.

Statements, such as Dedekind completeness, can be made within this first-order language. (Using the internal power-set function, and the internal membership operation)

Since Dedekind completeness is true in the standard model, it must also be true in the nonstandard model. (According to the aforementioned interpretation)

The resolution to the apparent paradox is that Dedekind completeness says the following:

Every internal Dedekind cut of R is generated by an element of R. (Why internal? Because we formulate it using the internal power-set operator)

When we pass to the nonstandard model, the above statement must be true. Since the converse is also true, we have the following external theorem:

A Dedekind cut of *R is internal if and only if it is generated by an element of *R.

Hopefully this exposition resolves your issues about completeness and the transfer principle!
------------------------------------------------------


We can, of course, study the ordered field *R without doing it the nonstandard way.

*R is ordered, so it has a topology. Its open sets are simply those of the form (a, b), where a and b are hyperreal numbers. For example, this means the definition of the limit of a function is entirely unchanged! (It's still given by the epsilon-delta formulation)


Among *R's topological properties is:

(1) It's totally disconnected. (Yuck)

(2) Every convergent countable sequence is eventually constant.

(3) Every countable subset of *R is bounded.


*Z is an integer part of *R. This means that it is a discrete ring with the property that for every element r of *R, there exists an n in *Z such that $n \leq r < n+1$.

*N is, of course, the nonnegative subset of *Z.



We have the following theorem:
Let F be any ordered field.
Let I be an integer part of F.
Define an $I^+$-indexed sequence by: $a_n := 1/n$.
Then, $\lim_{n \rightarrow +\infty} a_n = 0$.

Recall that, by definition, this means:

$$\forall \epsilon \in F^+ : \exists N \in I^+: \forall n \in I^+: (n > N) \implies (|a_n| < \epsilon)$$

Which is easy to prove, by the definition of "integer part". (There exists an M such that $M \leq 1/\epsilon < M+1$)



There are, of course, many different nonstandard models of R, and they have different external properties. For example, you can ask things like "What is the minimum length for an unbounded sequence?" or "What is the cardinality of *R?"

There are other interesting properties too. If I recall correctly, there exist Cauchy complete nonstandard models of R. However, a Cauchy sequence cannot be of countable length -- it must have an uncountable index set.

There's a refinement of Dedekind completeness that means the same thing, I think: the separation between the two parts of the Dedekind cut is not allowed to be bounded below by a positive number.

So, for example, the Dedekind cut:
(infinitessimals, and everything negative) : (positive finites and infinites)
is not fair game, because the gap between them is "too big".
---------------------------------------------------------------

These ideas extend into the algebraic domain in two ways.

The "baby version" of all of this is the theory of real closed fields -- it only concerns itself with the semi-arithmetic operators (+, *, <, etc), and doesn't try to model any set theory.

A good example of a real closed field is the set of real algebraic numbers.

A version of Dedekind completeness is valid for any real closed fields. (E.G. if you are only using the semi-algebraic operations, you'll never know you're missing pi)

Similarly, if you are willing to pretend that only semi-algebraic sets exist (things definable in terms of polynomial equations and inequalities), any real closed field is actually fairly nice, topologically. (This, I believe, is closely related to "internal sets" in nonstandard analysis, and is one of the issues that drives my current interests)



The other connection is that algebraists have defined something they call a hyperreal field, whose definition you can see at Wikipedia which are related to nonstandard models of the reals. (some of the hyperreal fields are some of the nonstandard models, and I suspect they all are, from the right perspective)

Ack, I'm rambling. I'll stop now!

 

[QuantumTheory]

That isn't how limits work. A limit is an exact number - what tends to infinity isn't the limit (actually, I'm not sure if it has a name, but it's the "x" below...).
Ie:
lim 1/(1-x) = 1
x-> 0 (sorry - hate latex)
The equals sign is for real - that limit really is exactly equal to 1. The "x" is approaching something (infinity), but the limit itself is exactly equal to 1.
An interesting link, using polygons inscribed in a circle: http://www.coolmath.com/limit1.htm
"n" approaches infinity, but the limit is exactly equal to the circle.
And that's the entire point of limits - to find exact values for things that are otherwise difficult to pin down (such as the slope of a curve at a single point).
I actually enjoyed calculus-I (yeah, I'm a freak), because when I got my arms around this concept (didn't just plug and chug, but actually understood the point of limits), calculus-I became a piece of cake. It also helped that I took it in conjunction with Newtonian physics and they are so related that both help you learn the other.

You're not a freak, I enjoyed math too, but calculus seems like ALOT more work. I always look ahead of my book several lessons and couldn't wait until the next lesson

You said you enjoy claculus, well I enjoy it too, but I can't do it or understand it. Why?

Mainly because I have a huge comprehension problemw hen reading, audio is not a problem, and when I can ask questionst o clear things up isn't a problem, but reading is, very often. Also homework is very difficult, except for math. It took me much longer and I regulary triple checked my work in math.

However, due to the stres of homework, in 10th grade I was in algebra I in a regular school. So I went to a charter school, never fully got though algebra (Although I did get through pre algebra, but I forgot everything! Yes, everything) never got through geometry, or trig, or pre calc. So as you can see, if I do these, will calculus and the limits become easier to understand? I really have a passion for learning it.

I now got my diploma from a charter school. I want to understand physics but can't without knowing claculus and trig.

 

[QuantumTheory]

Perhaps we are talking about 2 different things? 0.9... does not tend to 1 on the hyperreal line because the hyperreal line contains infinitesimals!

The picture argument. On the hyperreal line, There exist numbers between every number that disallow the step required for 0.9... to converge at 1. We cannot make the inductive jump. 0.9...9, 0.9...9 + e, 0.9...9 + 2e and so on.

The formal attacks. Consider the first proof given http://www.cs.uwaterloo.ca/~alopez-o/math-faq/node41.html#SECTION00531000000000000000".
The proof fails since it is possible to choose e > 0 but less than or equal to all *R. Then e cannot equal 10 ^ (-1/d). Indeed the proof ends up saying something meaningless like infinitely approximately less than zero.

Again in http://en.wikipedia.org/wiki/Proof_that_0.999..._equals_1" it states: Since there exist non zero infinitesimals we can only say that the series gets aproximately close to 1 since we are approximately close to 0 (there exists 0 < e ~= 0). But in fact the whole endeavoure is meaningless since the cuts that created the entire decimals should not have been possible! Since we have e, the gaps created should have been empty. There is no concept of inifinite precision, just a line riddled with gaps as we go through all c in our original set. In fact, the proof is not possible since a meaningful upper bound cannot be found in *R due to the existence of infinitesimals.

Suppose we rewrite the decimal number as a sequence, then because there is no completeness [axiom] in *R, it cannot be shown to converge or diverge etc. I end like so:

The archimdean property is equivelant to the statement s = {1/n} converges to 0. Any system which contains infinitesimals is non-archimedean. The hyperreals contain infinitesimal and thus s = {1/n} does not neccessarily converge to 0. This fact, coupled with the lack of gurantee of convergance on monotonic sequences and a lack of least upper bound property all point that the series 9 * sum(n, infinity,1/10^n) over *R, whose properties are accounted for in their totality cannot be shown to converge at 1.

Damn dude the R seems complicated I heard it means all real numbers whatever that means.

I'm 17 and quite frankly I am done with high school i should know calculus but I took the easy route and never needed to fully complete algebra, geometry, trig, pre calc, or calculus. Unfortunatey, it hurt me. Now i have to learn it somehow like with a tutor with no money (dont even have money for a book, car is sold, don't want to walk to work everyday, don't have any friends as I am a nerd)

Still, sounds confusing. I'm curious, Sir_Deenicus, do you have a PhD in math? Quite impressive. I think people here keep thinking I'm like 40 or so and have a PhD in mathematics because i may be able to understand limits one day but this...I doubt it.

 

[Sir_Deenicus]

Damn dude the R seems complicated I heard it means all real numbers whatever that means.
I'm 17 and quite frankly I am done with high school i should know calculus but I took the easy route and never needed to fully complete algebra, geometry, trig, pre calc, or calculus. Unfortunatey, it hurt me. Now i have to learn it somehow like with a tutor with no money (dont even have money for a book, car is sold, don't want to walk to work everyday, don't have any friends as I am a nerd)
Still, sounds confusing. I'm curious, Sir_Deenicus, do you have a PhD in math? Quite impressive. I think people here keep thinking I'm like 40 or so and have a PhD in mathematics because i may be able to understand limits one day but this...I doubt it.

Lol , I am glad you think me so skilled but I am far from it. I am only 3 years older than your 17 by a few months and am a first year maths student in University. In fact, I too did not learn any mathematics in high school (no geometry or calculus, only algebra). I reaped the public school system to its full benefit and did not learn much math through it. But at your age, 17, I decided to buckle down and teach my self a bunch of stuff since I am interested in fundamental questions. Between 17 and 19 i changed degrees a few times - from Computer science (1/2 a year) to computer engineering (1.5 years) before now settling with math. But yeah almost all my knowledge is self gained and taught. I am soon to begin study of algebraic topology...

I struggle to keep with Hurkyl and Matt's clear explanations are quite easy to follow. I have certainly learned much in these series of exchanges. Anyway, you should consider getting a job so you can buy books. You can never have too many books. Make some friends as well :), do not underestimate the value of human companionship.

 

[Sir_Deenicus]

Hurkyl your post is long and I was only able to skim but will read it more thorougly later. (it also got a bit disconnected as you went on :P). I have no problems with the internal sets agreeing, I see now that it is obvious. But I extended my issue not with them but to P(*R), which you later say is not fair game (exclude finites etc.) but state there exists an isomorphic statement to dedekind completeness in *R. I have to read more on this but from what i know DC is not fair game for extension. Sorry to seem so stubborn but I am quite certain of that correctness.

Another thing, if *R is totally disconnected then you can't define a concept of a topology for it! This is something my intuition strongly rebels agaisnt. Surely you must have made a mistatement that needs clarification there?

 

[matt grime]

Of course we can define a topology for it; if we couldn't then the statement 'is totally disconnected' would be meaningless since it is a statement about topological spaces (the cantor set is another such, as is any discrete topological space). If your intuition rebels against this and you're prepared to make such a strong statement of assertiveness then I will automatically dismiss any other suggestions you make.

 

[Hurkyl]

If it helps you internalize things, you should observe that Q is also a totally disconnected topological space.

("totally disconnected" means that every open set is disconnected)

 

[moose]

If there is no law stating whether .999999...=1 or not, than it's neither... I guess it depends on if you think it does or not... There is no way to prove that .9999...=1 without anything speculitive... It depends on how you view it, seriously...

 

[matt grime]

No, it depends upon how we have chosen to define the terms. There is no law stating anything if we choose not to accept definitions as being true, I am not even compelled to accept that 4/2=2 if I choose to accept different meanings from those commonly accepted. (No one is saying anything is absolutely true, only true within in the meanings we give things, by the way, and if you ignore those meanings that is your problem and puts you outside the realm of mathematics.)

 

[Sir_Deenicus]

("totally disconnected" means that every open set is disconnected)

Thanks, that one helped alot. My knowledge of topology is pretty cursory and gotten from a CS angle. I am soon to undertake a proper math treatment on it and its algebraic counterpart however. On NSA however, I am more confident on, having read more (first Keisler, then numbers, then an interest in proof systems) and so will continue to contend.

By the way, I've managed to find something that states explicitly what I've been arguing. More than just arguments from what may possibly be my misintepretation...

http://www.motionmountain.net/CAppendix.pdf" Apparantly the surreals have this property as well!

 

[Hurkyl]

I hate to take an authoritative stance, but that reference simply has it wrong.

I can only imagine three things the notation $0.99\bar{9}$ can possibly mean with regards to the surreals:

(1) It is a "hyperdecimal" -- that is, a *Z-indexed sequence of digits -- for which every place to the left of the decimal point is zero, and every place to the right of the decimal point is 9. This "hyperdecimal" is exactly equal to 1.

(2) It is trying to denote an "ordinary" decimal expansion of a hyperreal number -- but any nonterminating "ordinary" decimal does not converge. As I mentioned earlier, a countable sequence converges if and only if it is eventually constant. A strictly increasing sequence, such as 0.9, 0.99, 0.999, ... (that is, $\{ 1 - 10^{-n} \}_{n \in \mathbb{Z}^+}$), does not converge in *R, so it cannot denote any hyperreal number at all.

(3) The reals are embedded in the hyperreals. $0.99\bar{9}$ denotes a real number, which in turn is a hyperreal number. But this approach also clearly leads to $0.99\bar{9} = 1$.


The surreal case is even stickier -- the surreals are far "too big" for any sequence to converge in the ordinary sense... no matter how long the sequence was... unless it is eventually constant. So, I don't really have any idea what the author could possibly mean by trying to refer to a surreal number via a decimal expansion.

(of course, the real numbers can be embedded in the surreals -- so you denoting a surreal by a decimal expansion can make sense if you say that the decimal expansion denotes a real number, and then that real number is in the surreals... but then clearly that would say $0.99\bar{9} = 1$ in the surreals)



I suspect the author is simply confused by the fact that the sequence $\{ 1 - 10^{-n} \}_{n \in \mathbb{Z}^+}$ does not converge to 1 in either the hyperreals or the surreals. This is easy to see in the hyperreals because, for any positive infinitessimal e, we have that (1-e) is greater than every term in that sequence. (But, that's not true for the sequence $\{ 1 - 10^{-n} \}_{n \in {}^\star \mathbb{Z}^+}$)

Similarly for the surreals. We can construct a number "between" that sequence and 1 as being the surreal number:

s = { 0.9, 0.99, 0.999, ... :: 1 }

This defines a surreal number, which is strictly greater than each number in { 0.9, 0.99, 0.999, ... }, and yet strictly less than each number in { 1 }.

And then, I could look at { 0.9, 0.99, 0.999, ... :: s }, which is strictly less than s, and so forth.



However, this is a true statement in the surreals:
1 = {0.9, 0.99, 0.999, ... :: }

But this is misleading, because this is also true:

1 = {0.4, 0.49, 0.499, 0.4999, ... :: }


In both cases, 1 is the "simplest" number greater than everything in the left set, which gives us the equality.

 

[Maxwell]

Quite impressive. I think people here keep thinking I'm like 40 or so and have a PhD in mathematics because i may be able to understand limits one day but this...I doubt it.

What?

 

[Sir_Deenicus]

I hate to take an authoritative stance, but that reference simply has it wrong.

(2) It is trying to denote an "ordinary" decimal expansion of a hyperreal number -- but any nonterminating "ordinary" decimal does not converge. As I mentioned earlier, a countable sequence converges if and only if it is eventually constant. A strictly increasing sequence, such as 0.9, 0.99, 0.999, ... (that is, $\{ 1 - 10^{-n} \}_{n \in \mathbb{Z}^+}$), does not converge in *R, so it cannot denote any hyperreal number at all.

But it is your word vs his... Nonetheless I have no knowledge of surreal numbers beyond that they have to do with something called Conway Games. However on hyperreals, your second point, point 2, is exactly what I have been arguing!

Suppose we rewrite the decimal number as a sequence, then because there is no completeness [axiom] in *R, it cannot be shown to converge or diverge etc.
...

But then, since the hyperreals is not complete, a cut in *R would not allow one one to create a decimal concept would it? 9/10^n + 9/10^(n + 1) + 9 /10^(n + 2) + ... where all the numbers are not necessarily [simply] reals but [also] finite hyperreals, the series would not converge since there would be infinitesimals in the way [of precision]. Thus the question precludes itself.

Why am I still so certain of incompleteness of *R? I have 4 texts which claim so, (one by Keisler, , one a Springer Verlag Graduate text on Numbers, one on NSA with Automated Formal Systems) and http://www.cs.uiowa.edu/~stroyan/InfsmlCalculus/FoundInfsmlCalc.pdf" . Pages 11 and 14.