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The Compactness Theorem states: $F$ is satisfiable $\iff$ every finite subset of $F$ is satisfiable.

$\Rightarrow$: is trivial.

$\Leftarrow$: We assume every finite subset of $F$ is satisfiable, i.e. no finite subset of $F$ is unsatisfiable. Now we assume by the way of contradiction that $F$ is unsatisfiable, i.e. at least one formula in $F$ has to be a falsum. But this formula would be a finite subset of $F$ which contradicts the fact that by assumption no finite subset of $F$ is unsatisfiable. Therefore $F$ must be satisfiable. Done.

Why is this simple version wrong for I only find pretty sophisticated proofs of the compactness theorem.

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    What is the definition of "satisfiable"? What is "F"? You may want to add a bit more context for that? –  Oct 14 '19 at 20:38
  • Is this logics? If so, add this to the tags, please. – amsmath Oct 14 '19 at 20:53
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    The compactness theorem says that $F$ is satisfiable if and only if every finite subset of $F$ is satisfiable. The argument you use is wrong anyway. – egreg Oct 14 '19 at 21:01
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    @Jack "Satisfiable" is the standard technical term in logic - it's fine to use it in a logic question without defining it (similarly to how there's no problem using "scheme" without definition in algebraic geometry questions). – Noah Schweber Oct 14 '19 at 21:10
  • Well $\phi_1:= (\exists m: \forall n: m \le n)$ is satisfiable (for a language with a symbol $\le$ for an order, as an example) and so is $\phi_2:=(\forall n: \exists m: m \le n)$, but combined they are not satisfiable (and neither is falsum). So your idea can never work. – Henno Brandsma Oct 14 '19 at 21:11
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    @HennoBrandsma That example doesn't work - the two sentences are jointly satisfiable (e.g. the linear order with one element). – Noah Schweber Oct 14 '19 at 21:16
  • @NoahSchweber true, but then we just add a sentence saying there are at least 3 elements.. Good catch. – Henno Brandsma Oct 14 '19 at 21:18
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    @HennoBrandsma Still doesn't work: take $\le$ to hold between all pairs. (Even in linear orders it doesn't work, since any linear order with a least element satisfies both sentences because we can take $m=n$ in $\phi_2$.) – Noah Schweber Oct 14 '19 at 21:19
  • @NoahSchweber: Thanks for that. When I first read the post, the tag logic was not there. –  Oct 15 '19 at 01:11
  • @Jack Ah, my bad - my apologies! (I'm perhaps overly sensitive to this: I've seen a number of new posters get disproportionately harsh receptions for not defining standard terms in logic questions.) – Noah Schweber Oct 15 '19 at 01:52
  • @NoahSchweber: No worries. I appreciate your consideration for new posters. –  Oct 15 '19 at 02:18

1 Answers1

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As far as texts go, I recommend Kaye or chapters $9$-$10$ and $12$-$13$ of Boolos/Burgess/Jeffrey.

Ignoring the misstatement of the theorem in the first place ("subset" should be "finite subset" per egreg's comment), your mistake is implicit in the following sentence:

Now we assume by the way of contradiction that F is unsatisfiable, i.e. at least one formula in F has to be a falsum.

That "i.e." is false - unsatisfiability of $F$ doesn't mean that some specific sentence in $F$ is unsatisfiable, but rather that the sentences of $F$ can't all be true at once.

For example, consider $$F=\{c=d, c\not=d\}.$$ Each individual sentence is satisfiable, but the whole set $F$ is clearly unsatisfiable. (I'm assuming here the context is first-order logic; if you're looking at propositional logic, consider $F=\{p,\neg p\}$ instead.)

So in general a set of sentences could be unsatisfiable for a complicated reason. What you're trying to show is that unsatisfiability can't be too complicated: while we can't find a single culprit sentence in general, if $F$ is unsatisfiable then there is some finite subset of $F$ which is already unsatisfiable. In propositional logic this is already nontrivial, and in first-order logic it's genuinely hard.

Now how does the proof go?

Well, for both propositional and first-order logic, there are multiple different proofs of the compactness theorem. One option in the propositional setting is topological, the key step being Tychonoff's theorem; that's the one I'll outline here.

It's worth noting that this approach has a couple major drawbacks: it does not easily generalize to first-order logic, and it also has significant technical overhead. However, personally I've gradually come to view the finickiness of the topological argument as a positive: understanding precisely why it doesn't generalize gives a lot of insight into how both propositional and first-order logic work, and the additional overhead material is both worth learning on its own and a source of motivation for the study of nonclassical propositional logics and (for me at least) Stone duality.

Suppose I have a finitely satisfiable set $S$ of sentences in the propositional language $\{p_i:i\in I\}$ (that is, the propositional language built out of the sentence letters $p_i$ for $i\in I$). Let $Val$ be the set of all valuations in this language (that is, all maps from $\{p_i:i\in I\}$ to $\{0,1\}$, thinking of $0$ as false and $1$ as true); we want to show that there is some $f\in Val$ which makes $S$ true.

$Val$ can be thought of as the Cartesian product of $I$-many copies of $\{0,1\}$; as such, it has a natural topology, namely the product topology coming from the discrete topology on each of those sets. It's not hard now to show that in this topology, each propositional sentence $\sigma$ picks out a clopen set - that is, for every propositional sentence $\sigma$, the set of $f\in Val$ which make $\sigma$ true is both open and closed in this topology on $Val$; basically, just use the fact that Boolean combinations of clopen sets are clopen.

This tells us the following:

We can think of our set $S$ of sentences as being a set of clopen subsets of $Val$, and the finite satisfiability of $S$ says exactly that any intersection of finitely many of these clopen sets is nonempty.

But this topology on $Val$ was the product of compact topologies (every finite space is trivially compact), hence by Tychonoff's theorem is itself compact. In a compact space, any collection of closed sets with the finite intersection property (= any intersection of finitely many of them is nonempty) has nonempty intersection. Applying this to the set of clopen subsets corresponding to $S$ gives a(t least one) valuation $f$ which makes all of $S$ true at once.

Noah Schweber
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  • Thx Noah, I got it. Can u try to explain how the <= direction of this proof (in propositional logic) is done or at least explain the idea? Also: is there any book or link of math. logic for beginners that still covers the likes of compactness theorem & co. I am not new to logic, I know truth tables and stuff, first order logic and so on, but I have a hard time to understand math. logic for it seems one steep step further. can somebody recommend a book or video lecture (I'd even buy it)? –  Oct 18 '19 at 01:16
  • @Pippen I recommend Kaye's book. But this is a genuinely complicated subject. There are several proofs of the compactness of propositional logic, and they're each a bit sophisticated, and the compactness of first-order logic is even more deep. – Noah Schweber Oct 18 '19 at 02:01
  • @Pippen I've added a sketch of one correct proof of compactness in the propositional case. – Noah Schweber Oct 18 '19 at 02:36
  • What's your take on this: http://builds.openlogicproject.org/open-logic-complete.pdf ... seems to me the most complete book on logic plus it's kind of easy to understand. –  Oct 19 '19 at 03:10
  • @Pippen I'm not familiar with it so I can't vouch for its quality, but it certainly looks cool. I also recommend the section of Boolos/Burgess/Jeffrey on basic first-order logic - specifically, chapters 9-10 and 12-13 (chapter 11 is about computability theory, and can be skipped in this context) - for learning this material; it's what I used, and I don't know why I forgot about it when recommending Kaye's book (not that I don't also recommend Kaye, it's just that BBJ). – Noah Schweber Oct 19 '19 at 19:07
  • Getting back to the specific question here, though, do you have any other questions about the OP or this answer? – Noah Schweber Oct 19 '19 at 19:12