6

Is it possible to construct a formula or set of formulas using only equality that are satisfiable only in an infinite domain? I have seen such formulas but they all use a relation like greater than or less than instead of equality.

Mauro ALLEGRANZA
  • 99,247
Matt
  • 69
  • @AmitRajaraman No, I've read that post but it only gives such a formula without using equality instead of using equality – Matt Jun 21 '21 at 04:40
  • 1
    "Formula or set of formulas"? Yes, if you allow an infinite set of formulas. Otherwise no. – bof Jun 21 '21 at 04:43
  • 1
    @bof How could you construct such an infinite set? – Matt Jun 21 '21 at 04:45
  • 5
    For each $n$ you can construct a sentence $\sigma_n$ which is true if and only if there are more than $n$ elements in the domain. – bof Jun 21 '21 at 04:46
  • @bof Oh okay that makes sense. Why can you not construct a finite set that is only satisfiable in an infinite domain? – Matt Jun 21 '21 at 04:49
  • 2
    If you could do it with a finite set you could do it with a single formula; just take the conjunction. And there is very little you can say with a first-order formula with only $=$. You can say things like $x=y$ and $x\ne y$ and "there are at least $n$ things" and "there are at most $n$ things" and combinations of the like. The precise formulation and proof would be boring and tedious (for a non-logician like me) but basically trivial. – bof Jun 21 '21 at 06:04
  • If you were allowed predicates, you could create a FOPL version of the first 4 Peano Axioms with $N(x)$ corresponding to $x\in N$ and $S(x,y)$ corresponding to $S(x)=y$. I'm not sure why formulas and variables indexed by a pre-existing set of natural numbers are considered "logical" while non-numerical predicates are not, but I am told this is the case. – Dan Christensen Jun 22 '21 at 16:12
  • How does a two sentence PSQ Merit 6 upvotes, if not on behalf of the answerers? – amWhy Jul 04 '21 at 22:26

1 Answers1

5

The answers are already in the comments of bof, but let me make them precise (in particular the negative answer). To summarise, in just the language of equality we have the following:

  1. there is a set $\Sigma$ of formulas such that $M \models \Sigma$ if and only if $M$ is infinite,
  2. there is no formula $\phi$ such that $M \models \phi$ if and only if $M$ is infinite.

Note that the second point is equivalent to "there is no finite set of formulas $\Phi = \{ \phi_1, \ldots, \phi_n \}$ such that $M \models \Phi$ if and only if $M$ is infinite", just take $\phi$ to be the conjunction $\phi_1 \wedge \ldots \wedge \phi_n$.

Proof of claim 1. For $n \in \mathbb{N}$, let $\sigma_n$ express "there are at least $n$ elements". So for example, we could take $\sigma_n$ to be $$ \exists x_1 \ldots x_n (\bigwedge_{i \neq j} x_i \neq x_j). $$ Then $\Sigma = \{ \sigma_n : n \in \mathbb{N} \}$ is easily seen to have the property described in claim 1.

Proof of claim 2. Suppose, for a contradiction, that there is such a $\phi$. Note that we are working in just the language with equality, so the structures we consider are pure sets. By assumption we have that $M \models \neg \phi$ if and only if $M$ is finite. Now consider $T = \{\neg\phi\} \cup \Sigma$, where $\Sigma$ is as in the previous proof. Every finite subset $T_0 \subseteq T$ has a model. This is because it only contains a finite part of $\Sigma$ and so there is a maximal $n$ such that $\sigma_n \in T_0$. Then the set with $n+1$ elements is a model of $T_0$. By compactness there is then a model $M \models T$. So $M \models \neg\phi$, thus $M$ must be finite, but at the same time $M \models \Sigma$, so $M$ must be infinite. A contradiction, so we conclude that no such $\phi$ can exist.

Bruno Bentzen
  • 3,392
Mark Kamsma
  • 13,548
  • 1
  • 11
  • 35
  • Please remember the Enforcement of quality standards when dealing with answering very low quality "do this for me" questions. – amWhy Jul 04 '21 at 22:27
  • 1
    @amWhy I feel like OP clarified enough as to why they thought this question was worth asking and what their thoughts were on this. Part of that may be in the comments, but since the question itself was clear enough I felt like an answer would be appropriate. – Mark Kamsma Jul 04 '21 at 22:38
  • Not in the question. Please encourage the asker to include further comments in response to bof (not you, because you were already willing to answer the op), in the question post. And I do not think your "feeling like the OP"... clarified sufficiently is a sound bases to answer what was, at the time of your answer, a PSQ. – amWhy Jul 04 '21 at 22:43
  • 1
    @amWhy OP states a question, indicates a way they know how to solve it and wonder about a different way to solve it, in a precise manner. This, to me, is a sound basis to give an answer to an interesting question. If you think OP should improve the question, please educate them on how to concretely do this instead of just wondering how they get upvotes. – Mark Kamsma Jul 04 '21 at 22:53
  • Can this method be also used to prove that in a language with no predicates constants and functions with only =, to prove that any two infinite models are (elementary) equivalent. – Vivaan Daga Jun 06 '22 at 07:09
  • @VioletFlame No, tat requires a little more work. – Mark Kamsma Jun 06 '22 at 11:02
  • How would one prove that? – Vivaan Daga Jun 06 '22 at 12:23
  • @VioletFlame That should be a new question (or maybe there is already a question with an answer about this, I'm not sure). If you ask a new question, make sure to include your thoughts, what you've tried and which tools you have (e.g. which general theorems do you already know that might be relevant). – Mark Kamsma Jun 06 '22 at 12:57