Monthly Archives: November 2025

by | November 24, 2025 · 1:20 pm

Set Theory Jottings 19. GCH implies AC.

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Sierpiński’s Theorem: GCH implies AC

I had a look at the version of the proof in Cohen (§IV.12). Sierpiński was a clever fellow, and he came up with a few tricks that would be hard to motivate.

Here I will try to imagine how Sierpiński could have devised his proof. Cohen does offer one bit of intuition:

The GCH is a rather strong assertion about the existence of various maps since if we are ever given that A≤B≤P(A) then there must be a 1–1 map either from B onto A or from B onto P(A). Essentially this means that there are so many maps available that we can well-order every set.

Let A be the set we wish to well-order. Let’s write A≤B to mean there is an injection from A into B. GCH tells us that for any U,

A≤U≤P(A) implies U≡A or U≡𝒫(A)

If U is well-ordered, then U≡A and U≡𝒫(A) both imply that A can be well-ordered, the latter because A is naturally imbedded in 𝒫(A). But this is too simple an approach: A≤U already makes A well-ordered for a well-ordered U, so if we could show the antecedent we’d be done—we wouldn’t need GCH to finish the job.

Let’s not assume U is well-ordered, but instead suppose it contains a well-ordered set. Say we could show that

A≤W+A≤𝒫(A)

for a well-ordered set W, where ‘+’ stands for disjoint union. (That is, W×{0}∪A×{1}, or some similar trick to insure disjointness.) Then we’d have

W+A≡𝒫(A) or W+A≡A

Now, if W+A≡𝒫(A), then we ought to have W≡𝒫(A), just because A is “smaller” than 𝒫(A) (in some sense)—W should just absorb A, if W is “big enough”. Also, if W is “big enough” then that should exclude the other arm of the choice, where W+A≡A. And if W≡𝒫(A), then 𝒫(A) and so also A can be well-ordered, as we have seen.

At this point Hartog’s theorem shows up at the door. This gives us a well-ordered set W with

W≤𝒫4(A) and W≰A

So we have

A≤W+A ≤𝒫4(A)+A ?≡? 𝒫4(A)

(where ?≡? means that the equivalence needs to be proven). W≰A excludes W+A≡A, good. Let’s postpone the issue of the ‘?’. Deal first with the problem that the bounds are not tight enough for GCH to apply. We fix that by looking at:

𝒫3(A)≤W+𝒫3(A)≤𝒫4(A)+𝒫3(A) ?≡? 𝒫4(A)

So if W+𝒫3(A)≡𝒫4(A), we ought to have W≡𝒫4(A) and hence a well-ordering of A. What about the other case, W+𝒫3(A)≡𝒫3(A)? Ah, then we have

𝒫2(A)≤W+𝒫2(A)≤W+𝒫3(A)≡𝒫3(A)

and so we can repeat the argument: either W+𝒫2(A)≡𝒫3(A), which ought to make 𝒫3(A) well-ordered and hence also A well-ordered; or W+𝒫2(A)≡𝒫2(A), in which case we repeat the argument yet again. Eventually we work our way down to

A≤W+A≤W+𝒫(A)≡𝒫(A)

and W+A≡A is excluded since W≰ A, and we are done.

All this relies on the intuition that if W+M≡𝒫(M), then we should have W≡𝒫(M): we used this with M=𝒫n(A) for n=0,…,3. Well, we can prove something a little weaker.

Lemma: If W+M≡𝒫(M)×𝒫(M), then W≥𝒫(M).

Proof: Suppose h:W+M→𝒫(M)×𝒫(M) is a bijection. Restrict h to M and compose with the projection to the second factor: π2⚬(h↾M):M→𝒫(M). Cantor’s diagonal argument shows that this map cannot be onto. (The fact that π2⚬(h↾M) might not be 1–1 doesn’t affect the argument.) So for some s0∈𝒫(M), we know that h(x) never takes the form (−,s0) for x∈M. In other words, the image of h↾W must include all of 𝒫(M)×{s0}. Therefore 𝒫(M)×{s0} can be mapped 1–1 to a subset of W. qed.

The missing pieces of the proof now all take the form of absorption equations. We know that 𝒫(M)×𝒫(M)≡𝒫(M+M)—as an equation for cardinals, 2𝔪2𝔪=22𝔪. If we had 2𝔪=𝔪, that would take care of that problem. The ?≡? above also takes the form 2𝔪+𝔪 ?=? 2𝔪, for 𝔪 the cardinality of 𝒫3(A).

The general absorption laws for addition depend on AC. But we do have these suggestive equations even without AC:

𝔞+ω+1 = 𝔞+ω, 2𝔪+1 = 2·2𝔪

and so if 𝔞+ω=𝔪 and 2𝔪=𝔟, then 2𝔟=𝔟. So let’s say we set B=𝒫(A+ω). Then we have 2B≡𝒫(A+ω+1)≡B (where 2B, of course, is the disjoint union of B with itself). Also B≤B+1≤2B≡B, so B≡B+1 and so 𝒫(B)≡2𝒫(B). So if we replace A with B, then all gaps in the argument are filled and we conclude that B can be well-ordered. But obviously A can be imbedded in B, so A also can be well-ordered. QED.

Ernst Specker proved a “local” version of Sierpiński’s Theorem: if 𝔪 and 2𝔪 both satisfy CH, then 2𝔪=ℵ(𝔪).

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by | November 20, 2025 · 4:26 pm

Set Theory Jottings 18. The Axiom of Determinacy

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Just denying the axiom of choice doesn’t buy you much. If you’re going to throw away AC, you should add some powerful incompatible axiom in its place. The Axiom of Determinacy (AD) has been studied in this light.

Here’s one formulation. Let S be ℕℕ, i.e., the set of all infinite strings of natural numbers. Let G⊆S. Alice and Bob play a game where at step 2n, Alice chooses a number s2n, and at step 2n+1, Bob chooses a number s2n+1. If s0s1s2…∈G, Alice wins, otherwise Bob wins. We say elements of G are assigned to Alice, and elements not in G are assigned to Bob. We’ll call the infinite strings results (of the game). Rather than think of G as a set of results, think of it as a function G:S→{Alice,Bob}.

A strategy for Alice tells her how to play each move. Formally, it’s a function from the set of all number strings of finite even length to ℕ. Likewise, a strategy for Bob maps number strings of finite odd length to numbers. A game is determined if Alice or Bob has a winning strategy, i.e., if the player follows the strategy then that player will win. The Axiom of Determinacy says that each game is determined.

Interesting thing about the proof that AC → ¬AD: it’s much easier using the well-ordering theorem instead of Zorn’s lemma.

First note that there are c=ℵ0ℵ0 strategies (lumping together both Alice and Bob strategies), likewise c results. Assuming AC, well-order the strategies {Sα:α<ωc}. Here ωc is the least ordinal with cardinality c, so the set {α:α<κ} has cardinality less than c for each κ<ωc.

We construct a game G by inducting transfinitely through all the strategies, at step κ considering Sκ. Our goal is to assign some result to Alice or Bob that prevents Sκ from being a winning strategy. Say Sκ is an Alice strategy. Since we assign only one result at each step, fewer than c results have been assigned before step κ. However, there are c possible results if Alice follows Sκ, since Bob can play his numbers however he wants. So there exists a result where Alice follows Sκ but this result has not yet been assigned to either player. Assign it to Bob; this thwarts Sκ. If Sκ is a Bob strategy, just switch everything around. QED

The cardinality argument at the heart of this proof is harder to pull off with Zorn’s lemma (though possible, of course). (The exact same argument works with bit strings instead of strings of natural numbers, but for some reason AD is generally stated using ℕℕ instead of 𝒫(ℕ).)

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by | November 18, 2025 · 8:42 am

From Kepler to Ptolemy 22

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Libration Force

The Libration Force

Kepler coined the term “libration” for the oscillation of a planet’s distance from the Sun, approaching and receding.

He analyzed the libration for the eccentric-equant model, and found it unexpectedly complicated. Stephenson (p.78):

Many absurdities were involved in supposing that a planet could move, … non-uniformly, about the vacant center of the eccentric, with no guide except the apparent magnitude of the solar disk. Such complicated hypotheses, although designed to yield a perfectly simple eccentric circular path, were not physically credible…

Notice the remarkable thing that Kepler was doing here. He was analyzing motion on an eccentric circle, a model that had been in general use for nearly two millenia, apparently the simplest possible model with any empirical accuracy. He took apart this beautifully simple model and showed that as a physical process (and in the absence of solid spheres) it was really quite complicated, so complicated as to raise doubt about whether it could be real. He had performed so radical a reassessment by interpreting astronomy, for the first time, as a physical science.

Eventually Kepler achieved the elliptical orbit. Seeking a physical explanation, he hit on a magnetic force to produce the libration:

What if all the bodies of the planets are enormous round magnets? Of the earth (one of the planets, for Copernicus) there is no doubt. William Gilbert has proved it.

But to describe this power more plainly, the planet’s globe has two poles, of which one seeks out the sun, and the other flees the sun. So let us imagine an axis of this sort, using a magnetic strip, and let its point seek the sun. But despite its sun-seeking magnetic nature, let it remain ever parallel to itself in the translational motion of the globe…

—Astronomia nova, Chapter 57.

The figure at the top of this post (taken from the Epitome of Copernican Astronomy) shows how it works. (The figure in the Astronomia nova has extra clutter.) Kepler explains:

[When] the strip is at A and E, there is no reason why the planet should approach or recede, since it holds its ends at equal distance from the sun, and would undoubtedly turn its point towards the sun if it were allowed to do so by the force that holds its axis straight and parallel. When the planet moves [counterclockwise] away from A, the point approaches the sun perceptibly, and the tail end recedes. Therefore, the globe begins perceptibly to navigate towards the sun. After E, the tail end perceptibly approaches and the head end recedes from the sun. Therefore, by a natural aversion, the whole globe perceptibly flees the sun…

—Astronomia nova, Chapter 57. [I have changed the letters from C and F to A and E to match the diagram from the Epitome.]

Implicit: the magnetic force weakens with distance, so when the head
is closer to the Sun than the tail, the net force is attractive. And vice versa.

Kepler argued that this scheme gave the force a sinusoidal dependence on the longitude, and showed that this agreed with the libration for an elliptical orbit. Some aspects of this demonstration needed special pleading. Stephenson details the strong and the weak points of the reasoning (pp.110–117).

But: “The theory had one glaring flaw, however. The magnetic axis of the planet had to maintain a constant direction, perpendicular to the apsidal line.” (Stephenson, p.117.) The Earth’s rotational axis doesn’t come close to meeting this requirement. So why should we believe it holds for Mars? Kepler acknowledged the problem:

I will be satisfied if this magnetic example demonstrates the general possibility of the proposed mechanism. Concerning the details, however, I have my doubts. For when the earth is in question, it is certain that its axis, whose constant and parallel direction brings about the year’s seasons at the cardinal points, is not well suited to bringing about this reciprocation… And if this axis is unsuitable, it seems there is none suitable in the earth’s entire body, since there is no part of it that rests in one position while the whole body of the globe revolves in a ceaseless daily whirl about that axis.

As one possible out, Kepler appealed to a planetary mind.

Besides the radial libration, planets have a libration in latitude. This enmeshed the theory in further difficulties. Ever inventive, Kepler devised ad hockery around all these rough spots. But we have a contrast: we can trace a direct path from the whirlpool force to the area law. This cannot be said for Kepler’s libration theory. Kepler’s whirlpool speculations came years before the area law. The libration force came after the elliptical orbit.

There is a reason for this. You can justify the whirlpool force (more or less) using the conservation of angular momentum. Kepler’s libration force has no counterpart in Newtonian physics.

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by | November 13, 2025 · 12:01 pm

From Kepler to Ptolemy 21

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Nature of the Whirlpool Force

Kepler was unsure of the nature of the whirlpool force. Sometimes he compares it to light, sometimes to magnetism. Three successive marginal notes in Chapter 33 lay out the analogy with light: “The kinship of the solar motive power with light”; “Whether light is the vehicle of the motive power”; “The motive power is an immaterial species of the of the solar body”.

The Latin term species calls for a short discussion. Donahue (in his glossary) says:

This word, related to the verb “specio” (see, observe) has an extraordinarily wide range of meaning. Its root meaning is is “something presented to view”, but it can also mean “appearance”, “surface”, “form”, “semblance”, “mental image”, “sort”, “nature”, or “archetype” … I have therefore thrown up my hands, admitted defeat, and declined to translate it at all.

while Stephenson (p.68,footnote) writes:

“Image” is our rendering of Kepler’s species, which has for the most part been left untranslated in other accounts. As Kepler used it the word seems to mean the appearance or visible manifestation of the sun…

Gilbert’s De magnete furnished Kepler with another analogy, and a clue. Chapter 34 is titled “The body of the sun is magnetic, and rotates in its space.” A few quotes from it:

The magnet … has filaments (so to speak) or straight fibers (seat of the motor power) extended throughout its length. …it is credible that [the sun] … has circular fibers all set up in the same direction, which are indicated by the zodiac circle.

… It is therefore plausible, since the earth moves the moon through its species and is a magnetic body, while the sun moves the planets similarly through an emitted species, that the sun is likewise a magnetic body.

Kepler postulates filaments or fibers in the Sun as the source of the whirlpool force. These encircle the Sun along the circles of latitude. The Sun rotates, and the image of the moving fibers acts upon the planet to move it in the same direction.

Inverse Square vs. Inverse Linear

Taking a modern perspective, we have an inverse square law whenever we have a conservation law plus spherical symmetry. But with cylindrical symmetry (like the whirlpool force), we can have an inverse linear dependence.

A nice modern analogy: dipole radiation. Feynman discusses this in his Lectures, Chapter I-28:

The gradually discovered properties of electricity and magnetism … showed that these forces … fell off inversely as the square of the distance… As a consequence, for sufficiently great distances there is very little influence of one system of charges on another.

… Maxwell [to obtain a consistent system] … had to add another term to his equations. With this new term there came an amazing prediction, which was that a part of the electric and magnetic fields would fall off much more slowly with the distance than the inverse square, namely, inversely as the first power of the distance!

It seems a miracle that someone talking in Europe can, with mere electrical influences, be heard thousands of miles away in Los Angeles. How is it possible? It is because the fields do not vary as the inverse square, but only inversely as the first power of the distance.

He goes on to treat the dipole radiator. That is, an antenna. The key point: We have charges moving up and down the antenna. What matters is how that motion looks to a distant observer:

… all we have to do is project the motion on a plane at unit distance. Therefore we find the following rule: Imagine that we look at the moving charge … like a painter trying to paint a scene on a screen at a unit distance … We want to see what his picture would look like. So we see a dot, representing the charge, moving about in the picture. The acceleration of that dot is proportional to the electric field.

At root we have a simple matter of geometry. Substituting an eye for the painter, and the eye’s retina for the screen, we have the diagram below. We have a vertical antenna of height H at distance r from the eye’s lens. The image of the antenna on the retina has height I. The antenna has unit distance 1 from the lens. Thus:

From similar triangles, I/1=H/r. In other words, the length of the image is inversely proportional to the first power of the distance.

Note also that the symmetry is cylindrical and not spherical. If the line of sight is not perpendicular to the antenna, the image will be smaller, vanishing completely when the line of sight passes through the antenna.

Returning from Feynman’s Lectures to Kepler’s Astronomia nova, we can resolve the inverse-square problem. Instead of an antenna, we have the Sun’s circular fibers. Instead of the retina, we have the image (species) of those fibers moving the planet. The whirlpool force results from the sum total of those fiber images. Each image’s contribution diminishes inversely with distance, so the sum does too. For all the difference in the physics, the basic geometry remains the same for Feynman’s dipole and Kepler’s whirlpool.

Stephenson (p.75) puts it this way:

The composite motion was directed along a circle parallel to the sun’s equator (the resultant of the images of all the filaments) and it was therefore weakened—at any latitude—only as this circle expanded, in the simple proportion of distance.

This one sentence states the matter more clearly than Kepler does in Chapter 36. Nonetheless, the ingredients are all there, just scattered through the chapter.

One more thing: is the the whirlpool force is confined to the ecliptic? Some authors say that Kepler claimed this. Stephenson shows, with quotes, that this is not true. But something like it holds effectively. A planet above the solar pole would see the fibers moving in all directions, and the net effect would be complete cancellation. At intermediate latitudes, you get partial cancellation. Only at the ecliptic does the planet get the full effect. Figure 14 (p.75) of Stephenson illustrates this:

A final footnote. In 1645 Ismael Boulliau published his Astronomia philolaica. In it he noted, as Kepler had, that the whirlpool force should exhibit an inverse square dependence on distance. But he ignored Kepler’s solution to this problem. On the strength of this, the noted historian Thony Christie credits him with being The man who inverted and squared gravity. I am far less inclined to award him this accolade.

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