Re[2]: ​Mathematics with the potential infinite

Matthias matthias.eberl at
Fri Feb 3 09:51:11 EST 2023

Just a short remark to Zeno's paradoxes.

Since the potential infinite is actually a form of finiteness, from this 
perspective the number line or timeline does not have an (actual) 
infinite number of points, but it is infinitely divisible. So we can 
increase the number of points and intervals on the line indefinitely, 
and we can make the intervals as small as we want, but there are only 
finitely many of them at each step.  The number line has infinitely many 
points only in the sense that for any finite number, say n, we can 
increase the number of points and decrease the size of the intervals so 
that there are more than n (end)points.

Zeno's paradoxes existed for the ancient Greeks. But after Cantor, most 
mathematicians accepted actual infinities to a large extend, including 
the idea that the number line is identical to an uncountable set of 
points. This led to further paradoxes, most notably the Banach-Tarski 

Regards, Matthias

------ Originalnachricht ------
Von "Vaughan Pratt" <pratt at>
An fom at
Datum 01.02.2023 02:54:56
Betreff Re: ​Mathematics with the potential infinite

>So far it seems to me that every objection to my proposal (repeated 
>below) concerning gapless geodesics depends on the 
>assumption/axiom/whatever that a nonempty linear order with no greatest 
>(or least) element must be an actual infinity.
>To me, this has the flavor of Zeno's paradox, that movement from 1 to 
>0, say in units of meters, must be impossible because it must pass 
>through the infinitely many points 1/2, 1/4, 1/8, ...
>Now it is certainly a fair point that obtaining zero as the limit as n 
>goes to infinity of 2^-n is the limit of an infinite sequence.  What is 
>bothering me is the idea that because the sequence is infinite, it is 
>therefore not something we can experience.
>Now if we were to spend one second at each of those points, we'd be 
>less than one Planck length away from zero in about two minutes.
>Yet we still wouldn't be right at zero, not even a billion years later. 
>  (Math is not physics.)
>The existence of a model in which we can reach 0 in one second by 
>traveling at a constant speed of 1 meter per second shows that there is 
>nothing inconsistent about dropping the axiom that a nonempty linear 
>order with no least element is impossible to experience in practice.  
>And we reach uncountably many other points along the way, with the 
>transcendental ones a set of Lebesgue measure 1, all in unit time.
>Cantor is not to blame for the paradoxes arising from viewing the unit 
>interval as consisting of infinitely many points.  After all, Zeno was 
>from the fifth century BCE.  As far as paradoxes go, the fact that 
>Cantor identified more points than Zeno is somewhat of a red herring.
>We can find Zeno's paradox elsewhere, namely in the transcendence of 
>pi, which is how we prove pi is not a constructible number, i.e. not a 
>quantity that can be "computed" with a straightedge and compass.  
>(Which might seem overkill, surely it should be easier just to prove 
>that if pi is algebraic it must be of degree at least 3, yes?)
>Today we can define the Euclidean plane very slickly as the 
>two-dimensional vector space over the field of reals, made an inner 
>product space realized with the usual dot product, a bilinear 
>In Book I of his *Elements*, Euclid accomplishes this bilinear aspect 
>of his geometry by working with quantities of two types, lengths and 
>areas.  For example the penultimate Proposition 47 concerning 
>right-angled triangles asserts the Pythagorean Theorem, that the square 
>on the hypotenuse equals the sum of the squares on the other two sides, 
>while the last, 48, asserts the converse, that such an equality implies 
>a right angle.
>The paradox here is that even though pi is transcendental, an area of 
>size pi can be drawn with a compass set to unit radius, and moreover 
>the perimeter of that circle is 2 pi.
>In this case we can approach pi as a length with a simple infinite 
>construction due essentially to Archimedes, though starting with a 
>square rather than a hexagon, and only computing the outer polygons, 
>which takes about the same effort as the mean of the outer and inner 
>Start with X = (1,0), O = (0,0), and Y = (0,1).  XOY is a right 
>isosceles triangle, four of which can be distributed evenly about O to 
>make a four-slice pie with a square boundary.
>At step n = 1, move Y to the midpoint of XY, namely (x,y) = (1/2,1/2),  
>and shorten OX (along itself without moving O, i.e. move X left) to 
>make XOY isosceles again.  X is now at (sqrt(x^2 + y^2), 0) = 
>(0.7071..., 0). However the angle XOY has now been halved, to 45 
>degrees.  2^{n+2} = 8 copies of this prototype triangle can be 
>distributed evenly about O to make an eight-slice pie with an octagonal 
>At step n = 2, halving the 45-degree isosceles prototype in the same 
>way produces a 22.5-degree isosceles triangle, 2^{n+2} = 16 of which 
>make a regular 16-agon.  And so on, with the base of the isosceles 
>prototype becoming more vertical at every step.  The length of the base 
>therefore tends to 2^-n, and the perimeter of the 2^{n+2}-agon 
>therefore tends to 4.
>In the limit we obtain a circle of perimeter 4.
>Now given a flexible line of length 4, if we bend it into a circle and 
>connect its endpoints so as to have constant curvature throughout, we 
>have constructed a circle of radius 2/pi.  We were able to do this with 
>a Zeno-like approach involving infinitely many steps with just a 
>compass and straightedge, but with this flexible line as an alternative 
>to a compass for creating a circle, we were able to construct 2/pi (or 
>4/pi if we prefer to measure the diameter) in finite time.
>In Cartesian coordinates, the midpoint of XY is a point (x,y).  
>Observing that y halves at each step, we can take y_n = 2^-n.  We can 
>then define the x coordinate of the successive midpoints as
>         x_1 = 1/2
>  x_{n+1} = (x_n + sqrt(x_n^2 + (2^-n)^2))/2
>For n from 1 to 28, the first 16 digits of 2/x_n after the decimal 
>point are as follows.
>(Newton computed pi to this precision in 1665, though with a method 
>much more different from Archimedes' than the above.)
>At n = 2,  x_2 = (1 + sqrt(2))/2 = 1.20710678118...  .  Now  2/x_n is 
>constructible for all finite n, but the limit is the transcendental 
>quantity pi, with the precision increasing by almost exactly two bits 
>at each halving of our isosceles triangle.
>The form taken by Zeno's paradox here is that with infinitely many 
>constructible numbers, presumably taking infinite time, we can 
>construct a radius r = 2/pi such that sweeping out a circular arc of 
>length 1 constructs a right angle, and of length 4 a circle, of radius 
>2/pi.  Or we can do it in finite time with a flexible line of length 4, 
>as our two-dimensional counterpart of smooth motion from 1 to 0.
>On Sat, Jan 28, 2023 at 12:13 AM Vaughan Pratt <pratt at> 
>>Define a *geodesic* to be a nondegenerate linearly ordered group (G, 
>>*, 0, <=).  (Although G is not assumed abelian, the linear order makes 
>>it abelian.)  Examples include the integers, the dyadic rationals, 
>>every field between the rationals and the reals, and many 
>>non-Archimedean extensions thereof.
>>Call a geodesic G *gapless* when (i) it is dense, and (ii) for every 
>>nonempty suborder (U, <=) of (G, <=) having no least element, and 
>>every nonempty suborder (L, <=) of (G, <=) with L < U and having no 
>>greatest element, such that there is at most one element of G between 
>>L and U; then there exists an element of G between L and U.
>>I claim that every gapless geodesic is isomorphic to R with the above 
>>(Proof outline: Take any element x of G with 0 < x and pair 0 and x 
>>with 0 and 1 in R.  Pair the integers in R with the subgroup of G 
>>generated by x, cyclic and therefore abelian.  Repeatedly divide the 
>>intervals in (n, n+1) in G into two equal parts and pair the results 
>>with the dyadic rationals in (0,1), a dense set.  Pair each dyadic 
>>irrational q in R with the unique x given by the gaplessness condition 
>>for any L and U in G whose counterpart in R converges to q from each 
>>side.  Lastly, G must be Archimedean or there would be an empty gap 
>>between the finite and infinite elements of G.)
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