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A formula $\phi(x, y)$ defined a function on a set X if for each $x \in X$, there is exactly one set y such that $\phi(x,y)$ holds. We then say $\phi$ is a functional formula, in the sense that it defines a function.

Now I wonder, can function in set theory, which is just some special subsets in set theory be written as a functional formula?

In the first answer of this post about well-orderings of the reals, it was mentioned that no formula in the language of set theory defined a well-ordering of the reals. Since a well ordering is just a function $f: \mathfrak{c} \to \mathbb{R}$, and the axiom of choice assumed its existence, does this mean that such a function can never be written as a functional formula?

Now in the axiom of replacement, I saw many places used $\phi(x,y)$, the functional formula. Does this mean some functions can never be used in the axiom of replacement?

patchouli
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2 Answers2

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There is an important distinction here about whether parameters are allowed in our functional formulas.

If $f$ is a function (i.e. a set of input/output pairs), then $f$ can be trivially defined by a functional formula using $f$ as a parameter: take $\phi(x,y)$ to be the formula $(x,y)\in f$. But there is no reason to expect that an arbitrary $f$ should be definable by a functional formula without parameters.

The axiom schema of replacement, in its usual formulation, applies to functional formulas with parameters. That is, it says "for all $p_1,\dots,p_n$, if $\phi(x,y,p_1,\dots,p_n)$ is a functional formula, then ...". Now it happens that if we assume the axiom schema of replacement for all functional formulas without parameters, we can also prove every instance of it for functional formulas with parameters, but this is non-trivial to prove.

Finally, regarding "no formula in the language of set theory defines a well-ordering of the reals". The same discussion about parameters applies here. In any model of ZFC, since there does in fact exist a well-ordering $\prec$ of $\mathbb{R}$, this well-ordering is trivially definable using $\prec$ as a parameter: again, take $\phi(x,y)$ to be the formula $(x,y)\in {\prec}$. But there is no reason to think that $\prec$ should be definable by a formula without parameters.

The more precise formulation of your quoted statement is that there is no formula $\phi(x,y)$ in the language of set theory (without parameters) such that ZFC proves that $\phi(x,y)$ defines a well-ordering of the reals. In fact, there are formulas $\phi(x,y)$ which define well-orderings of the reals in some models of ZFC, but not in others. Michael Weiss has addressed this in the other answer.

Correspondingly, there is no functional formula $\phi(x,y)$ without parameters such that ZFC proves that $\phi$ defines a bijection between an ordinal and $\mathbb{R}$. But of course there is one if we allow parameters.

Alex Kruckman
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The answer you link to answers this question, at least implicitly. But since you've already read that, perhaps a rephrasing will help.

Gödel defined a subclass of the universe, the so-called constructible sets. This is denoted $L$. He also showed how to construct a functional formula mapping the class of all ordinals 1-1 onto $L$. It's easy enough to restrict this so that we have a 1-1 correspondence, defined by a formula, between the constructible reals and an ordinal. Or equivalently, there is a definable well-ordering of the reals.

So if $V=L$, i.e., all sets are constructible, then there is a definable well-ordering of the reals. Gödel showed that it is consistent with ZF to assume that $V=L$. (This was the basis of his famous pair of results that Con(ZF) $\Rightarrow$ Con(ZFC) and Con(ZFC+GCH).)

Therefore we cannot prove in ZF (or even in ZFC) that there isn't a definable well-ordering of the reals.

On the other hand, Cohen constructed a model of ZFC in which there is a non-constructible real. It turns out that in this model, there is no definable well-ordering of the reals. So we cannot prove that there is such a definable well-ordering.

As you point out, asking if there is a definable well-ordering of $\mathbb{R}$ is equivalent to asking if there is a definable bijection between the $\mathfrak{c}$ and $\mathbb{R}$.

Summing up, we have a candidate formula, but we can't prove it does or doesn't do the job.

Turning to the Axiom (Schema) of Replacement, this allows one to use parameters. (All my mentions of "definable" above should be read as, "definable without parameters".) That is, the functional formula $\phi$ that occurs in it is $\phi(x,y,w_1,\ldots,w_n)$. If we have some function $f$ (i.e., a set of ordered pairs satisfying the usual functionality constraint), that can be one of the parameters $w_i$, even if we have no formula defining $f$.

So any function can be used in Replacement, provided that the function is a set of ordered pairs, and not a proper class.

Michael Weiss
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