Question 1:

In the same language, let $ X $ be a nonempty set, and let $ \{ (\forall x_{x(i)} f(i)) \ | \ i \in X \} $ be a set of formulas. We use $ x(i) $ to denote the index of the variable on which the universal quantifier acts in the formula $\forall x_{x(i)} f(i) $. Let $ \equiv $ denote that two assignments have the same value for the free variables of the formula indexed by $ \equiv $ (In its lower right corner). Let $ U $ be an ultrafilter on $ X $, and $ (M, \nu) $ be a model structure in this language.

The question is:

Statement 1: For all assignments $ \mu $ in the structure $ M $, if $ \{ i \ | \ \nu \equiv_{(\forall x_{x(i)} f(i))} \mu \} \in U, $ then $ \{ i \ | \ (M, \mu) \models f(i) \} \in U. $

And Statement 2: $ \{ i \ | \ \text{for all assignments } \mu \text{ in } M, \nu \equiv_{(\forall x_{x(i)} f(i))} \mu, \text{ then } (M, \mu) \models f(i) \} \in U. $

Are these two statements equivalent?

Question 2:

Is there a simpler formula that can equivalently replace this: $ \{i \mid \{j \mid \phi(i,j) \} \in U\} \in U \quad \text{and} \quad \{j \mid \{i \mid \phi(i,j) \} \in U\} \in U, $ where $U$ is an ultrafilter?

I understand that we might express it using some commutative statement of generalized quantifiers, such as $\forall^* i \forall^* j \phi(i,j)$, but I still hope to get some concrete content rather than just a different representation.

Additionally, I feel that the content of these two questions is quite different, but the difficulty I encounter when thinking about them feels very similar. What is it? I currently do not have the ability to abstract what this common difficulty is.

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    $\begingroup$ I'm confused by your notation. For each $i\in X$, $f(i)$ is a formula which has $x_{x(i)}$ as a free variable? Could it have more free variables? Did you really mean to write "the formulas indexed by $\equiv$"? $\endgroup$ Commented Jul 7 at 17:14
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    $\begingroup$ Also, is $f(i)$ a formula, or have you valuated one of the variables of formula $f$ at the individual $i$? And you mention $\mu$ and $\nu$ in the hypothesis of statement 1, but only $\mu$ appears in the conclusion, so I am confused about the intended quantification for that. $\endgroup$ Commented Jul 7 at 17:46
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    $\begingroup$ Your equivalence $\nu\equiv\mu$ seems to be only about same-value of the valuations, not truth of the formula at those valuations, whereas the conclusion of statement 1 depends on the truth of the formula. So isn't this obviously wrong? Perhaps you are missing a hypothesis about $(M,\nu)$? $\endgroup$ Commented Jul 7 at 17:49
  • $\begingroup$ I am sorry that my incorrect translation has caused reading difficulties. I translated the Chinese content using gpt, and the effect is not good. I have modified part of it. $ x_x(i)$ is only a variable in the language. It may not be a free variable of $f(i)$, and it may not even appear in $f(i)$. $∀x_x(i)f(i)$ is only the universalization of a variable in the language. $f(i)$ is only a formula, and $f$ is a function $ \{ (i, f(i))| f(i)\text{ is a formula in the language and } i ∈ X \}$ $\endgroup$ Commented Jul 7 at 20:25
  • $\begingroup$ Let me try again to understand. Does $\mu\equiv_{\forall x_{x(i)} f(i)} \nu$ mean that $\mu$ and $\nu$ agree on all the free variables in $f(i)$ except possibly for $x_{x(i)}$? $\endgroup$ Commented Jul 7 at 20:34

1 Answer 1


First, since I found your notation confusing, I hope you don't mind if I rewrite your Question 1 in more standard notation.

Fix a language $L$. Let $I$ be a non-empty set, and let $(\varphi_i)_{i\in I}$ be a family of $L$-formulas, all of which have free variables from a set $V=\{x_0,x_1,x_2\dots\}$. For each $i\in I$, let $V_i\subseteq V$ be the finite set of variables which are free in $\varphi_i$. We also pick a variable $x_{k_i}\in V$ for each $i\in I$, which may or may not be free in $\varphi_i$.

Let $M$ be an $L$-structure. An assignment $a = (a_x)_{x\in V}$ is a family of elements from $M$ indexed by $V$ (so the variable $x$ gets assigned to $a_x\in M$). Given assignments $a$ and $b$, we write $a\equiv_i b$ if $a_x = b_x$ for all $x\in (V_i\setminus \{x_{k_i}\})$.

Fix an assignment $a$ and an ultrafilter $U$ on $I$.

Statement 1: For all assignments $b$, if $\{ i \mid a \equiv_i b \} \in U$, then $\{ i \mid M\models \varphi_i(b) \} \in U. $

Statement 2: $ \{ i \mid \text{for all assignments } b , \text{if }a \equiv_i b, \text{then } M\models \varphi_i(b)\} \in U.$

Are these two statements equivalent?

Next, to make the logical structure of the statements more transparent, I'll rewrite the statements using the quantifier $\forall^*$, which means for "almost all" $i$ in the sense of the ultrafilter.

Statement 1: $\forall b\, ((\forall^* i\, a\equiv_i b)\rightarrow (\forall ^* i\, M\models \varphi_i(b)))$

Statement 2: $\forall^* i \forall b\, (a\equiv_i b\rightarrow M\models \varphi_i(b))$

Now the nice thing about the ultrafilter $\forall^*$ quantifier is that it commutes with / distributes over all propositional connectives. So Statement 1 is equivalent to $\forall b\forall^* i\,(a\equiv_i b\rightarrow M\models \varphi_i(b))$.

Unfortunately, the quantifier $\forall^* i$ does not commute with $\forall b$. Statement 2 obviously implies Statement 1, since $\forall^* i\forall b$ is stronger than $\forall b\forall^* i$. But the converse may not hold.

I'll give a counterexample to the implication from Statement 1 to Statement 2.

Let $L$ be the empty language (the language of equality). Let $I = \omega$ and $U$ any nonprincipal ultrafilter on $\omega$.

For each $i\in \omega$, let $\varphi_i$ be $\bigwedge_{0\leq j<k\leq i+1} x_j = x_k$. Note that the set of free variables in $\varphi_i$ is $V_i = \{x_0,\dots,x_{i+1}\}$. Let $k_i = i+1$, so the specified variable $x_{k_i}$ is $x_{i+1}$.

Let $M$ be any set with at least two elements, and let $0$ and $1$ be distinct elements of $M$. Let $a$ be the constant assignment with value $0$, i.e., $a_k = 0$ for all $k$. Note that for an assignment $b$ and $i\in \omega$, $a\equiv_i b$ if and only if $b_j = a_j = 0$ for all $j\leq i$ (since $V_i\setminus \{x_{k_i}\} = \{x_0,\dots,x_i\}$).

Now with this setup, Statement 1 is true. Let $b$ be an assignment, and suppose $\{i\mid a\equiv_i b\}\in U$. I claim that $b = a$. Indeed, for all $j\in \omega$, since $\{i\mid a\equiv_i b\}$ is infinite, there exists $i\geq j$ such that $a\equiv_i b$, and hence $b_j = a_j = 0$. Thus $b$ is the constant assignment with value $0$, so $\{i\mid M\models \varphi_i(b)\} = \omega \in U$.

But Statement 2 is false. Fix $i\in \omega$, and let $b$ be the assignment with $b_j = 0$ for all $j\leq i$ and $b_j = 1$ for all $j>i$. Then $a\equiv_i b$, but $M\not\models \varphi_i(b)$, since $b_{i+1} = 1 \neq 0 = b_i$.

Thus $\{ i \mid \text{for all assignments } b , \text{if }a \equiv_i b, \text{then } M\models \varphi_i(b)\} = \varnothing \notin U$.

Regarding Question 2, you probably know that if $U$ and $V$ are ultrafilters on $X$, then $U\otimes V = \{Y\subseteq X^2\mid \{i\mid \{j\mid (i,j)\in Y\}\in V\}\in U\}$ is an ultrafilter on $X^2$. So you can rewrite your statement as $\{(i,j)\mid \phi(i,j)\}\cap \{(j,i)\mid \phi(i,j)\}\in U\otimes U$. But this is just different notation. I doubt there is a conceptually simpler way to describe the concept.

  • $\begingroup$ Thanks for your reply. Your understanding is correct. I will add some background to my question, which arises from my personal interest (and my attempt to secure a postgraduate admission offer). I am trying to apply ultrafilter to formulas, specifically in reasoning. This is why I am using a function (a selection function) as an unusual main object($[f]$ is the equivalence class obtained by naturally quotienting the selection function $𝑓$ by the equivalence relation.). To clarify, I will define that for an assignment $\mu$, $(M, \mu) \models [f]$ iff $\{i \mid (M, \mu) \models f(i)\} \in U$ $\endgroup$ Commented Jul 9 at 6:01
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    $\begingroup$ @Stanleysun I have updated my answer with a counterexample to the equivalence. $\endgroup$ Commented 2 days ago
  • $\begingroup$ Very clever construction, thank you! I understand. $\endgroup$ Commented 2 days ago

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