1 0
_______________________________ _________________________________
1 0 1 0
____________ ______________ ______________ ______________
1 0 1 0 1 0 1 0
_____ ____ _____ ______ ______ _____ ______ ______
This fractal graph continues with the bottom row always containing 2n cells, where n is the row number.
One may follow any downward path such that the path does not cross spaces between cells.
If n is finite, we see that the legal paths constitute a finite binary number (where the binary extension point is not pictured) of all numbers that have n cells. That is, we have all non-repeating rationals of cell length n. Further, our system exactly orders the paths (or numbers) from greatest to least as we scan right to left.
As 2n represents the power set of n, each graph represents every set, and number, of length n. For example, the null set corresponds to the rightmost path 000.
We may design an algorithm, such as a Turing machine, that exactly rolls out this graph to a denumerable infinity of rows.
Every path -- if we accept its existence -- represents a real. All paths are implied by the algorithm. So you know that if you make, say three right forks and one left fork and then the remainder of forks right that this number will be greater than the path of four right forks and one left and the remainder right.
Now it is true that one can't actually specify three adjacent reals from this end of the algorithm. But, the fractal implies three adjacent paths at "actual infinity." So we suggest that the reals appear to be well-ordered without actually being able to count them.
Every path (real number) logically follows from the graph and, by convention, the ordering method is simply that as you travel down a path every fork is either less (right) or more (left), or that is 0 or 1, in base 2.
Notice that the logical consequence of following the fractal is that there is one and only one path ε; adjacent to the all-0 path. This path ε would, Platonically, represent a least element of R\{0} (the reals from 0 to 1, minus 0). But, constructively, it can't be reached because there is always a "future" left fork possible.
If one accepts the Platonic existence of ε, then perhaps the reals are well-ordered. But otherwise, the fractal fails to establish a least element of the reals (unless the axiom of choice is invoked).
If the reals exist as a Cantorian actual infinity, rather than a Brouwerian potential infinity, then I would argue that ε exists. And if ε exists, then so does ε', which would be adjacent to ε, and so on.
On the other hand, the number of rows is a denumerable infinity, which in the Cantorian perspective is also an actual infinity. So we are faced with the issue of the "last possible" left fork occurring at the alleged penultimate row of this infinity, N, representing the set of naturals. So N - 1 would then represent the set of rows minus the last row. But of course both N -1 and N have cardN. And so, if positing the existence of ε, we might be accused of a contradiction.
If the noncomputable reals are said to exist, in that case every path in the fractal exists "all the way." That is, by the axiom of choice the path composed of a denumerable number of lefts "followed by" one right must exist. Hence, that path represents the least element of R\{0} in the interval [0,1]. So, if one accepts this reasoning, then the axiom of choice implies well-ordering.
This view is further buttressed by the fact that it is possible to stand this graph on its head.
If an actual infinity is accepted, then we can turn around the fractal. We start with a a row containing an infinity of alternating 0s and 1s, where the first horizontal 0 is placed on the x axis to the left of the origin (to be consistent with our conventions).
So consider
...010|
The first 0 to the left of origin intersects an infinity of upward paths, including the path composed of all 0s. The first 1 to the left of origin also intersects an infinity of upward paths, including the path composed of all 0s but one. Hence, from this perspective, that path expresses the number ε that is immediately adjacent to 0.
(If we were to require that the alternating 0s and 1s extend infinitely on either side of the origin, then we would not achieve a well-ordering.)
Of course, if we do invert the fractal, we would have to concede that the set of 0s and 1s on the x axis has the cardinality of the reals. And yet we see that, from the origin, the set is countable. So this militates against the idea of inversion. Cantor's anti-diagonal proof here discloses a paradox in our presentation.
Also, note that if we look at the fractal right side up, we can see that it is possible to count the cells in row 1, followed by those in row 2, followed by... indefinitely. So, from such a perspective, we see a means of enumerating every binary element in a set with cardinality 2n. That is, we have
Σ 2i, where i runs from 0 to denumerable infinity.
We have not counted the reals. But we have counted the finite subsets of the reals.
Even had we been unambiguously successful in well-ordering the reals without resort to the axiom of choice, we find that the axiom cannot be avoided when examining this fractal.
The computable paths are those that can be notionally traveled via a finite algorithm determining at row n whether to take the left or right fork. But, as we very well know, most paths cannot be so traversed. Hence, in order to traverse any of the Turing-noncomputable paths we require some rationale for so doing. That rationale is that when confronted with a fork, some choice of left or right is always notionally possible.
We might say that a super mind could make such choices, regardless of determinism. We might say that some random process -- such as a fictitious Geiger counter -- can be invoked to make choices, though it is not absolutely certain that the random procedure wouldn't hit upon a Turing-computable path.
Or we might argue that if algorithms can have a denumerable infinity of logic gates, it might be possible to hit every Turing-noncomputable, though this proposition fails on ground that one would first require a Turing-noncomputable path for the logic gates.
However, now that we can see a method to systematically imply an x <= y <= z ordering for every real, we have a strong reason for upholding the axiom of choice, insofar as that axiom is used to require that a decision, even if fictitious or "Platonic," can always be made at an arbitrary fork.
Let us consider the computable paths. If a path is computable, there is a finite Turing machine for it. This means there is a mechanical process for deciding on left or right as one travels down the rows. So we have three cases:
i. A single unique string of finite length followed by an indefinite sequence of zeros.
ii. A single unique string of finite length that repeats indefinitely.
iii. A single unique string of infinite length over a denumerable number of cells.
In each the first two cases, a halt signal occurs at the end of the unique string and says either repeat the string indefinitely or follow with repeating 0s indefinitely.
In the first case, disregarding the "repeat" command, the set of finite paths of cardinality 2n is one-to-one and onto with the set of natural numbers of cardinality 2n. Hence, all such paths are computable, because every natural less than or equal to 2n is computable.
In the second case, we have, in effect, a discrete recursive function that says fk(T) = kT, where k is a positive integer and T is the period. The computability here is obvious.
In the third case, it is evident that we have an aperiodic recursive function which progressively computes finite substrings.
Any of these computables must be amenable to an induction proof. That is, an induction proof is equivalent to a proof of computability.
So, if a fork is determined at a finite distance along a path, that fork can be computed. And, a noncomputable requires that there is no mechanism for determining some forks. Hence a noncomputable has one or more forks that are at a denumerably infinite distance from the top row. Further, a noncomputable is held to exist in spite of the fact that an induction proof is ruled out.
Now there is a question as to whether we want to admit the set of noncomputables (which I call W, for the set of wilds) onto the real number line.
Cantor's proof, to modern eyes, relies on the idea that it would be possible to define a wild string if the reals were well-ordered. So he is requiring the admission of all "possible" binary strings. And yet, as we have shown, only wild strings require an axiomatic choice of branch at the nth row when n is infinite. So is there a good reason to accept such strings? I suggest it may be a matter of personal preference.
Note that W is bijective with R. So the continuum hypothesis amounts to whether there is a subset W' of W that is neither bijective with N nor with R. As there is no means of computing w' ε W', there is no test as to whether W' is or is not bijective with N or R. So we can easily appreciate the results of Goedel and Cohen.
At any rate, we can see intuitively and directly Zermelo's solution to the well-ordering problem: infinitudes of arbitrary choices.
If we accept that the fractal represents the power set of the naturals, or 2N, we also see directly that 2N = kN, because the base 2 number system is equivalent to the base k system, where k is some positive integer.
So if we say that the power set of the naturals exists, should we not accept the existence of the 2N paths, including the 2Nth row of alternating 0s and 1s at the "bottom" of the fractal? In that case, the power set of the reals is obtained by continuing the paths past the "bottom" row, yielding a "secondary" infinitude that can be dubbed (2N)2N. That would identify the power set of the reals.
Of course, one may well object that the power set axiom does not require a spatio-temporal order. On the other hand, we see that the fractal up to 2N implies not only an exact ordering of the reals, but a precise definition of the power set of N. So what better way to view the power set of R than as a continuation of the fractal "past" lim n --> infinity 2n?
An axiom stating that the bottom row of the fractal exists with cardinality 2N is no more extraordinary than the power set axiom of standard ZFC theory. Our axiom can be generalized to assert that power sets can be made sequential thus: 2[2N...2N].
From Gregory H. Moore's translation of five letters on the axiom of choice published in France in 1905. The letters appear in Zermelo's Axiom of Choice: It's Origins, Development and Influence, (Springer-Verlag 1982, Dover reprint 2013).
Henri Lebesgue to Jacques Hadamard: "The question comes down to this, which is hardly new: Can one prove the existence of a mathematical object without defining it?"
Lebesgue continues:
"You allude to the following argument: 'To well-order a set, it suffices to choose one element from it, then another, and so on.' Certainly this argument presents enormous difficulties which are even greater, at least in appearance, than are Zermelo's."
The step-by-step process seems to have been Cantor's intuitive concept of well-ordering of any set, an assumption he dropped once challenged to prove well-ordering of the reals.
If we agree that the wilds exist, then we may select any arbitrary w thus: R\{w} = X. So we have cX = {w}. We may arbitrarily number this subset's element as w1. This permits us to select and arbitrarily number some w for a denumerable infinity of cases (call it Y). So, assuming 2N exists, we have a contradiction as we cannot exhaust all the w's by this means. Hence, one may wonder whether the set Y exists, even granting the denumerable version of the axiom of choice.
You may say that that's what's wrong with a spatio-temporal interpretation. However, we have that the axiom of choice permits us to withdraw an arbitrary w from R, and so we can certainly withdraw a second, and a third, ad infinitum.
This apparent contradiction arises even though we have not, in this instance, well-ordered the purported set Y. Arbitrarily numbering the w's does not disclose the value of some w. The values, though said to exist, cannot be defined.
If we simply refuse to accept any infinity beyond what can be computed by a Turing machine, either the power set of N does not exist or cardP(N) = cardN. Of course, as Moore writes, such a denial would invalidate numerous proofs in topology and other disciplines.
We may see cause for such a denial by looking at the bottom of our fractal, with its nondenumerable en toto but denumerable consecutively set of alternating 01's.
As a thought experiment, consider that we may construct an "inverse computable" by beginning at some 0 or 1 a finite number of cells from origin, and then defining an upward path for some finite number of forks followed by a denumerable infinity of 0's or 1's, or an aperiodic path defined as a computable function whereby an ordinary induction proof says the path will zigzag over a denumerable infinity.
In most cases, we cannot bridge the bottom-up infinity to link an inverse computable with a corresponding top-down computable.
One may follow any downward path such that the path does not cross spaces between cells.
If n is finite, we see that the legal paths constitute a finite binary number (where the binary extension point is not pictured) of all numbers that have n cells. That is, we have all non-repeating rationals of cell length n. Further, our system exactly orders the paths (or numbers) from greatest to least as we scan right to left.
As 2n represents the power set of n, each graph represents every set, and number, of length n. For example, the null set corresponds to the rightmost path 000.
We may design an algorithm, such as a Turing machine, that exactly rolls out this graph to a denumerable infinity of rows.
Every path -- if we accept its existence -- represents a real. All paths are implied by the algorithm. So you know that if you make, say three right forks and one left fork and then the remainder of forks right that this number will be greater than the path of four right forks and one left and the remainder right.
Now it is true that one can't actually specify three adjacent reals from this end of the algorithm. But, the fractal implies three adjacent paths at "actual infinity." So we suggest that the reals appear to be well-ordered without actually being able to count them.
Every path (real number) logically follows from the graph and, by convention, the ordering method is simply that as you travel down a path every fork is either less (right) or more (left), or that is 0 or 1, in base 2.
Notice that the logical consequence of following the fractal is that there is one and only one path ε; adjacent to the all-0 path. This path ε would, Platonically, represent a least element of R\{0} (the reals from 0 to 1, minus 0). But, constructively, it can't be reached because there is always a "future" left fork possible.
If one accepts the Platonic existence of ε, then perhaps the reals are well-ordered. But otherwise, the fractal fails to establish a least element of the reals (unless the axiom of choice is invoked).
If the reals exist as a Cantorian actual infinity, rather than a Brouwerian potential infinity, then I would argue that ε exists. And if ε exists, then so does ε', which would be adjacent to ε, and so on.
On the other hand, the number of rows is a denumerable infinity, which in the Cantorian perspective is also an actual infinity. So we are faced with the issue of the "last possible" left fork occurring at the alleged penultimate row of this infinity, N, representing the set of naturals. So N - 1 would then represent the set of rows minus the last row. But of course both N -1 and N have cardN. And so, if positing the existence of ε, we might be accused of a contradiction.
If the noncomputable reals are said to exist, in that case every path in the fractal exists "all the way." That is, by the axiom of choice the path composed of a denumerable number of lefts "followed by" one right must exist. Hence, that path represents the least element of R\{0} in the interval [0,1]. So, if one accepts this reasoning, then the axiom of choice implies well-ordering.
This view is further buttressed by the fact that it is possible to stand this graph on its head.
If an actual infinity is accepted, then we can turn around the fractal. We start with a a row containing an infinity of alternating 0s and 1s, where the first horizontal 0 is placed on the x axis to the left of the origin (to be consistent with our conventions).
So consider
...010|
The first 0 to the left of origin intersects an infinity of upward paths, including the path composed of all 0s. The first 1 to the left of origin also intersects an infinity of upward paths, including the path composed of all 0s but one. Hence, from this perspective, that path expresses the number ε that is immediately adjacent to 0.
(If we were to require that the alternating 0s and 1s extend infinitely on either side of the origin, then we would not achieve a well-ordering.)
Of course, if we do invert the fractal, we would have to concede that the set of 0s and 1s on the x axis has the cardinality of the reals. And yet we see that, from the origin, the set is countable. So this militates against the idea of inversion. Cantor's anti-diagonal proof here discloses a paradox in our presentation.
Also, note that if we look at the fractal right side up, we can see that it is possible to count the cells in row 1, followed by those in row 2, followed by... indefinitely. So, from such a perspective, we see a means of enumerating every binary element in a set with cardinality 2n. That is, we have
Σ 2i, where i runs from 0 to denumerable infinity.
We have not counted the reals. But we have counted the finite subsets of the reals.
Even had we been unambiguously successful in well-ordering the reals without resort to the axiom of choice, we find that the axiom cannot be avoided when examining this fractal.
The computable paths are those that can be notionally traveled via a finite algorithm determining at row n whether to take the left or right fork. But, as we very well know, most paths cannot be so traversed. Hence, in order to traverse any of the Turing-noncomputable paths we require some rationale for so doing. That rationale is that when confronted with a fork, some choice of left or right is always notionally possible.
We might say that a super mind could make such choices, regardless of determinism. We might say that some random process -- such as a fictitious Geiger counter -- can be invoked to make choices, though it is not absolutely certain that the random procedure wouldn't hit upon a Turing-computable path.
Or we might argue that if algorithms can have a denumerable infinity of logic gates, it might be possible to hit every Turing-noncomputable, though this proposition fails on ground that one would first require a Turing-noncomputable path for the logic gates.
However, now that we can see a method to systematically imply an x <= y <= z ordering for every real, we have a strong reason for upholding the axiom of choice, insofar as that axiom is used to require that a decision, even if fictitious or "Platonic," can always be made at an arbitrary fork.
Let us consider the computable paths. If a path is computable, there is a finite Turing machine for it. This means there is a mechanical process for deciding on left or right as one travels down the rows. So we have three cases:
i. A single unique string of finite length followed by an indefinite sequence of zeros.
ii. A single unique string of finite length that repeats indefinitely.
iii. A single unique string of infinite length over a denumerable number of cells.
In each the first two cases, a halt signal occurs at the end of the unique string and says either repeat the string indefinitely or follow with repeating 0s indefinitely.
In the first case, disregarding the "repeat" command, the set of finite paths of cardinality 2n is one-to-one and onto with the set of natural numbers of cardinality 2n. Hence, all such paths are computable, because every natural less than or equal to 2n is computable.
In the second case, we have, in effect, a discrete recursive function that says fk(T) = kT, where k is a positive integer and T is the period. The computability here is obvious.
In the third case, it is evident that we have an aperiodic recursive function which progressively computes finite substrings.
Any of these computables must be amenable to an induction proof. That is, an induction proof is equivalent to a proof of computability.
So, if a fork is determined at a finite distance along a path, that fork can be computed. And, a noncomputable requires that there is no mechanism for determining some forks. Hence a noncomputable has one or more forks that are at a denumerably infinite distance from the top row. Further, a noncomputable is held to exist in spite of the fact that an induction proof is ruled out.
Now there is a question as to whether we want to admit the set of noncomputables (which I call W, for the set of wilds) onto the real number line.
Cantor's proof, to modern eyes, relies on the idea that it would be possible to define a wild string if the reals were well-ordered. So he is requiring the admission of all "possible" binary strings. And yet, as we have shown, only wild strings require an axiomatic choice of branch at the nth row when n is infinite. So is there a good reason to accept such strings? I suggest it may be a matter of personal preference.
Note that W is bijective with R. So the continuum hypothesis amounts to whether there is a subset W' of W that is neither bijective with N nor with R. As there is no means of computing w' ε W', there is no test as to whether W' is or is not bijective with N or R. So we can easily appreciate the results of Goedel and Cohen.
At any rate, we can see intuitively and directly Zermelo's solution to the well-ordering problem: infinitudes of arbitrary choices.
If we accept that the fractal represents the power set of the naturals, or 2N, we also see directly that 2N = kN, because the base 2 number system is equivalent to the base k system, where k is some positive integer.
So if we say that the power set of the naturals exists, should we not accept the existence of the 2N paths, including the 2Nth row of alternating 0s and 1s at the "bottom" of the fractal? In that case, the power set of the reals is obtained by continuing the paths past the "bottom" row, yielding a "secondary" infinitude that can be dubbed (2N)2N. That would identify the power set of the reals.
Of course, one may well object that the power set axiom does not require a spatio-temporal order. On the other hand, we see that the fractal up to 2N implies not only an exact ordering of the reals, but a precise definition of the power set of N. So what better way to view the power set of R than as a continuation of the fractal "past" lim n --> infinity 2n?
An axiom stating that the bottom row of the fractal exists with cardinality 2N is no more extraordinary than the power set axiom of standard ZFC theory. Our axiom can be generalized to assert that power sets can be made sequential thus: 2[2N...2N].
From Gregory H. Moore's translation of five letters on the axiom of choice published in France in 1905. The letters appear in Zermelo's Axiom of Choice: It's Origins, Development and Influence, (Springer-Verlag 1982, Dover reprint 2013).
Henri Lebesgue to Jacques Hadamard: "The question comes down to this, which is hardly new: Can one prove the existence of a mathematical object without defining it?"
Lebesgue continues:
"You allude to the following argument: 'To well-order a set, it suffices to choose one element from it, then another, and so on.' Certainly this argument presents enormous difficulties which are even greater, at least in appearance, than are Zermelo's."
The step-by-step process seems to have been Cantor's intuitive concept of well-ordering of any set, an assumption he dropped once challenged to prove well-ordering of the reals.
If we agree that the wilds exist, then we may select any arbitrary w thus: R\{w} = X. So we have cX = {w}. We may arbitrarily number this subset's element as w1. This permits us to select and arbitrarily number some w for a denumerable infinity of cases (call it Y). So, assuming 2N exists, we have a contradiction as we cannot exhaust all the w's by this means. Hence, one may wonder whether the set Y exists, even granting the denumerable version of the axiom of choice.
You may say that that's what's wrong with a spatio-temporal interpretation. However, we have that the axiom of choice permits us to withdraw an arbitrary w from R, and so we can certainly withdraw a second, and a third, ad infinitum.
This apparent contradiction arises even though we have not, in this instance, well-ordered the purported set Y. Arbitrarily numbering the w's does not disclose the value of some w. The values, though said to exist, cannot be defined.
If we simply refuse to accept any infinity beyond what can be computed by a Turing machine, either the power set of N does not exist or cardP(N) = cardN. Of course, as Moore writes, such a denial would invalidate numerous proofs in topology and other disciplines.
We may see cause for such a denial by looking at the bottom of our fractal, with its nondenumerable en toto but denumerable consecutively set of alternating 01's.
As a thought experiment, consider that we may construct an "inverse computable" by beginning at some 0 or 1 a finite number of cells from origin, and then defining an upward path for some finite number of forks followed by a denumerable infinity of 0's or 1's, or an aperiodic path defined as a computable function whereby an ordinary induction proof says the path will zigzag over a denumerable infinity.
In most cases, we cannot bridge the bottom-up infinity to link an inverse computable with a corresponding top-down computable.
