Infinitesimals in the Cantor set

[nomadreid]

I am not sure into which rubric to put this, but since there is some Model Theory here, I am putting it in this one.

First, I define the Cantor set informally:
A(0) = [0,1]
A(n+1) = the set of closed intervals obtained by taking out the open middle third of each interval contained in A(n), for natural numbers n.
The Cantor set = the points not removed at any step.

It is tempting to put
The Cantor set = the intersection of all A(n), but this would cause problems, as follows:

I concentrate on p = (the number corresponding to) the right end-point of the left-most interval of the Cantor set. If we stayed in the real numbers, we would have p=0, which is not what we want. So the intersection definition is not adequate. (Even if it were adequate, the following reasoning would still hold. Just covering my bases.)

Rather, p fulfills the following description:

for all natural numbers n, 0<p<1/3^n.

By the Archimedean property of the real numbers, p is not a real number, but by the Compactness Theorem, we assume the existence of a model which includes both real numbers and p, and all such infinitesimals. This new model can contain all the points of the Cantor set.

So far my reasoning. However, everywhere I look, the Cantor set is considered a subset of the real numbers. What is wrong here?

 

[Rasalhague]

I concentrate on p = (the number corresponding to) the right end-point of the left-most interval of the Cantor set.

Is there a "left-most interval"?

 

[nomadreid]

I expressed that rather sloppily. Let me make that more precise.
There are two series of points
p(n) = the right-hand (largest-value) endpoint of the interval A(n).
Let Can = the intersection of all A(n) = the Cantor set (I realized my statement that the Cantor set was not the intersection was wrong, but that does not change my basic problem)

Now, for each n, let
S(n) is the statement "There exists c(n) such that 0<c(n)<p(n)"
For any finite subset K of N (the set of natural numbers), Can serves as a model of
{S(n)|n in K}.

By the compactness theorem (of Model Theory), there is then a Model M which fulfills
{S(n)|n in N}.

My question thus is: how are M and Can related? That is, although there is no interval of non-zero length, can we say that there are "intervals" of infinitesimal (and hence, in the real numbers, having zero length, but in R U {set of infinitesimals}, not being points)?

 

[Preno]

Now, for each n, let
S(n) is the statement "There exists c(n) such that 0<c(n)<p(n)"
For any finite subset K of N (the set of natural numbers), Can serves as a model of
{S(n)|n in K}.

By the compactness theorem (of Model Theory), there is then a Model M which fulfills
{S(n)|n in N}.

My question thus is: how are M and Can related? That is, although there is no interval of non-zero length, can we say that there are "intervals" of infinitesimal (and hence, in the real numbers, having zero length, but in R U {set of infinitesimals}, not being points)?

Well, Can does satisfy this infinite set of sentences. Of course, the c(n) will be different for each n. What I presume you wanted to say was more like: add a constant c to our language. Each finite subset of {S(n) | n in N}, where S(n) is "0 < c < p(n)" (not "there is a c_n such that 0 < c_n < p(n)"), is satisfiable, hence (by compactness) the whole set is satisfiable (has a model, M).

Now you don't really know anything about M other than the fact that it satisfies this set of sentences. However, you do know that Can does not satisfy this set of sentences (under any interpretation). So M != Can, and that's about all you can say.

Note that this has very little to do with infinitesimals and the Cantor set and everything to do with the impotence of first-order logic (even first-order set theory) to express finitude and infinitude. You can construct a perfectly analogical argument concerning the natural numbers: let S(n) be the sentence "c > n", then by the same argument the set of sentences {S(n) | n in N} is satisfiable, so we have a model of first-order arithmetic containing an element larger than all the naturals. First-order theory is incapable of preventing this, because it holds that if a first-order theory has models of arbitrarily large finite size, it also has an infinite model.

 

[nomadreid]

Well, Can does satisfy this infinite set of sentences. Of course, the c(n) will be different for each n. What I presume you wanted to say was more like: add a constant c to our language. Each finite subset of {S(n) | n in N}, where S(n) is "0 < c < p(n)" (not "there is a c_n such that 0 < c_n < p(n)"), is satisfiable, hence (by compactness) the whole set is satisfiable (has a model, M).

Right. I have been away from mathematics too long; I've become really sloppy.

Now you don't really know anything about M other than the fact that it satisfies this set of sentences. However, you do know that Can does not satisfy this set of sentences (under any interpretation). So M != Can, and that's about all you can say.

Note that this has very little to do with infinitesimals and the Cantor set and everything to do with the impotence of first-order logic (even first-order set theory) to express finitude and infinitude. You can construct a perfectly analogical argument concerning the natural numbers: let S(n) be the sentence "c > n", then by the same argument the set of sentences {S(n) | n in N} is satisfiable, so we have a model of first-order arithmetic containing an element larger than all the naturals. First-order theory is incapable of preventing this, because it holds that if a first-order theory has models of arbitrarily large finite size, it also has an infinite model.

Very good point. That helps. Thanks very much.