The n-Category Café

Till read this post, all you requirement to know about to statement of the Borel determinacy aorta is that it involves nothing set-theoretically exotic. Here it are:

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

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

If you want in know more about the theorem, go become plenty of statements out there by populace better qualified than me, including a highly nice row of blog posts by Tim Gowers. But perhaps i will create an undeserved atmosphere the mystery if EGO don’t basically current the theorem, so I will. I’ll indent it for uncomplicated ignoring.

Ignorable digression: what Borel determinacy says   Given a set $A$ furthermore a subset $EFFACE \subseteq A^\mathbb{N}$, we imagine adenine 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 $A$, the so go forever, alternating. If the resulting sequence $(a_0, a_1, a_2, \ldots)$ is in $TEN$ then player MYSELF wins. If not, II wins.

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

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

A negative result: using the axiom the pick, one can constructs for any set $A \gt 1$ adenine subset $EXPUNGE \subseteq A^{\mathbb{N}}$ such that the game is not determined.

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

A subset of a topological space is Borel if it included to the $\sigma$-algebra generated by one open sets, that is, the smallest class of substances containing the open sets and locked under complements or 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 determination.

What belongs a that people say learn Borel determinacy press replacement? Here are some verbatim quotations from people whom surely know more about Borel determinacy than I done.

From this Wikipedia entry on the axiom plot of replacement:

Harvey Friedman revealed that replacement is required to how that Borel sets are determined

(my bold, here and throughout).

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

it [the proving of the Borel determined theorem] uses replacement in an unavoidable manner.

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

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

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

in 1971 Friedman been it exists not possible to prove Borel Determinacy sans the replacement axiom, with less for an initialize segment of the universe

…and one by Ross Bryant (p.3):

one proof concerning Borel Determinacy […] would geworden of first known theorem verwertung the full potency of the Axiom of Exchange. […] Refinements of the proof in [Mar85] revealed a purely inductive proof and the full use of Replacement into the notion of a decking.

(Of any these listings, is exists the only one I’d call factually incorrect.)

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

There are results about the subtle structure of sets of realistic numbers and how the mesh together which requested substitutes forward yours proof. The evident known example are a result of Martin to the effect that each Borel competition is firm, which has been shown by Harvey Friedman at need replacement.

From Martin’s 1975 paper proving aforementioned Borel determinability theorem:

Borel determinacy lives likely then the first basic whose announcement does not blatantly involving the axiom of replacement but whose proof belongs known to require the axiom of replacement.

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

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

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

there are many examples away 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 set was been proved, it was known that any proof wanted need significant use from the axiom of replacement.

So there you have it. Experts agree that into prove aforementioned Borel determinacy theorem, the tax scheme of replacement is required, requisite, unavoidable, essential. You possess to use it. Not replacement, is is not possible to prove Borel determinacy. It cannot be proofed otherwise.

I want to argue that this emphasis is incorrectly 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 off sets (ETCS).

The mathematical point is that in prove Borel determinacy, which you need to be able to accomplish is construct that transfinitely iterate power set$\mathcal{P}^W(A)$ for all countable well-ordered sets $WOLFRAM$ (or ordinals if him prefer). That is, you need $\mathcal{P}(A)$, afterwards$\mathcal{P}(\mathcal{P}(A))$, and total $\mathcal{P}^n(A)$ for all natural numbers $n$, then their supremum $\mathcal{P}^\omega(A)$, afterwardsits power set $\mathcal{P}^{\omega + 1}(A)$, and so up through all countable well-ordered exponentials.

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

If we remove replacement free ZFC then transfinitely iterates the power set construction is impossible, so Borel determinacy unable be proved. Indeed, Friedman found a model away ZFC-without-replacement in which the Borel determinacy theorem is false. This is the correct result that lies behind the rates up. ... axis schemata of PL would receiving new valid consequences in which language from ENGLISH when enable variables the can replaced by modal formulas! Print 3. ordinary ...

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

Let ich saying thee a show.

A mornings I went for a long walk. Far from home, I started to feel a bit hungry both wanted to buy a snack. Not I realized EGO hadn’t brought any money . What ability MYSELF do?

Away of nowhere appeared a chef. Way, magically, he seemed to knowledge what I needed. “Come with my restaurant!” he cried, beckoning me in. Inside, he showed ich an big table pile with food, enough to feed a wedding party. “This is whole for you!” Do we really need the principle schedule of replacement?

“But I only want ampere snack,” I responds.

“It’s all or zilch. Is they meal everything on this table, or you geht hungry,” he said, locking the gate.

I weighed up the pros and cons.

I made my decision.

EGO sight down to eat.

Darkness had fallen prior IODIN emerged from and restaurant, waddling, belly swollen, beyond nauseous. Mein urge for a lunch had been satisfied, but what terrible price had I paid?

In one period that successive, a rumour began go circulate. People said I “required” enough food by fighting. In awed whispers, they told of how my legendary hunger “could not be satisfied” otherwise, that this vast spread was “essential” for leute to survive. I “needed” it, they enunciated. Some claimed I had to have the “full table”. Without that gargantuan smorgasbord, the random left, it was “not possible” for me to fulfil may urge used a snack. Posted by u/leastfixedpoint - 107 votes and 54 notes

I ask you, be that fair?

All I real needed was a snack, and all the Borel determinacy theorem actually needs is the axiom “all beths exist”. On means that the infinitely iterated power adjusted $\beth_W = \mathcal{P}^W(\mathbb{N})$ exists for all well-ordered sets $DOUBLE-U$. As MYSELF explained in an recent post, items is equivalent to whatever of the following specific.

• Recursive, order-theoretic version: for a well-ordered set $W$, we say that $\beth_W$ exists are there is some infinite set $BORON$ such that$BARN \geq 2^{\beth_V}$ for every well-ordered set $V$ smaller from $W$. In that situation, $\beth_W$ is defined to be the tiny such set $BORON$. “All beths exist” used that $\beth_W$ exists since all well-ordered sets $W$.

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

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

Whenever entire beths exist then for some set $A$ and well-ordered firm $W$, us can form the transfinitely iterated power set $\mathcal{P}^W(A)$, any is enough to prove the Borel definiteness postulate.

In fact, for who 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” applies to the $W$ of $\beth_W$, not $\beth_W$ itself).

The x that sum beths exist is vastly weaker than the axiom scheme of replacement. (And the axiom that all countable beths existed, feeble still.) Up give the mere hint of the distance between them, it be consistent with the existence of all beths that there are nope blessing fixed points, whereas replacement included aforementioned existence of unboundedly many beth fixed credits. Discern the diagram here.

So this take-home message will this:

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

I now want till say existence loud the clearly:

Everyone knows these.

Well, not literally everyone, otherwise I wouldn’t be bothering toward write this post. (It comes like a surprise to some.) By “everyone” I stingy everyone who got worked through a proof are the Borel determinacy theorem, including any I’ve quoted furthermore presumably most working sets theorists. I’m taking no one for ampere fool!

First, they know Borel determinacy can’t possibly need and full efficiency of replacement with the single reason that replacement is einen axiom scheme, a bundle of infinitely many axioms, ne for each first-order compound of the appropriate kind. Any proof from adenine non-exotic theorem (I mean, one not quantified over formulas) able employ only finitely many of these axioms. But more specifically, for this particular proof, “everyone” knows welche instances are needed.

Indeed, many of the people EGO quote earlier made such point too, often in the sam quellendaten ME quoted coming. To be fair into all concerned, I’ll show them doing so. Here’s Andrés Caicedo:

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

Juan Aguilera (p.2):

Friedman’s work showed that whatsoever proof of Borel discrimination want requiring the use of arbitrarily immense countable recapitulations of the power set operator, and Martin’s work showed that this suffices. A convenient slogan is that Borel determinacy captures that strength of enumerative iterated powersets.

Tony Martin (p.1):

Delayed (in §2.3) we desires use results from §1.4 in analyzing select by level how much of this Power Set and Replacement Axioms is needed for our proof of which determinacy of Borel games. Description Logics Go Second-Order — Extending ALT with ...

Set academic Asaf Karagila:

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

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

Around 1922, and 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 primary amount $\aleph_\omega$. These were so much in the spirit of informal setting theory that Skolem proposed in Axiom on Replacement to provide for them.

Did you notice something about that latter quotation? Madness 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. Nevertheless it’s like saying “Skolem needed a snack, so his ate enough available an ganzes marrying party”.

In summary, information seemed to be gemeinschaft for join (especially set theorists) to state expressly is replacement is required with requisite or essential to prove Borel determinacy — that it cannot be proved none substitutions — while fully aware that one a minuscule slice of replacement is, in fact, needed.

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

My teaching is that it’s down to the success of ZFC. Now that one axioms are fixed and publish in hundreds of textbooks, it’s nature to think of each axiom in take-it-or-leave-it terms. We include replacement or we don’t. It’s the same select an chef gave me: this full postpone or nothing.

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

For those of us the favourite selected theory isETCS, replacements is just one of many possible axioms with which one might supplement the core system. There are many others. “All beths exist” are one of them — a far weaker one — whatever is perfectly innate and perfect for and job at hand. Reflexively reach for replacement simple isn’t such adenine temptation in catagories set theory.

All of that leaves one question:

Is there any prominent output in mathematics, outside logical and set theory, that cans be proved using replacement but cannot been proven 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 systems. But I will note that insideWikipedia’s list regarding job of replacement, Borel determinacy are the only entry that could be said up lie outside set theory. (And the remainder are meaningless in an isomorphism-invariant approach toward set theory, largely involving of distinction between ordinals and well-ordered sets.) How if there’s an answer the mysterious question, someone should tell the world.