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Show that doesn't exists $\Delta$ such that it is satisfable iff assigment has finite number of $1's$. It means that we assign only finitely many $1's$ for variables and from this fact we would like to conclude that $\Delta$ is satisfable. And if $\Delta$ is satisfable then assigment constains only finite unmber of $1's$. Number of variables is infinite: $a_1,a_2,...$. We are in predicate-logic (no quantifiers).

My idea is following:
Let's suppose that such $\Delta$ exists. Lets consider any finite subset of $\Delta_0\subseteq_{\text{fin}}\Delta$ such that it may be contradictory - finite subset with this property must exists - if it doesn't exists $\Delta$ is satisfable regardless to valuation of variables - then we have contradiction and show that such $\Delta$ doesn't exists.

So, we have finite $\Delta_0$ such that it may be contradictory - we force contradictory - we use only finitely many values $1$ because subset is finite. Rest of variables we set to $0$. However, $\Delta$ is unsatisfable becuase there exists unsataisfable finite susbet (compacntess theorem). Although, we assign only finite number of ones, $\Delta$ is unsatisfable.

Am I ok ?

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    How do you get a contradiction from the existence of a finite set of sentences with the desired property? You haven't argued that at all. – 2017-02-17
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    Do you mean first part ? – 2017-02-17
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    Frankly, I mean all of it - I don't actually see an argument here. But the main point that's missing is, why *can't* you have a finite set of sentences which axiomatizes the set of assignments with finitely many $1$s? This is indeed a compactness problem, but you're not using compactness in the right way (in particular, compactness isn't needed to show that if a finite subset of a theory is inconsistent, then so is the theory). – 2017-02-17
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    Firstly, if you agree with it: Lets get finite subset. Let assign variables in such way that this subset is contradictory - we use only finite number of ones. Then from compactness we conclude that this set can't be satisfied – 2017-02-17
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    "Let assign variables in such way that this subset is contradictory - we use only finite number of ones. " What does that mean? How do you know you can do it? – 2017-02-17
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    If I can't then this finite subset is tautology – 2017-02-17
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    You're missing my emphasis - how do you know there's an assignment falsifying $\Delta_0$ *which uses only finitely many $1$s*? (Of course this statement is *true*, but you need to *justify* it . . .) – 2017-02-17
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    because $\Delta_0$ is finite – 2017-02-17
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    So? I'm not convinced - what *consequence* of $\Delta_0$ being finite are you using here? (Again, the statement is true; I'm just saying you haven't clearly justified it.) – 2017-02-17
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    $\Delta_0$ is finite, hence it contains only finite number of variables$ – 2017-02-17
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    And if such subset which may by falsified doesn't exists then each finite subset is satisfable regardless of valuation, hence only with infinitely many ones – 2017-02-17
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    Alright, so you've shown that no finite $\Delta_0$ axiomatizes the set of assignments with finitely many variables. So how are you going to apply compactness here? – 2017-02-17
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    I apply compactness on finite susbet which may be falsified. – 2017-02-17
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    And in particular I falsify them, variable absent in this subset are set to $0$ - I got set with finitely many ones and unsatisfable - unsatisfable because it contains contradictory susbet – 2017-02-17
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    "I apply compactness on finite susbet which may be falsified." I don't understand what this means. Compactness states: "If $\Gamma$ implies $\varphi$, then some finite $\Gamma_0\subseteq\Gamma$ implies $\varphi$. Presumably your $\Gamma$ is $\Delta$ - what's your $\varphi$? – 2017-02-17
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    My commapctnes states: set if satisfable iff each finite subset is satisfable. The same is: set is unsatisfable iff there exists infinite subset that is unsatisfable. In particular exists finite subset that it is unsatisfable. However your version is conclusion (I think so) of my version – 2017-02-17
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    "set is unsatisfable iff there exists *infinite* subset that is unsatisfable." I think you mean *finite*. Regardless, *how are you applying compactness here*? – 2017-02-17
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    @NoahSchweber look: I got finite subset $\Delta_0$. I make it contradictory. In this moment I have in $\Delta$ contradictory finite subset $\Delta_0$. So using compactness theorem $\Delta$ is unsatisfable. – 2017-02-17
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    You haven't made it *contradictory*, you've shown that it doesn't characterize "finitely many $1$s", and you've shown that it *can* be *not satisfied*; but this isn't an immediate contradiction, and it doesn't mean $\Delta_0$ is *unsatisfiable.* Unsatisfiable and can-be-unsatisfied are different things! (And, **again,** you don't need compactness to conclude that a theory with an unsatisfiable subset is unsatisfiable; that's basically immediate from the definition.) – 2017-02-17
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    Wait, we said that there exists finite subset that may be falsified, yeah ? (if not exists then we got contradiction, so we assume that it exists) – 2017-02-17
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    I set variables in way that this subset is unsatisfable(=contradictory) – 2017-02-17
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    So after this valuation, $\Delta$ contains finite subset which contradictory(unsatisfable) because a minute ago I set variables in proper way – 2017-02-17
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    @NoahSchweber Why I am wrong ? Look at my last comments. – 2017-02-17
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    It looks like I was misunderstanding what you're getting at, since you are using the wrong terms. Saying that a set of sentences is *unsatisfiable* means that there is *no* assignment making it true; you need to distinguish between "unsatisfiable" and "unsatisfied *by this assignment*." In general, you need to be much clearer about what you're actually doing - I was genuinely confused by your entire argument. I've posted a cleaned-up version of your argument; I hope you see how this is clearer. (In particular, compactness is not used here.) – 2017-02-17

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From the comment thread, I think a correct solution (or rather, a clear-ish statement of the idea in your post, which is correct) has emerged; but let me state it cleanly. In particular, we can avoid the Compactness Theorem entirely.

You argue as follows:

  • Suppose $\Delta$ characterizes the assignments with finitely many $1$s. Then consider the assignment $\nu$ which sets all variables to $1$; this assignment has infinitely many $1$s, so we can find some element $\varphi$ of $\Delta$ not satisfied by $\nu$.

  • Alright, so look at this $\varphi$. Any assignments which agree on the variables in $\varphi$, agree on whether they satisfy $\varphi$ or not. Since $\varphi$ is a single sentence, it uses only finitely many variables; so, letting $\mu$ be the assignment which agrees with $\nu$ on those variables (i.e. sets them to $1$) and sets all other variables to $0$, since $\nu$ doesn't satisfy $\varphi$ we must also have that $\mu$ doesn't satisfy $\varphi$.

  • But then $\mu$ doesn't satisfy $\Delta$, since $\varphi\in \Delta$; a contradiction, since $\mu$ has only finitely many $1$s.

Note that this argument did not use compactness at all. This was one of the points that confused me in the comment thread - I didn't understand how compactness was entering into the proof.

There is also an argument using compactness; I suspect this is the intended solution.

Given such a $\Delta$, consider the theory $$\Delta\cup\{a_i: i\in \mathbb{N}\}.$$ This is finitely satisfiable (why?), hence has a model; but that model has infinitely many $1$s, contradicting the assumption on $\Delta$.

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    Yes, your simplification of my solution is great and crucial. And you was right - idea is ok indeed, but the main problem was mixin unsatisfable with unassigment values :D. And when it comes to second solution. I can't understand why I didn't come up with it. Thanks!! – 2017-02-17
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    @HaskellFun I mean, your solution is better in a meaningful sense - it doesn't invoke any unnecessary theorems! – 2017-02-17