
#1
Jan214, 10:18 PM

P: 76

I'm just thinking aloud on this one. So let's imagine a hypothetical function S_r that's the real analogue of S_n (the successor function over the naturals). Now let's say a, b in R are defined as follows: b=S_r(a); b is the real successor of a. Now of course if such a function existed then ba must equal zero. If it didn't, then we can always find a c in R that's between a and b which would then make c a new candidate for a's successor. But if ab is equal to zero, then a must necessarily equal b and so b can't be its successor. Such an argument seems to show that consecutive reals cannot exist, or at least can't be expressed using real numbers themselves.
On the other hand, if we are working with the reals under the assumption that consecutive reals do not exist, then there is no intuitive way to refer to the "next" real number. Certainly there's no way to define it using reals themselves as above. But if one says that there actually and certainly is no next real number then how are we moving along the field at all? How do you move between one and two if there's no conceptual way of referring to the real number that comes immediately after the number 1 in the set of reals. If ba must equal zero then you can add all the zeroes you want in the world to a but it won't change a thing unless you give infinity some mystical power to turn those zeroes into real values. I'm somewhat familiar with Cantor's diagonal argument now and I understand that that is precisely the whole point: that there exists no bijection between the reals and the naturals, and so no consecutive real by definition. But while I submit that such a thing can't be formalized in anyway using current language, doesn't the above reasoning at least justify the existence of consecutive real numbers in some as of yet tobedefined form? 



#3
Jan214, 11:37 PM

Mentor
P: 4,499

A couple of points:
You can choose an ordering on the real numbers which is different from the one we normally give it such that you can in fact have a well defined successor function for each real number. This ordering is probably terrible and doesn't correspond to what we think of when we think of real numbers being larger or smaller than each other. That said, the problem of defining successive reals is independent of the ability to "move along" the real number line. What does that mean mathematically? Most people think of it as "I give you a position for each point in time". Time typically takes any real number as a value, so to say "where am I at the next point in time" is a meaningless question to begin with. More generally you will be hard pressed to give a good definition of moving along the real number line such that an actual contradiction can exist. 



#4
Jan214, 11:40 PM

P: 76

Consecutive Reals 



#5
Jan214, 11:44 PM

P: 326

The phenomenon that you are investigating is more related to the fact that the normal order on the reals is what is called a dense linear order. It's the dense part that makes it seems that there is no "successorlike function". The rationals are also densely ordered, and so there is no function on the rationals which is successorlike and respects the normal order. So cardinality (and thus Cantor's diagonal argument) is really not the issue.
It turns out that, assuming the Axiom of Choice/WellOrdering Principle, there is a successorlike function for every set. This isn't so hard believe to if you believe that every set has a cardinality in the sense that it can be put into 11 correspondence with a cardinal number. 



#6
Jan314, 12:06 AM

P: 76





#7
Jan314, 10:17 AM

P: 76





#8
Jan314, 12:17 PM

P: 748

What is provable (given the Axiom of choice) is that there exists a "well ordering" of the reals. However, the proof of this is not constructive. That is to say that it is not possible to explicitly exhibit such an ordering. An "order" (or sometimes "partial order") is a binary relation, "<" between pairs of elements that is transitive, irreflexive and antisymmetric. Transitive means that if a < b and b < c it follows that a < c. Irreflexive means that is it never the case that a < a. Antisymmetric means that if a < b it follows that it is not the case that b < a. A "total order" is an order that also satisfies trichotomy. That is, given any pair of elements, either a < b, b < a or a = b. A "well ordering" is a total order that also has the property that any nonempty set has a smallest member. The natural numbers are wellordered by their usual ordering. The rational numbers are not wellordered by their usual ordering. For instance, the set of all rationals has no smallest member. But if you map each rational to a natural number then the rational numbers are wellordered by the order that is induced by that mapping. The real numbers cannot be mapped (one to one) to the integers. So that trick does not work for them. Given a well ordering... For every point x "between" [in the sense of the well ordering] 1 and 2 there would be a "next point" [the smallest real that is greater than x and less than or equal to 2]. That's guaranteed by what it means to have a well order. But that doesn't mean that you could get from 1 to 2 a step at a time, even with infinitely many steps. It could be that some numbers "between" 1 and 2 are not successors of any other number. 



#9
Jan314, 12:38 PM

P: 76





#10
Jan314, 01:15 PM

P: 748

The definitions I am familiar with involve epsilons and deltas and "less than". How do you do "less than" without an ordering? If you change out the ordering, you've changed out the definition of continuity. 



#11
Jan314, 02:30 PM

P: 76

Oh nvm, I should've looked up what standard ordering means first. But I still don't understand why invoking a wellordering on the reals removes standard ordering. This is probably over my head, but I'll do my best to follow.




#12
Jan614, 03:41 PM

P: 76

Having said that, are you essentially saying (in laymen's terms) that since I'm basically quantizing the field, I can't then base that quantized field on the concept of continuity, the definition of which requires that the field not be quantized? 



#13
Jan614, 03:50 PM

Mentor
P: 4,499





#14
Jan614, 09:00 PM

P: 391

To find a formal disproof of the existence of a successor in the standard ordering, see, e.g., the (constructive) argument that between two Rationals there is a Rational; basically between any two Reals there is a Rational, and there is an Irrational number. Or look at the decimal representation of any two Reals, that differ in the nth decimal place, you can also go farther back than the nth place to find an inbetween Real.




#15
Jan614, 09:21 PM

P: 76

Ok, please stop me if I'm just putting words in your mouths, but when I take this:




#16
Jan614, 09:37 PM

P: 76





#17
Jan714, 01:10 PM

P: 748





#18
Jan714, 02:46 PM

P: 76

I'm asking because I've been playing around with this idea of consecutive reals onandoff (just realized I posted about them here 2 years ago) and I figured if there was anything to this, then it should be possible to infer their existence with available terminology. Though at this point I get the feeling that nothing short of directly challenging Cantor's arguments on infinity would cut it if one's trying to will such a concept into existence. 


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