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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.

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    $\begingroup$ 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. $\endgroup$ Dec 18, 2020 at 16:17
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    $\begingroup$ 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. $\endgroup$
    – user436658
    Dec 18, 2020 at 16:28
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    $\begingroup$ 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. $\endgroup$
    – hardmath
    Dec 18, 2020 at 16:30
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    $\begingroup$ You might find the first part of this article helpful: arxiv.org/pdf/1105.4597.pdf $\endgroup$ Dec 18, 2020 at 17:03
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    $\begingroup$ 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". $\endgroup$
    – Asaf Karagila
    Dec 18, 2020 at 17:06

1 Answer 1

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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 $(\dagger)_{set}$.

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