## July 23, 2021

### Borel Determinacy Does Not Require Replacement

#### Posted by Tom Leinster

Ask around for an example of ordinary mathematics that uses the axiom scheme of replacement in an essential way, and someone will probably say “the Borel determinacy theorem”. It’s probably the most common answer to this question.

As an informal statement, it’s not exactly wrong: there’s a precise mathematical result behind it. But I’ll argue that it’s misleading. It would be at least as accurate, arguably more so, to say that Borel determinacy does not require replacement.

For the purposes of this post, it doesn’t really matter what the Borel determinacy theorem says. I’ll give a lightning explanation, but you can skip even that.

Thanks to David Roberts for putting me onto this. You can read David’s recent MathOverflow question on this point too.

To read this post, all you need to know about the statement of the Borel determinacy theorem is that it involves nothing set-theoretically exotic. Here it is:

Theorem (Borel determinacy)   Let $A$ be a set and $X \subseteq A^{\mathbb{N}}$. If something then something.

Neither “something” is remotely sneaky. The highest infinity mentioned is $\mathbb{N}$. Moreover, the set $A$ is often taken to be $\mathbb{N}$ too.

If you want to know more about the theorem, there are plenty of explanations out there by people more qualified than me, including a very nice series of blog posts by Tim Gowers. But perhaps it will create an undeserved aura of mystery if I don’t actually state the theorem, so I will. I’ll indent it for easy ignoring.

Ignorable digression: what Borel determinacy says   Given a set $A$ and a subset $X \subseteq A^\mathbb{N}$, we imagine a two-player game as follows. Player I chooses an element $a_0$ of $A$, then player II chooses an element $a_1$ of $A$, then player I chooses an element $a_2$ of $A$, and so on forever, alternating. If the resulting sequence $(a_0, a_1, a_2, \ldots)$ is in $X$ then player I wins. If not, II wins.

A strategy for player I is a function that takes as its input an even-length finite sequence of elements of $A$ (to be thought of as the moves played so far) and produces as its output a single element of $A$ (to be thought of as the recommended next move for player I). It is a winning strategy if following it guarantees a win for player I, regardless of what II plays. Winning strategies for player II are defined similarly.

Does either of the players have a winning strategy? Clearly they can’t both have one, and clearly the answer depends on $A$ and $X$. If one of the players has a winning strategy, the game is said to be determined.

A negative result: using the axiom of choice, one can construct for any set $A \gt 1$ a subset $X \subseteq A^{\mathbb{N}}$ such that the game is not determined.

Positive results tend to be stated in terms of the product topology on $A^\mathbb{N}$, where $A$ itself is seen as discrete. For example, it’s fairly easy to show that if $X$ is open or closed in $A^\mathbb{N}$ then the game is determined.

A subset of a topological space is Borel if it belongs to the $\sigma$-algebra generated by the open sets, that is, the smallest class of subsets containing the open sets and closed under complements and countable unions.

The Borel determinacy theorem, proved by Donald A. (Tony) Martin in 1975, states that if $X \subseteq A^\mathbb{N}$ is Borel then the game is determined.

What is it that people say about Borel determinacy and replacement? Here are some verbatim quotations from people who surely know more about Borel determinacy than I do.

Harvey Friedman showed that replacement is required to show that Borel sets are determined

(my bold, here and throughout).

From a nice math.stackexchange answer by set theorist Andrés Caicedo:

it [the proof of the Borel determinacy theorem] uses replacement in an unavoidable manner.

From the opening post of that splendid series by Tim Gowers:

In order to iterate the power set operation many times, you have to use the axiom of replacement many times.

Gowers begins by citing two helpful sources, one by Shahzed Ahmed, who writes (p.2):

in 1971 Friedman showed it is not possible to prove Borel Determinacy without the replacement axiom, at least for an initial segment of the universe

…and one by Ross Bryant (p.3):

a proof of Borel Determinacy […] would become the first known theorem realizing the full potency of the Axiom of Replacement. […] Refinements of the proof in [Mar85] revealed a purely inductive proof and the full use of Replacement in the notion of a covering.

(Of all these quotes, this is the only one I’d call factually incorrect.)

From a paper by philosopher Michael Potter (p.185):

There are results about the fine structure of sets of real numbers and how they mesh together which require replacement for their proof. The clearest known example is a result of Martin to the effect that every Borel game is determined, which has been shown by Harvey Friedman to need replacement.

From Martin’s 1975 paper proving the Borel determinacy theorem:

Borel determinacy is probably then the first theorem whose statement does not blatantly involve the axiom of replacement but whose proof is known to require the axiom of replacement.

From philosopher of mathematics Penelope Maddy (Believing the axioms I, p.490):

Recently, however, Martin used Replacement to show that all Borel sets are determined […] Earlier work of Friedman establishes that this use of Replacement is essential.

And from a recent paper by logician Juan Aguilera (p.2):

there are many examples of theorems that cannot be proved without the use of the axiom of replacement. An early example of this was that of Borel determinacy. […] Even before Martin’s Borel determinacy theorem had been proved, it was known that any proof would need essential use of the axiom of replacement.

So there you have it. Experts agree that to prove the Borel determinacy theorem, the axiom scheme of replacement is required, needed, unavoidable, essential. You have to use it. Without replacement, it is not possible to prove Borel determinacy. It cannot be proved otherwise.

I want to argue that this emphasis is misplaced and misleading. It is especially misplaced from the point of view of categorical set theory, or more specifically Lawvere’s Elementary theory of the category of sets (ETCS).

The mathematical point is that to prove Borel determinacy, what you need to be able to do is construct the transfinitely iterated power set $\mathcal{P}^W(A)$ for all countable well-ordered sets $W$ (or ordinals if you prefer). That is, you need $\mathcal{P}(A)$, then $\mathcal{P}(\mathcal{P}(A))$, and generally $\mathcal{P}^n(A)$ for all natural numbers $n$, then their supremum $\mathcal{P}^\omega(A)$, then its power set $\mathcal{P}^{\omega + 1}(A)$, and so on through all countable well-ordered exponents.

When I say you “need to be able to do” this, I mean it in a strong sense: Friedman showed in the early 1970s that Borel determinacy can’t be proved otherwise (even, I think, in the case where $|A| = 2$). And when Martin later proved Borel determinacy, his proof used this construction and no more.

If we remove replacement from ZFC then transfinitely iterating the power set construction is impossible, so Borel determinacy cannot be proved. Indeed, Friedman found a model of ZFC-without-replacement in which the Borel determinacy theorem is false. This is the precise result that lies behind the quotes above.

So what’s the problem? It’s that replacement is vastly more than necessary.

Let me tell you a story.

One morning I went for a long walk. Far from home, I started to feel a bit hungry and wanted to buy a snack. But I realized I hadn’t brought any money . What could I do?

Out of nowhere appeared a chef. Somehow, magically, he seemed to know what I needed. “Come into my restaurant!” he cried, beckoning me in. Inside, he showed me a big table heaped with food, enough to feed a wedding party. “This is all for you!”

“But I only want a snack,” I replied.

“It’s all or nothing. Either you eat everything on this table, or you go hungry,” he said, locking the door.

I weighed up the pros and cons.

I sat down to eat.

Darkness had fallen before I emerged from the restaurant, waddling, belly swollen, beyond queasy. My urge for a snack had been satisfied, but what terrible price had I paid?

In the months that followed, a rumour began to circulate. People said I “required” enough food for fifty. In awed whispers, they told of how my legendary hunger “could not be satisfied” otherwise, that this vast spread was “essential” for me to survive. I “needed” it, they said. Some claimed I had to have the “full table”. Without that gargantuan smorgasbord, the rumour went, it was “not possible” for me to fulfil my urge for a snack.

I ask you, is that fair?

All I actually needed was a snack, and all the Borel determinacy theorem actually needs is the axiom “all beths exist”. This means that the transfinitely iterated power set $\beth_W = \mathcal{P}^W(\mathbb{N})$ exists for all well-ordered sets $W$. As I explained in a recent post, it is equivalent to any of the following conditions.

• Recursive, order-theoretic version: for a well-ordered set $W$, we say that $\beth_W$ exists if there is some infinite set $B$ such that $B \geq 2^{\beth_V}$ for every well-ordered set $V$ smaller than $W$. In that case, $\beth_W$ is defined to be the smallest such set $B$. “All beths exist” means that $\beth_W$ exists for all well-ordered sets $W$.

• Non-recursive, order-theoretic version: “all beths exist” means that for every well-ordered set $W$, there exist a set $X$ and a function $p: X \to W$ with the following property: for $w \in W$, the fibre $p^{-1}(w)$ is the smallest infinite set $\geq 2^{p^{-1}(v)}$ for all $v \lt w$.

• Non-recursive, non-order-theoretic version: “all beths exist” means that for every set $I$, there exist a set $X$ and a function $p: X \to I$ such that for all distinct $i, j \in I$, either $2^{p^{-1}(i)} \leq p^{-1}(j)$ or vice versa.

If all beths exist then for any set $A$ and well-ordered set $W$, we can form the transfinitely iterated power set $\mathcal{P}^W(A)$, which is enough to prove the Borel determinacy theorem.

In fact, for the Borel determinacy theorem, we only need this when $W$ is countable. Moreover, one often assumes that the set $A$ on which the game is played is countable too. In that case, we only need countable beths to exist (where “countable” refers to the $W$ of $\beth_W$, not $\beth_W$ itself).

The axiom that all beths exist is vastly weaker than the axiom scheme of replacement. (And the axiom that all countable beths exist, weaker still.) To give the merest hint of the distance between them, it is consistent with the existence of all beths that there are no beth fixed points, whereas replacement implies the existence of unboundedly many beth fixed points. See the diagram here.

So the take-home message is this:

Borel determinacy does not require replacement. It only requires the existence of all beths.

I now want to say something loud and clear:

Everyone knows this.

Well, not literally everyone, otherwise I wouldn’t be bothering to write this post. (It comes as a surprise to some.) By “everyone” I mean everyone who has worked through a proof of the Borel determinacy theorem, including everyone I’ve quoted and presumably most working set theorists. I’m taking no one for a fool!

First, they know Borel determinacy can’t possibly need the full power of replacement for the simple reason that replacement is an axiom scheme, a bundle of infinitely many axioms, one for each first-order formula of the appropriate kind. Any proof of a non-exotic theorem (I mean, one not quantified over formulas) can use only finitely many of these axioms. But more specifically, for this particular proof, “everyone” knows which instances are needed.

Indeed, many of the people I quoted earlier made this point too, often in the same sources I quoted from. To be fair to all concerned, I’ll show them doing so. Here’s Andrés Caicedo:

the [set needed] is something like $\mathcal{P}^\alpha(\mathbb{N})$ if the original Borel set was at level $\alpha$.

Juan Aguilera (p.2):

Friedman’s work showed that any proof of Borel determinacy would require the use of arbitrarily large countable iterations of the power set operator, and Martin’s work showed that this suffices. A convenient slogan is that Borel determinacy captures the strength of countably iterated powersets.

Tony Martin (p.1):

Later (in §2.3) we will use results from §1.4 in analyzing level by level how much of the Power Set and Replacement Axioms is needed for our proof of the determinacy of Borel games.

Set theorist Asaf Karagila:

Even the famous Borel determinacy result (which is the usual appeal of Replacement outside of set theory […]) only requires $\beth_{\omega_1}$ which is far, far below the least [beth] fixed point.

And Penelope Maddy (p.489, shortly before she gets on to Borel determinacy specifically):

Around 1922, both Fraenkel and Skolem noticed that Zermelo’s axioms did not imply the existence of

$\{ N, \mathcal{P}(N), \mathcal{P}(\mathcal{P}(N)), \ldots \}$

or the cardinal number $\aleph_\omega$. These were so much in the spirit of informal set theory that Skolem proposed an Axiom of Replacement to provide for them.

Did you notice something about that last quotation? Maddy goes straight from the existence of $\aleph_\omega$ and $\beth_\omega$ to the full axiom scheme of replacement. To be fair, she’s recounting history rather than doing mathematics. But it’s like saying “Skolem needed a snack, so he ate enough for an entire wedding party”.

In summary, it seems to be common for people (especially set theorists) to state explicitly that replacement is required or necessary or essential to prove Borel determinacy — that it cannot be proved without replacement — while fully aware that only a minuscule fragment of replacement is, in fact, needed.

I’ve said that I think this is misleading, but it also strikes me as curious. Those who do this, many of them experts, are eliding the difference between a light snack and a health-threateningly huge food mountain. Why would they do that?

My theory is that it’s down to the success of ZFC. Now that the axioms are fixed and published in hundreds of textbooks, it’s natural to think of each axiom in take-it-or-leave-it terms. We include replacement or we don’t. It’s the same choice the chef gave me: the full table or nothing.

Membership-based set theorists do study fragments of replacement, including in this context. This conversation began when David Roberts pointed out Corollary 2.3.8 and Remark (i) after it in Martin’s book draft, which concern exactly this point. But there are orders of magnitude fewer people who read technical side-notes like Martin’s than who read and repeat the simple but misleading message “Borel determinacy requires replacement”.

For those of us whose favourite set theory is ETCS, replacement is just one of many possible axioms with which one could supplement the core system. There are many others. “All beths exist” is one of them — a far weaker one — which is perfectly natural and perfect for the job at hand. Reflexively reaching for replacement just isn’t such a temptation in categorical set theory.

All of that leaves a question:

Is there any prominent result in mathematics, outside logic and set theory, that can be proved using replacement but cannot be proved using “all beths exist”?

I don’t know. People ask questions similar to this on MathOverflow from time to time (although I don’t think anyone’s asked exactly this one). I haven’t been through the answers systematically. But I will note that in Wikipedia’s list of applications of replacement, Borel determinacy is the only entry that could be said to lie outside set theory. (And the remainder are meaningless in an isomorphism-invariant approach to set theory, largely involving the distinction between ordinals and well-ordered sets.) So if there’s an answer to my question, someone should tell the world.

Posted at July 23, 2021 6:32 PM UTC

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### Re: Borel Determinacy Does Not Require Replacement

As you explicitly asked: In Etale cohomology of diamonds, I end up using beth-fixed points. (See Section 4; the somewhat weird conditions of Lemma 4.1 are equivalent to saying that $\kappa$ is a beth fixed point of uncountable cofinality. When I wrote this section, I didn’t really know what I was doing…)

That said, actually I recently thought again about why I wanted such large $\kappa$, and it is likely that merely having all beths exist is enough. If you really care, I might try to figure it out.

By the way, thanks also from me for this nice series of posts!

Posted by: Peter Scholze on July 23, 2021 9:31 PM | Permalink | Reply to this

### Re: Borel Determinacy Does Not Require Replacement

Ah, great — that’s a fantastic answer! Thanks.

I don’t think I’m super-invested in whether everything can be done with just the existence of all beths, or whether we need beth fixed points too. I don’t really feel the difference between the two. What would feel different is if there was some vastly more general principle that turned out to be necessary, like replacement.

Posted by: Tom Leinster on July 23, 2021 9:49 PM | Permalink | Reply to this

### Re: Borel Determinacy Does Not Require Replacement

For the record, let me state here the lemma implicit in Peter’s parenthetical comment. It’s that a nonempty set $X$ is a beth fixed point if and only if:

for all $Y \lt X$, there exists a strong limit $Z \lt X$ such that $Y \lt cf(Z)$

(“below $X$, cofinalities of strong limits are unbounded”).

I didn’t know that, and found it pleasantly challenging to prove.

Posted by: Tom Leinster on July 24, 2021 1:31 AM | Permalink | Reply to this

### Re: Borel Determinacy Does Not Require Replacement

For what it’s worth, I really care!

Posted by: David Roberts on July 26, 2021 12:21 AM | Permalink | Reply to this

### Re: Borel Determinacy Does Not Require Replacement

I expect another reason people talk about replacement rather than axioms like “beths exist” is that it’s easier to justify why one should “believe” it, as a “true” statement about “the real universe of sets”. The intuitive justification of replacement is fairly straightforward: any way of constructing a bunch of sets should yield a family of sets. But is there any reason to believe “beths exist” that doesn’t also justify believing the rest of replacement?

Posted by: Mike Shulman on July 24, 2021 6:14 AM | Permalink | Reply to this

### Re: Borel Determinacy Does Not Require Replacement

My best reply to that question was here. Though I’m as wary as I think you are of talk about “believing” axioms.

You may well be right about why people talk about replacement here. It’s dangerous to speculate on personal motivations and psychology, and irresistible too :-)

Posted by: Tom Leinster on July 24, 2021 12:36 PM | Permalink | Reply to this

### Re: Borel Determinacy Does Not Require Replacement

Allow me to re-quote Cantor here:

if there is some determinate succession of defined whole real numbers, [i.e. ordinals] among which there exists no greatest, on the basis of this second principle of generation a new number is obtained which is regarded as the limit of those numbers, i.e. is defined as the next greater number than all of them.

Instead of ordinals here (which in the context of this series gives “all alephs exist”, if I haven’t got my wires crossed), one could take instead iterated power sets, giving “beths exist”.

I guess the subtlety is how one decides to which processes this principle should or shouldn’t apply. In the context of ETCS, ‘power set’ is one of the baked-in operations, so I think taking “beths exist” as an instance of Cantor’s Second Principle is natural. Further instances, leading to larger cardinals, seem to me less close to the raw stuff of ETCS.

Oh, and one can uncharitably call Replacement a fix for the fact that from a purely structural view, ZFC needs it so as to be closed under isomorphism, among other things. As seen in the draft book of Martin linked in the post, $\Sigma_1$-replacement is superficially needed so that the usual constructions of ordinal and cartesian products go through. This aspect is clearly not needed in a structural setting. However , this is not meant to dunk on ZFC: its richness as a set theory comes from the fact it is not purely structural.

Posted by: David Roberts on July 26, 2021 12:14 AM | Permalink | Reply to this

### Re: Borel Determinacy Does Not Require Replacement

I have trouble reading that quote as saying anything nontrivial that’s less than full replacement. Shouldn’t any family of sets uniquely specified by a formula be a “determinate succession”?

Posted by: Mike Shulman on July 27, 2021 3:43 AM | Permalink | Reply to this

### Re: Borel Determinacy Does Not Require Replacement

I see it as more of a transfinite recursion principle for ordinal-indexed constructions. Certainly I strongly suspect Cantor was only ever thinking about ordinal indexing. Maybe for generic first-order logic we can get everything, but restricting ourselves to the operations available in ETCS, then at least superficially it’s not clear to me we get everything possible with Replacement. But I can see your point, and haven’t thought it through.

Posted by: David Roberts on July 27, 2021 4:01 AM | Permalink | Reply to this

### Re: Borel Determinacy Does Not Require Replacement

Well, as Tom told us in Part 12, transfinite induction is equivalent to replacement. And I’m not sure what it would mean here to restrict ourselves to the ETCS operations – since we’re talking about specifying a set, is it really meaningful to restrict the formula specifying it to use only bounded quantifiers?

Posted by: Mike Shulman on July 27, 2021 4:17 AM | Permalink | Reply to this

### Re: Borel Determinacy Does Not Require Replacement

Ah, yes. I knew I was treading a fine line. I didn’t go back and check to see what I was committing myself to.

Posted by: David Roberts on July 27, 2021 5:06 AM | Permalink | Reply to this

### Re: Borel Determinacy Does Not Require Replacement

A loosely related question I was asking myself yesterday: what’s the strongest natural axiom that’s both implied by replacement (R) and implied by “there are unboundedly many inaccessibles” (UI)? I have this diagram in mind.

I inserted the informal word “natural” because, of course, the literal answer is “R or UI”. But I wonder whether there’s some more compelling axiom with intuitive appeal.

“There are unboundedly many beth fixed points” is an axiom implied by R and also implied by UI, but it’s not the strongest such.

The reason why I was thinking about this yesterday is that I came across an exercise in a course by Benedikt Löwe as follows: prove that every regular set is the cofinality of some beth fixed point. The result follows if one can construct the $W$th beth fixed point $B_W$ for each well-ordered $W$. (For then given a regular set $X$, we have $cf(B_{I(X)}) \cong X$ where $I(X)$ is $X$ with an initial well order.) I can construct $B_W$ assuming replacement, and I can construct $B_W$ assuming UI. So “$B_W$ exists for all $W$” is an axiom implied by replacement and implied by UI, and in turn it implies “there are unboundedly many beth fixed points”.

One could keep going, looking at $B$-fixed points, and so on, to get stronger axioms still — all implied by R or, alternatively, UI.

I guess this has been investigated, partly because of the existence of the theory of normal functions, and partly because of the connection between inaccessibles and Grothendieck universes.

Posted by: Tom Leinster on July 27, 2021 10:34 AM | Permalink | Reply to this

### Re: Borel Determinacy Does Not Require Replacement

Tom wrote

And the remainder are meaningless in an isomorphism-invariant approach to set theory, largely involving the distinction between ordinals and well-ordered sets.

This puts me in mind of my quotation of Lawvere, which begins

Andrej Bauer asked whether large cardinals other than inaccessible ones have a natural definition in topos theory. Indeed, like most questions of set theory which have an objective content, this too is independent of the a priori global inclusion and membership chains which are characteristic of the Peano conception that ZF formalizes. Various kinds of “measurable” cardinals arise as possible obstructions to simple dualities of the type considered in algebraic geometry.

It seems that there are two criteria for a cardinal concept here: that which makes sense from a structural, category-theoretic point of view and that which occurs in mainstream “real” mathematics, such as algebraic geometry. We might imagine the latter is somewhat stricter than the former, and both are much stricter that what makes sense from a ZF approach.

Do we have cases, or, if not, is it imaginable, that there are ideas concerning large cardinals that arise from a structural point of view that would make little sense to a material set theorist? (I recall Mike saying something to the effect that synthetic homotopy theory practiced by HoTT-theorists could not have been envisaged from a material set-theoretic outlook.)

Posted by: David Corfield on July 24, 2021 3:53 PM | Permalink | Reply to this

### Re: Borel Determinacy Does Not Require Replacement

Various kinds of “measurable” cardinals arise as possible obstructions to simple dualities of the type considered in algebraic geometry.

One of the frustrations of reading Lawvere is that he’s always tossing out these references to entire subject areas, not saying what he means, let alone actually, you know, giving a reference. What, in fact, is he talking about here? What dualities in algebraic geometry? I can try to guess, but it’s annoying to be forced into guessing games, and it makes me feel like I’m interpreting some obscure religious text rather than doing science. I’ve gained enormously from reading Lawvere over the years — I mean, I just wrote a 13-post series about the set theory he proposed, and an entirely different insight of Lawvere’s is crucial to my whole magnitude project. And this particular passage isn’t too bad, comparatively (I more or less managed to make it through). But sometimes…

that which makes sense from a structural, category-theoretic point of view and that which occurs in mainstream “real” mathematics, such as algebraic geometry. We might imagine the latter is somewhat stricter than the former

When you say “stricter”, do you have in mind things like measurability that are reasonably natural in categorical set theory but have only the most tangential relevance outside logic and set theory?

I’d certainly agree that that’s the case. One point (I guess exaggerating a bit) is that “what’s categorically natural” could be seen as a timeless criterion, whereas “what occurs in mainstream mathematics” is very much about the present-day world, historical accident, power structures in academia, etc. My tongue is slightly in my cheek when I say “timeless”, but I think there’s a point here.

I’m not sure whether the following is a good comparison, but I often think about how the Australian school of category theory has been so consistently ahead of its time. So much of what they were doing decades ago must have seemed like the most obscure, abstract-for-its-own-sake stuff to the vast majority of mathematicians at the time. And so much of it has paid off and come to be seen as quite mainstream now.

Posted by: Tom Leinster on July 24, 2021 4:29 PM | Permalink | Reply to this

### Re: Borel Determinacy Does Not Require Replacement

When you say “stricter”, do you have in mind things like measurability that are reasonably natural in categorical set theory but have only the most tangential relevance outside logic and set theory?

Yes, that sort of thing, to the extent that a case of measurability is of at most tangential relevance. But in fact measurability is what Lawvere turns to, as in the quotation above, and also the passage I quoted here:

The study of such examples is always related to double-dualization monads and to the failure of reasonable geometrical theorems in case so-called measurable cardinals are admitted into the category of small sets. This suggests to me that first of all “small” should not be identified with “member of some class”, but should explicitly exclude the measurable cardinals. The category of all small sets is an object we frequently use and if it itself is a measurable cardinal that should not dismay us any more than that the category of all finite sets is not finite.

There he’s after bornology to turn functional analysis into ‘algebraic geometry’, and he returns to bornology in this recent preprint

• Toposes generated by codiscrete objects in combinatorial topology and functional analysis, Reprints in Theory and Applications of Categories, No. 27 (2021) pp. 1-11, pdf,

where we get a brief mention of large cardinals on p. 11, following on from a comparison of the topological topos and the bornological topos.

All of which brings us back (see this comment) to the work of the first commentator on this post.

Posted by: David Corfield on July 24, 2021 8:42 PM | Permalink | Reply to this

### Re: Borel Determinacy Does Not Require Replacement

Has anyone studied an axiom of countable Replacement?

This would be much weaker than the full axiom and also quite enough for Borel Determinacy.

Posted by: Daniel Grubb on December 1, 2022 2:27 PM | Permalink | Reply to this

### Re: Borel Determinacy Does Not Require Replacement

Zermelo considered the axiom of countable Replacement in the late 1920s, but I don’t know if anyone else since has taken that seriously.

Posted by: David Roberts on December 3, 2022 11:09 PM | Permalink | Reply to this

### Re: Borel Determinacy Does Not Require Replacement

Gah, it might not be Zermelo, it might have been Fraenkel! I’m trying to track down old comments I’ve made to see if I source any claims, and I don’t see that I have.

Note also that Randall Holmes mentions countable replacement briefly in the notes Separating Hierarchy and Replacement.

Posted by: David Roberts on December 4, 2022 11:14 AM | Permalink | Reply to this

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