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The Math-tea argument says there must be some real numbers that cannot be specified because we only have countably many definitions.

Why is that wrong?

Is there a simple explanation?

I don't understand this blog.

https://alexanderpruss.blogspot.com/2020/11/the-math-tea-argument.html

I don't understand what does this mean:

For instance, there is no guarantee that there is a set of all the definable real numbers.

Why does it matter that there's no set that contains the enumeration?

There is no set of all sets, yet we can enumerate through every set of sets that has a finite description by considering all combinations of the following characters from least length to greatest length: { },,

{}
{{}}
{{{}}}
{{{{}}}}
{{},{{}}}
{{{{{}}}}}
{{{},{{}}}}
{{{{{{}}}}}}
{{{{},{{}}}}}
{{},{{},{{}}}}

etc.

The axioms of set theory tell us that for any predicate F in the language of set theory there is a set of all the numbers that satisfy F.

I think this makes sense - we can specify a predicate that corresponds to a subset of real numbers that satisfy the predicate

I don't follow the logic here:

Elegant as this argument is, it has crucial set-theoretic flaws. For instance, there is no guarantee that there is a set of all the definable real numbers. The axioms of set theory tell us that for any predicate F in the language of set theory there is a set of all the numbers that satisfy F. But the predicate "is definable" is in English, not in set theory.

Jesus is Lord
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    It of course depends on what you mean by “defined,” but given most definitions, this is true - there must be real numbers which can’t be defined. – Thomas Andrews Dec 18 '20 at 16:17
  • If by "defined" you mean something like "computed by a some program" then, sure, there must be undefined real numbers. But I expect the issue here is with the definition of "defined", so to speak. – lulu Dec 18 '20 at 16:18
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    It's not wrong. There are only countably many reals we can construct algorithmically. So? That's still an infinite quantity, sufficient for the next few centuries or so. –  Dec 18 '20 at 16:28
  • if $D$ is the set of definitions, there is no injection that maps $D$ to the set of reals – Vasili Dec 18 '20 at 16:28
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    Perhaps one should distinguish between a definition of an individual real number and a definition of all real numbers. We don't want to approach the latter task by defining real numbers one by one. – hardmath Dec 18 '20 at 16:30
  • Perhaps "specified" would be a better word than "defined". – saulspatz Dec 18 '20 at 16:53
  • @MatthewTowers I added what I was confused about in the blog – Jesus is Lord Dec 18 '20 at 17:02
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    You might find the first part of this article helpful: https://arxiv.org/pdf/1105.4597.pdf – Chris Eagle Dec 18 '20 at 17:03
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    There's a reason why this is called a "math tea argument", it's an easy to understand argument. Like dropping a stone into a well. But much like that, the solution of removing the stone from the well is subtle, intricate, and requires some grasp of more than just "a well is a hole, stone falls downwards". – Asaf Karagila Dec 18 '20 at 17:06
  • @ProfessorVector I'm arguing with someone that there exists a universal verifier such that every (theorem + purported proof) can be verified. Someone rejects my argument, saying we cannot enumerate through every (theorem + purported proof) because the Math-tea argument is wrong. – Jesus is Lord Dec 18 '20 at 18:05
  • Maybe I'm missing something but I don't see a reference to the math-tea argument in the linked post? – Noah Schweber Dec 18 '20 at 23:03
  • @NoahSchweber: Post "The Math Tea argument is...that there are real numbers that can’t be defined. The idea is this: there are only countably many definitions of real numbers ... and uncountably many real numbers, so there are real numbers that have no definitions." vs "The Math-tea argument says there must be some real numbers that cannot be specified because we only have countably many definitions." Don't those say the same thing? – Jesus is Lord Dec 18 '20 at 23:22
  • @Jared I'm talking about your most recent comment above to ProfessorVector, referring to here - where the math tea argument doesn't seem to be mentioned. I don't see where someone brings up the math tea argument there. – Noah Schweber Dec 18 '20 at 23:23
  • @NoahSchweber Oh - someone said something like "if there's a universal verifier, then you can enumerate through all possible proofs. If that were the case, the math-tea argument would be valid, which it's not, therefore you can't enumerate through all possible proofs, therefore your universal verifier does not exist." – Jesus is Lord Dec 18 '20 at 23:24
  • @Jared According to find-text, the phrase "math-tea" doesn't occur anywhere there. In fact, searching for "tea" calls up three hits: the word "instead" in the OP, the word "tea" in the link to this post under "linked post," and the word "teams" at the bottom of the page. – Noah Schweber Dec 18 '20 at 23:25
  • @NoahSchweber I'm discussing with someone on Discord, so I came here - math-tea isn't on the question I linked – Jesus is Lord Dec 18 '20 at 23:26
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    @Jared Well, without knowing more about what the person you're talking to says, it's hard to give a relevant answer to the question. – Noah Schweber Dec 18 '20 at 23:27
  • Read my post about this. – user21820 Sep 08 '21 at 11:12

1 Answers1

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I would describe the situation as follows: it's not that the math tea argument is wrong, so much as that it is easy to misapply it. Specifically, here's a theorem which on the face of it may seem to contradict the math tea argument:

$(@)$ Assuming $\mathsf{ZFC}$ is consistent, there is a model $M$ of $\mathsf{ZFC}$ such that every element of $M$ is parameter-freely definable in $M$. In particular, every real number in $M$ is parameter-freely definable in $M$.

(Really it's more convenient to talk about $\omega$-models here, but I'll ignore this since it makes things more technical and isn't the key point.)

The issue is that we have to be very careful about what system the math tea argument takes place in. There's a version of the math tea argument which we can run inside $M$, but it is substantially weaker than one might expect.

Specifically, for $n\in\mathbb{N}$ consider the sentence

$(*)_n$: "There is a real which is not parameter-freely $\Sigma_n$-definable over $M$.

Each $(*)_n$ is outright provable in $\mathsf{ZFC}$ via essentially the math tea argument, and so holds in $M$ in particular. However, the following sentence

$(\dagger)$: "There is a real which is not definable over the universe"

is not even expressible in the language of set theory - this is a consequence of Tarski's undefinability theorem - and so we can't talk about whether or not $M$ thinks it's true.

We can (express and) prove in $\mathsf{ZFC}$ the statement

$(\dagger)_{set}$: "For every model $A$ of $\mathsf{ZFC}$, there is a real $r$ which is not definable over $A$,"

again using the math tea argument. This is because we can talk about truth within a set, rather than over the universe. However, note that the real $r$ is not guaranteed to live in $A$. So e.g. the fact that $(@)$ is a $\mathsf{ZFC}$-theorem doesn't contradict the fact that $\mathsf{ZFC}$ proves each $(*)_n$ and $(\dagger)_{set}$.

Noah Schweber
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