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I've heard the result before that naive set theory is consistent in infinite-valued Łukasiewicz logic. This answer contains a citation. In this logic, every connective is continuous (w.r.t the product topology when necessary). Additionally, $[0, 1]$ equipped with the standard topology has the property that any continuous unary function has a fixed point.

I'm curious whether this last property is enough to make naive set-theory consistent.

I'm also curious whether almost-naive set theories that restrict comprehension to formulas with continuous connectives only are consistent.

What follows is an explanation and my attempt to understand the material.

I'm interested in the following first-order fuzzy logic.

Let $r : \mathbb{R} \to [0, 1]$ be defined in the following way:

$$ r(x) = 1 \;\;\text{iff}\;\; x \ge 1 \\ r(x) = 0 \;\; \text{iff}\;\; x \le 0 \\ r(x) = x \;\; \text{otherwise} $$

Let $[a]$ be the truth value associated with the well-formed formula $a$ in the given context. The set of all truth values is $[0, 1]$ and $1$ is the designated truth value.

I'm taking the following connectives from infinite-valued Łukasiewicz logic, $\to, \lnot, \leftrightarrow$.

$$ [a\to b] = r(1-[a]+[b]) \\ [a \leftrightarrow b] = 1-\mathrm{abs}([a]-[b]) \\ [\lnot a] = 1-[a] $$

Additionally, I add the following connective called strong negation, written $!a$. Strong negation exchanges designated and undesignated truth values.

$$ [!a] = 0 \;\;\text{iff}\;\; [a] = 1 \\ [!a] = 0 \;\; \text{otherwise} $$

Next, I will extend this logic to a first-order setting by adding a notion of relations and a notion of constants.

Let $M$ be a metric space $(M_0, d)$ with the additional constraint that the maximum distance between any two elements is $1$. Let $M$ additionally be equipped with an interpretation of each constant symbol, and let $M$ be equipped with a function of type $M^n \to [0, 1]$ for each relation symbol.

Let $a_1, \cdots, a_n$ be terms, the truth value of $R(a_1, \cdots, a_n)$ is the truth value returned by the interpretation of $R$ in $M$ when applied to the tuple $(a_1, \cdots, a_n)$. The truth value of $a_1 = a_2$ is $1-d(a_1, a_2)$ where $d$ is the distance function.

Additionally, the truth value of $\forall x \mathop. \varphi(x)$ is the greatest lower bound of $\varphi(w)$ for all $w$ in M. The truth value of $\exists x \mathop. \varphi(x)$ is the least upper bound of $\varphi(w)$ for all $w$ in $M$.

I will define equality $=$ as an ordinary predicate that is constrained to be a congruence with respect to all other predicates, i.e.

$$ \forall \vec{x} \vec{y} \mathop. (\vec{x} = \vec{y} \to (R(\vec{x}) \leftrightarrow R(\vec{y}))) $$

The above statement includes an abuse of notation; it cannot be expressed in our logic without a $\land$-like connective representing the minimum of two truth values. $\vec{x} = \vec{y}$ is defined as $1 - [\text{the maximum distance in any component between $\vec{x}$ and $\vec{y}$}]$. The connectives $\to$ and $\leftrightarrow$ are the Łukasiewicz connectives.

So, here is our axiomatization of almost-naive set theory.

We have the axiom of extensionality. Extensionality is sort of bizarre because of the presence of intermediate truth values, it constrains the value of $=$ even when the sets in question are not equal (more specifically, $[a=b]$ must be greater than or equal to $[\forall x \mathop. x \in a \leftrightarrow x \in b]$.)

$$ \forall a \mathop. \forall b \mathop. ((\forall x \mathop. x \in a \leftrightarrow x \in b) \to a = b) $$

We have the axiom schema of continuous comprehension. Let $\varphi$ be constrained not to include any mention of the connective $!$. The double strong negation $!!$ is there to stop a sequence that approximates $\{x : \varphi(x)\}$ more and more closely from satisfying the axiom.

$$ \exists x \mathop. !!(\forall u \mathop. u \in x \leftrightarrow \varphi(x)) $$

In this setting, the Russell set $\{x : \lnot (x \in x)\}$ simply assigns a truth value of $\frac{1}{2}$ to the statement expressing self-membership. $\{x : !(x \in x)\}$ produces a genuine contradiction, but comprehension does not promise us this particular set.

This got me thinking, continuous functions from $[0,1]$ to $[0,1]$ must always have a fixed point, and the presence of these fixed points seems to defuse a broad range of possible paradoxes. Are almost-naive set theories in fuzzy FOL with the continuous-only restriction given above consistent? Is the particular theory given above consistent?

Greg Nisbet
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    I have two comments. Firstly, you should know that the consistency proof mentioned in your first link has been called into question. (See my comments under that answer.) Secondly, the double negation in your comprehension schema isn't necessary. If you were able to show the consistency of the weaker version of the theory, a fairly standard model-theoretic argument would show that a sufficiently saturated model of the weak theory satisfies your theory. I could write this out explicitly if you want, but it doesn't actually answer your question. – James E Hanson Jan 14 '23 at 19:31
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    Actually my second comment depends on an aspect of your question. Are you requiring that $\in$ be uniformly continuous with regards to the metric? In other words, what form of the axioms of equality are you assuming? – James E Hanson Jan 14 '23 at 19:42
  • @JamesHanson I added an explicit axiom of equality, but my understanding of nonclassical first-order logics is pretty weak. When I wrote this question I was not thinking about the continuity of predicates (w.r.t. the induced topology on the domain) at all. – Greg Nisbet Jan 14 '23 at 21:08
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    Isn't your informal description backwards? Your equality axiom is saying that $|R(\vec{x})-R(\vec{y})| \leq d(\vec{x},\vec{y})$ (where $d$ is the max metric on tuples), right? – James E Hanson Jan 14 '23 at 21:13
  • @JamesHanson ... Yes, you are completely correct. I need to think about this and probably rewrite the question. I picked an axiom of equality that I have seen before in other settings, but its behavior here is odd. Thank you for commenting though, I didn't realize that the domain had a topology that I need to pay attention to, or that I had failed to specify how equality was supposed to behave, or that the proof I originally indirectly cited through Noah's answer has been challenged, or that the double negation thing was not necessary. – Greg Nisbet Jan 14 '23 at 21:31
  • The fact that that proof has been called into question doesn't seem to be widely known; Terui never published his note. Also, regarding the double negation thing, while I said the argument is 'fairly standard,' I certainly wouldn't consider it obvious. – James E Hanson Jan 14 '23 at 21:38

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