Advanced Infinity

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    We have examined infinity, the mathematical concept of endlessness, as it applies to the set of the natural numbers and to various of its subsets. Now I want to extend that examination to the set of the real numbers and I want to do so in part to illustrate the importance of the connection between semantics and mathematics.

    When I began accumulating the notes for this essay I hoped to discover the description of a set whose cardinality equals Aleph-Three, the third of the grand infinities that Georg Cantor inferred to lie beyond the ordinary infinity of the natural numbers. But then I discovered that Cantor had made an error in the very foundation of his transfinite mathematics, one whose correction makes the whole realm of transfinite numbers evaporate. So now I want to look at what Cantor did, prove and verify the statement that he did go wrong, and then analyze how he went wrong.

    How many elements comprise the set of the decimal extensions; that is, how many unique decimal fractions can we put between zero and one? Because the digits of a decimal fraction can extend an infinite number of places to the right of the decimal point, we assume, as Cantor did, that the set contains an infinite number of elements. Following Cantor, we list all of those elements, in no particular order, in a column and claim that we have a complete list. Then, using Cantor's diagonalization procedure, we create an element that we did not put on that list.

    In that new number the first digit differs from the first digit in the first digit in the first number on our list; the second digit differs from the second digit in the second number on our list; the third digit differs from the third digit in the third number on our list; and so on. (The "and so on" represents a kind of induction or an invitation to an induction). We can generate an endless sequence of such numbers and we know that each of them differs in at least one place from all of the numbers on our list; therefore, none of these new numbers sat on our original list, though they clearly belong on it. That fact means that our list could not have been complete, as we assumed, even though we had an infinite number of entities on it. From that reductio ad absurdum Cantor inferred that the magnitude of the set of the decimal extensions coincides with an infinity larger than the infinity of the natural numbers.

    We can prove and verify easily that Cantor's inference is false to mathematics. Take a decimal extension, such as 0.31532061, and reflect it through the decimal point to obtain a unique natural number, 16023513 in the present example. We can apply that procedure to each and every element of the set of the decimal extensions and thereby associate each and every decimal extension uniquely with one and only one natural number. We can also reflect any natural number, such as 7182956, through the decimal point to obtain a unique decimal extension, 0.6592817000... in the present example. We can apply that procedure to each and every element of the set of the natural numbers and thereby associate each and every natural number with one and only one decimal extension. Thus we have an ubijoto (Unique, BIJective, One-To-One) match between the sets of the natural numbers and the decimal extensions, which necessarily means that the two sets contain the same number of elements.

    We can confirm that result through a simple description of each of the sets. The set of the decimal extensions comprises all permutations of ten symbols on a line of spaces extending toward infinity to the right of the decimal point. The set of the natural numbers comprises all permutations of ten symbols on a line of spaces extending toward infinity to the left of the decimal point. This reference to our base-ten, place-value-notation number system emphasizes the fact that the two sets are mirror images, each of the other, with the decimal point serving as the mirror. That fact remains true to mathematics if we use a number system with another base, noting that the use of a base implies the use of place-value notation.

    So how did Cantor go wrong? The short answer says that he forgot the definition of infinity. If he had remembered that infinity denotes the mathematical concept of endlessness, he might have noticed and understood the absurdity of endlessnesses more endless than endlessness and dismissed his hierarchy of Alephs as false to mathematics.

    The long answer expands that analysis. In setting up his diagonalization proof, Cantor asks us to assume that we have the complete set of the decimal fractions, a set comprising an infinite number of elements, on our list. But the phrase "infinite number" denotes something that does not and cannot exist. Infinite does not denote a number of any kind, but rather denotes the impossibility of producing a number; specifically, a last number in an open-ended sequence. Cantor's diagonalization procedure proves and verifies just that proposition and nothing more: it gives us an analogue of the addition proof and verification of the proposition that the natural numbers comprise an infinite sequence.

    But decimal is not the only form in which we have fractions: we also have the rational fractions; that is, fractions of the form N/M, in which N and M represent integers. If we take any natural number (e.g. 137) and match it with its reciprocal (in this case 1/137), then we get an ubijoto match between the set of the natural numbers and the set of the unit fractions. But the unit fractions comprise a subset of the set of irreducible rational fractions: the full set contains other irreducible rational fractions, such as 4/137 or 11/123. Matching all of the natural numbers ubijotoly with the unit fractions leaves those other fractions unmatched, so we seem to have demonstrated that the set of the irreducible rational fractions contains more elements than does the set of the natural numbers. Does this mean that we must assert that the set of the irreducible rational fractions, which we expect to have fewer elements than does the set of the decimal fractions, has a magnitude greater than the infinity of the natural numbers?

    No, of course it doesn't. It means simply that we have found the rational fractions' analogue of the Hilbert's Hotel paradox. To see how that works in this case we need to take a new approach to creating the desired ubijoto match with the set of the natural numbers.

    From some point in a plane draw a straight line vertically without end and draw a second line at a right angle to the first line, extending to the right without end. Subdivide the area above the horizontal line and to the right of the vertical line into a uniform array of square cells, organized in straight rows and columns. Below the horizontal line draw the set of the natural numbers in such a way that each column has one number associated with it and on the left side of the vertical line draw the set of the natural numbers in such a way that each row has one number associated with it. We can identify each cell in that array by the number C of its column and the number R of its row. In each cell we put the rational fraction C/(C+R): thus the first column (C=1) contains all of the unit rational fractions, beginning with 1/2; the second column contains the fractions 2/3, 2/4, 2/5, etc.; and so on. In that way we fill the matrix with all of the rational fractions. Some of those fractions duplicate others (for example, 3/6 duplicates 1/2), but we accept that fact because we know that the set of the irreducible rational fractions contains fewer elements than does the set of all rational fractions. Now we need to figure out a way in which we can create an ubijoto match between the cells of the matrix (and, therefore, of their contents) with the set of the natural numbers.

    The cells comprising Column-1 contain all of the unit fractions. Each of those cells heads a row that extends to the right toward (but not to) infinity. We might want to try converting the matrix into a linear array that we can match with the natural numbers by putting its rows end to end. But we know that we cannot do that. If we try to put the left end of Row-2 against the right end of Row-1, we run afoul of the fact that Row-1 has no right end. The cells in any given row comprise an infinite set, so, having a definite end on the left, they cannot have an end on the right. We cannot put the left end of a row against a right end that does not, indeed cannot, exist. We must, therefore, find some other way to reorganize the cells of our square matrix into a well-ordered array.

    Let's start with the single cell in the lower left corner of the matrix (C=1, R=1) and define it as Chevron-1. Three cells touch the upper side, right side, and upper right corner of that cell; define those three cells to comprise Chevron-2. For convenience let's define the upper side, the right side, and the upper right corner of a chevron to comprise that chevron's outer boundary. Five cells touch the outer boundary of Chevron-2; define them to comprise Chevron-3. Seven cells touch the outer boundary of Chevron-3; define them to comprise Chevron-4. Continuing that process, we have Chevron-N comprising 2N-1 cells. In this way we divide the set of cells comprising the matrix into subsets that each have two definite ends (at C=1 and at R=1).

    Next we assign an index number to each cell. To the cell that comprises Chevron-1 we assign the index number 1. To the cells that comprise Chevron-2 we assign the index numbers 2, 3, and 4, proceeding from left to right and from up to down. To the cells that comprise Chevron-3 we assign the index numbers 5, 6, 7, 8, and 9. Continuing in that way, we create an ubijoto match between the cells (and their contents) and the set of the natural numbers. In effect we get the same result that we would have obtained from straightening out the chevrons and placing them end to end, putting the left end of Chevron-2 against the right end of Chevron-1, putting the left end of Chevron-3 against the right end of Chevron-2, and so on, numbering the cells as we go. In that equivalent linear array the unit rational fraction 1/R occupies the cell in the position bearing the index number (R-1)2+1. So now we have an ubijoto match between the set of the rational fractions and the set of the natural numbers.

    Does that match remain when we remove from the matrix the duplicate rational fractions? Note that each column in the matrix represents an infinite set and remove from each column the duplicates. From Column-2 we remove every second entry, from Column-3 we remove every third entry, from Column-4 we remove every second and fourth entry (because two is one of the factors of four), and so on. In that way we remove from each column a well-ordered subset of the column's entries. Removing a well-ordered subset from an infinite set leaves an infinite set, so we can simply shift the un-removed entries in each column down to re-occupy the empty cells and we still have a full infinite matrix. Thus we have an ubijoto match between the set of the irreducible rational fractions and the set of the natural numbers.

    Consider the real numbers, the set whose elements comprise every possible combination of a natural number with a decimal fraction. Surely that set must contain vastly more elements than does the set of the natural numbers. We can prove and falsify that statement by constructing and analyzing a square matrix as we did above. Number all of the rows and columns of the matrix with the natural numbers as we did above, except this time begin each set of natural numbers with zero instead of one. Reflect all of the row numbers through the decimal point to convert them into their associated decimal fractions. The row number and the column number of each cell now combine to form a real number, so all of the cells in the matrix carry all of the elements of the set of the real numbers. We know that we can create an ubijoto match between the cells of the matrix and the set of the natural numbers by way of our chevronization process, so now we know that we have an ubijoto match between the set of the real numbers and the set of the natural numbers.

    Can we create an ubijoto match between the set of the rational numbers, the combination of the natural numbers with the rational fractions, and the set of the natural numbers? Label the rows and columns of the square matrix with the natural numbers beginning with one. Each cell contains the rational number R/C and the whole matrix contains all of the rational numbers because it displays all possible ratios among the natural numbers. Chevronization then gives us our ubijoto match.

    Before we go any further we must adjust our matrix. Draw a unit interval horizontally and at its left end draw a unit interval vertically with its lower end coinciding with the horizontal line's left end. Those lines define a unit square. Label the points on the horizontal line with the set of the decimal fractions and label the points on the vertical line with the set of the decimal fractions: because the unit interval extends from zero to one, we know that it must contain enough points to accommodate the entire infinite set of the decimal fractions. Reflect the decimal fractions on the horizontal line through the decimal point to convert them into their associated natural numbers. With the full set of the natural numbers on the horizontal line and the full set of the decimal fractions on the vertical line, we now know that the points that comprise the unit square correspond to the full set of the real numbers. Because we have already created an ubijoto match between the set of the real numbers and the set of the natural numbers, we now know that the unit square contains the same number of points as does the unit interval. Our chevronization process thus works as well on arrays of points as it does on arrays of square cells.

    Consider the unit cube. We can represent it as an infinite set of unit squares placed face to face. Each unit square contains the same number of points as does the unit interval, so we can replace those unit squares with unit intervals, thereby converting the unit cube into an infinite set of unit intervals set side by side. But that set of unit intervals constitutes a unit square, which contains the same number of points as does a single unit interval. Thus, if we have a set whose elements we can match ubijotoly with the points in a unit cube, then we know that we can create an ubijoto match between that set and the set of the natural numbers.

    Consider the unit tesseract, the unit right prism in four dimensions. We can represent the unit tesseract as an infinite set of unit cubes placed body to body (however that works in four dimensions). Each unit cube contains the same number of points as does the unit square, so we can replace those unit cubes with unit squares, thereby converting the unit tesseract into an infinite set of unit squares placed face to face. But that set of unit squares constitutes a unit cube, which we can collapse into a unit square, which we can collapse into a unit interval without losing any points. So now we have proven and verified the proposition that the unit tesseract contains the same number of points as does the unit interval.

    We can continue that process of looking at unit right prisms of ever higher dimensions. We can represent the N-dimensional unit right prism as an infinite set of (N-1)-dimensional unit right prisms. We assume that we have already shown that the (N-1)-dimensional unit right prism contains the same number of points as does the (N-2)-dimensional unit right prism, so we can replace the infinite set of (N-1)-dimensional unit right prisms with an infinite set of (N-2)-dimensional unit right prisms without changing the number of points in the object. But an infinite set of (N-2)-dimensional unit right prisms, in the present circumstance, constitutes a single (N-1)-dimensional unit right prism. Thus we prove and verify the proposition that an N-dimensional unit right prism contains the same number of points as does the (N-1)-dimensional unit right prism. We repeat that procedure N-2 times to prove and verify the proposition that the N-dimensional unit right prism contains the same number of points as does the unit interval.

    We don't actually carry out that procedure for all possible values of N, so we have an inductive, rather than a deductive, proof of the proposition. Recall that in mathematical induction we discern a pattern in some subset of mathematical objects and assert that the pattern applies to the entire set. We base mathematical induction upon Occam's Razor, the principle of parsimony (entities are not to be multiplied without necessity). In so doing we actually put the greater burden of proof onto anyone who wants to falsify the induced proposition. If someone wants to falsify my statement pertaining to the number of points in an N-dimensional unit right prism, they will have to prove and falsify my theorem that the unit square contains as many points as does the unit interval OR they must prove and verify that the abstract infinite-dimensional space in which we place our analytic geometry has at least one dimension in which the unit right prism does not consist of an infinite set of the unit right prisms of the next lower dimension. Of that latter possibility we inevitably ask Why that particular dimension and not some other (or others)? If we don't get a suitably compelling answer to that question, then we can maintain our proposition, which we derived from a minimum of premises.

    For clarity's sake, consider another example of an induction that we have made. Someone offers up the set of the natural numbers and claims that it contains a finite number of elements. Because the natural numbers, by definition, comprise an ordered set, then if the set has a finite size, it must contain a largest/last element. We feign having identified that element and then we add one to it to generate a new largest/last natural number. We believe that we can repeat that process without end, so by induction (by assuming that the pattern continues) we infer that the natural numbers comprise an endless (infinite) set. If our someone wants to maintain that the natural numbers comprise a finite set, then they must prove and verify a proposition to the effect that at some number the process of addition no longer works. Until such a proposition can come to us a properly tested theorem we can continue to accept addition as a process that we can apply to any number and thereby preserve our determination that the natural numbers comprise an infinite set.

    In neither of the above cases did we obtain our result through axiomatic-deductive logic, so we must accept our propositions not as necessary truths of mathematics, but rather truths contingent upon the non-existence of any factor that would change the logic in at least one place in our chain of reasoning. Although we feel confident that such a factor does not exist, we do not have an axiomatic-deductive proof of that non-existence, so we must classify the proofs offered above as inductive, those in whose truth to mathematics we have confidence but not certainty. So now we know, with confidence but not certainty, that if we create an ubijoto match between the elements of some set and the points of an N-dimensional unit right prism, then we can create an ubijoto match between the elements of that set and the elements of the set of the natural numbers.

    Before I offer two examples of such sets I want to look again at the unit interval in the light of another of Cantor's proofs that some sets extend beyond infinity.

    In December 1873 Cantor worked out a proof that some sets have non-denumerable (i.e. uncountable) extent, representing infinities beyond the denumerable (i.e. potentially countable (!?)) infinity of the natural numbers. He presented that proof with the subdivision of an interval, such as the unit interval. According to Cantor we have a non-denumerable set if we have a set that possesses the following four properties:

1. It is linearly ordered. By that statement we mean that among the elements of the set we find a relationship of comparison (greater than/less than or before/after) such that for any two elements (X and Y) of the set only one of three possible relations can exist; X is less than or comes before Y (X<Y), X equals or coincides with Y (X=Y), or X is greater than or comes after Y (X>Y). Further, the relationship among the elements of the set is transitive; that is, if we have X>Y and Y>Z, then we must necessarily have X>Z.

2. It must contain at least two elements. Of course it will contain very many more.

3. It is densely ordered. That statement means that between any two members of the set another member must exist. So between A and B we must find C, between A and C we must find D, between A and D we must find E, and so on endlessly. This criterion makes the set infinite.

4. It is complete. That means that the set contains all of the elements necessary to fulfill the above criteria. So if we divide the set into two non-empty subsets, P and Q, such that every element of P is less than every element of Q, then completeness requires that the set include a boundary element R such that every element less than R is an element of P and every element greater than R is an element of Q.

    For Cantor's proof we assume that the presumably non-denumerable set consists of the points that comprise the unit interval. Making explicit our implicit analogy between the unit interval and the region between zero and one on the number line, we assign to each point a locator number equal to the decimal fraction that corresponds to the point's position on the line segment between zero and one. With Cantor we re-conceive the points of the unit interval as a sequence of indexed elements x-i. Within that sequence we define two sub-sequences with elements p-i and q-i such that p-1<q-1. Properties 1 and 2 above allow us to make that statement. By property 4 we must have a point R not an element of p-i or q-i, but lying between them. Cantor wants us to prove and verify that, although R cannot be an element of p-i or q-i (by property 4), it must be an element of p-i or q-i.

    Identify p-2 as the first element in the sequence xi lying next to p-1 between p-1 and q-1. Identify q-2 as the first element in the sequence x-i lying next to q-1 between p-2 and q-1. Identify p-3 as the first element in the sequence x-i lying next to p-2 between p-2 and q-2. Continue that process of absorbing elements x-i into the subsets p-i and q-i, thereby extending those subsets progressively closer to R. Cantor tells us that, in order to be an element of the set, R must correspond to some element x-j, so when our expansion of the subsets P and Q reaches the index number j we must perforce absorb R into one subset or the other, contrary to our statement that the set fulfills the criterion of completeness. Cantor dissolved the contradiction by asserting that the elements x-i comprise a non-denumerable set, one in which the index-based counting used above is impossible because the set contains vastly more elements than does the set of the natural numbers. By asserting the existence of a set that we cannot put into an ubijoto match with the natural numbers, Cantor believed that he had discovered an infinite realm beyond the ordinary infinity of the natural numbers.

    In this proof Cantor erred primarily in using natural numbers for his indices. We can get a clearer picture of how he went wrong if we use the locator numbers defined above as indices on the points in the subsets P and Q. For simplicity let's assign to R the index 0.5000.... To the first point we put into P we assign the index 0.4900... and to the first point we put into Q we assign the index 0.5100.... As you will see, our place-value numbering system actually makes Cantor's procedure impossible, so instead of taking points adjacent to the first points in P and Q as the second points in those subsets, I will take points separated by small jumps in the locator numbers. So the next point that I draft into P has the index number 0.4990... and the next point that I draft into Q has the index number 0.5010.... The third points drafted into P and Q take as index numbers 0.49990... and 0.50010... respectively. We can continue that process endlessly without ever drafting R into either subset P or subset Q. We can guarantee that statement by noting that in our place-value system, the available places to the right of the decimal point comprise an infinite set.

    That latter fact tells us why Cantor's proof doesn't work. First, in order to find the A first element in the sequence x-i lying next to p-1 between p-1 and q-1" when p-1 has the index number 0.4900... we must go to the right end of the line of decimal places and draw a one. But the line of decimal places, as an infinite set, has no right end, so we can't even take the steps that Cantor specified in his proof. That's why I drafted points into my subsets in little jumps, incorporating into the subset by implication the infinite set of points lying between the jumped-from point and the jumped-to point. Even so, we discovered that we could not reach the point R and incorporate it into one or the other of the subsets in a finite number of jumps. And we found that we could infer all of that without invoking a non-denumerable infinity: the ordinary, denumerable infinity of the natural numbers gives us all the endlessness that we need.

    Now consider the set of all single-valued functions y=f(x), in which x and y represent real numbers. The symbol f() represents a rule or set of rules that determine the value of the number that y represents in accordance with the value of the number that x represents. Thus, for example, in the case of y=x2 for every value of x we associate a value of y that equals the square of that value of x. If we lay a grid comprising a uniform array of small squares onto an infinite plane and designate one direction parallel to one set of lines on the grid as the x-direction and the other direction as the y-direction, then we can blacken every point that has the coordinates (x, y=f(x)) and thereby trace in the plane a curve that corresponds to the function y=f(x). Every function corresponds to a curve in that plane and every curve in that plane represents some function, however complicated we must make it. Indeed, some of the things that mathematicians call curves don't look so curvy to the average person.

    We can describe the set of all possible single-valued functions in a very simple way. Pick a real-number value for x. For that value of x we have an infinite set of possible values, both positive and negative, for y. We associate those values with the points on unit interval. Pick a second value for x. Again we have an infinite set of possible values for y, which values we again associate with the points comprising a unit interval. That infinite set goes with each possible value of y for the first value we chose for x, so we draw a copy of that second unit interval away from each and every point in the first unit interval. Thus we find that all possible values of y associated with two values of x match the points in a unit square. For every value of x that we add to our list, we must add one dimension to the unit right prism to generate the correct number of points that corresponds to all the possible values that y can take for the values of x that we have selected. If we list all possible values for x, then we must represent all possible values for y - that is, all possible single-valued functions y=f(x) - as an infinite-dimensional unit right prism. But we have already proven and verified as true to mathematics the fact that the unit right prism of any dimensionality contains exactly as many points as does the unit interval and we discern no reason that would alter that fact if the dimensionality of the unit right prism increases endlessly; therefore, we infer that we can make an ubijoto match between the set of the single-valued functions and the set of the natural numbers.

    We can apply the same analysis to the set of the double-valued functions, the functions that associate two values of y with every value of x. In that case every new value of x will add two dimensions to the unit right prism representing the possible values of y associated with our list of values of x. We end up with the same result that we get for the single-valued functions. And we will get the same result for the triple-valued functions, the quadruple-valued functions, and so on.

    Further, the double-valued functions, instead of representing paired curves in the plane (two values for the y-coordinate for every value for the x-coordinate), can also represent a single curve in a three-dimensional space (one value for the y-coordinate and one value for the z-coordinate for every value for the x-coordinate). Higher-valued functions can thus represent single curves or multiple curves in higher dimensions. Our analysis of those curves yields the same result as does our analysis of the single-valued functions; no set of functions/curves will contain a number of elements that exceeds the infinity of the natural numbers.

    Finally let's consider the power set of the natural numbers, the set that comprises all of the subsets of the set of the natural numbers. We include the empty set and the full set of the natural numbers as elements of the power set, so that gives us two elements. Next we have the two power subsets that comprise all of the natural numbers taken one at a time (each number constituting a subset of the natural numbers) and all versions of the full set of the natural numbers with one number removed. We can make an ubijoto match between each of those power subsets and the points in the unit interval. For convenience we absorb the empty set and the full set of the natural numbers into each of those power subsets respectively.

    Our next two power subsets comprise those subsets of the natural numbers that we create by taking the natural numbers two at a time (e.g. {1,2}, {1,3}, {2,5}, etc.) and those subsets of the set of the natural numbers that we create by removing pairs of numbers from the full set of the natural numbers. We first naively assume that the elements of the first power subset correspond to the points in a unit square. We obtain that assumption by labeling the points on the lower edge and on the left edge of a unit square with the full set of the natural numbers and then taking the label on each point in the square ({X,Y}) to constitute a subset of the natural numbers that we want to include in our power subset. But that model counts too many points. We certainly want to eliminate from our count all points with labels of the form {X,X}, because counting the same number twice cannot give us a proper subset of the natural numbers; thus we remove from our unit square the points comprising the diagonal line extending from the lower left corner to the upper right corner, noting that those points comprise a set fully equal to the unit interval. We must also eliminate from our count duplicate subsets; for every subset of the form {X,Y} we must discard the one of the form {Y,X}. Thus our power subset of number pairs corresponds to half of a unit square minus a unit interval.

    That same analysis applies as well to the power subset comprising all examples of the full set of the natural numbers from which we have removed pairs of numbers, so it also corresponds to a half unit square minus a unit interval. If we combine that power subset with the one above and add in the previous two power subsets, we get an array of elements that corresponds to the points in a unit square.

    Our fifth power subset comprises all of the subsets of the natural numbers made by taking the numbers three at a time and our sixth power subset consists of all of the versions of the set of the natural numbers from which we have removed trios of numbers. Again we make a naive assumption: this time we assume that the elements in one of those power subsets correspond to the points in a unit cube. And again we must remove from that figure the points that correspond to impossible and duplicate subsets of the set of the natural numbers.

    Impossible subsets consist of trios of numbers in which one of the numbers appears twice or thrice; that is, subsets of the form {X,X,X}, {X,X,Y}, {X,Y,X}, and {Y,X,X}. The first of those subsets corresponds to the points that make up a straight line extending from the lower left front corner of the cube to the upper right rear corner of the cube. The second of those subsets corresponds to the points that constitute a plane that extends from the cube's front left edge to its rear right edge. The third of those subsets corresponds to the points that lie in a plane that extends from the cube's lower left edge to its upper right edge. And the fourth of those subsets corresponds to the points that comprise a plane that extends from the cube's lower front edge to its upper rear edge. Those three planes intersect each other on the line defined by the first of the impossible subsets. We must remove the points of that line from each of the planes and then remove from the unit cube the line and all three modified planes.

    Duplicate subsets enter our calculation because of the way in which we imagine creating the subsets that comprise the fifth power subset. We imagine letting the values that we assign to X, Y, and Z vary independently over the entire range of the natural numbers and remove from the resulting set of number trios the impossible subsets described above. In whatever we have left we also have for every subset {X,Y,Z} the subsets {X,Z,Y}, {Y,X,Z}, {Z,X,Y}, {Y,Z,X}, and {Z,Y,X}. We don't make our power set from ordered subsets, so we have six times as many subsets as we need. Thus we associated one-sixth of our modified unit cube with our fifth power subset.

    We apply the same analysis to our sixth power subset, whose elements we create by removing trios of numbers from the full set of the natural numbers, and we get the same result. Again we naively begin with a unit cube and, after removing the points representing impossible subsets and duplicate subsets, end up with one sixth of a unit cube from which we have removed one diagonal line and three planes each missing a diagonal line.

    If we combine those two power subsets and absorb the previous power subsets into them, then we obtain a number of elements that corresponds to one-third of a unit cube.

    We apply the same analysis to the seventh and eighth power subsets, whose elements we create, respectively, by taking the natural numbers four at a time and by removing numbers four at a time from the full set of the natural numbers. After we remove the impossible subsets and the duplicate subsets, combine the two power subsets, and then incorporate the elements of the previously defined power subsets we have a number of elements that corresponds approximately to the number of points in 1/12 of a unit tesseract.

    As we continue that analysis we find that all of the elements comprising the power subsets up to and including the 2N-th power subset correspond to approximately the number of points in the fraction 2/N! Of the N-dimensional unit right prism. We thus infer that the power set of the natural numbers contains vastly fewer elements than does the infinite-dimensional unit right prism, so that power set does not oblige us to assert the existence of any infinity beyond the infinity of the natural numbers.

    We certainly should have suspected such a result from our definition of infinity as denoting the mathematical concept of endlessness. Cantor's hierarchy of ever-larger infinities, his family of Alephs, would give us endlessnesses more endless than endlessness. The absurdity of that phrase does not give us a proof of the statement that entities larger than the infinity of the natural numbers do not exist in the realm of mathematics, but it does give us probable cause to offer the statement as a conjecture and to go looking for a proper proof. That task I will leave to others.

    Now, because of the infinity of the set of the real numbers, we know that the set of all numbers has the property of closure with respect to multiplication and division. We know that we have enough numbers to solve the equations AX=B and X/A=B for any numbers A and B.

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