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Often sentences that are false, are nonetheless "true almost everywhere."

Example 0. The integers do not satisfy the cancellation law:

$$\forall a\forall xy(ax = ay \rightarrow x=y)$$

This is because we can find a counterexample, namely $a=0$ (e.g. take $x=1,y=2$).

However, the way I see it, the tuple $(a,x,y)$ lives in a $3$-dimensional free module, namely $\mathbb{Z}^3$. Since the set of all counterexamples to the cancellation law forms a $2$-dimensional subspace, hence there are only "trivially-many" counterexamples, and therefore it would be reasonable to deem the cancellation law "true almost everywhere" in $\mathbb{Z}.$

Example 1. The rational numbers do not satisfy the reciprocation law:

$$\forall x\exists y(xy=1)$$

This is because we can find a counterexample, namely $x=0$. However, since $\{x \in \mathbb{Q} \mid \neg \exists y(xy=1)\}$ forms a $0$-dimensional subspace of the $1$-dimensional vector space $\mathbb{Q},$ hence it would be reasonable to deem the reciprocation law "true almost everywhere" in $\mathbb{Q}.$

Example 2. Here's a third example. Let $P$ denote the set of all points of the projective plane, and $L$ denote the set of all lines. Then for any two points $p,q \in P$, there is a unique line $p \vee q \in L$ that passes through both points. And for any two lines in the projective $l,m$, there is a unique point $lm$ at which they intersect. This defines two binary functions, $$\vee : P \times P \rightarrow L, \quad \times : L \times L \rightarrow P.$$

They satisfy the following interesting-looking identities.

$$\frac{\forall p \in P \quad \forall r,s \in P}{(p\vee r)(p\vee s) = p}\qquad \frac{\forall l \in L\quad \forall m,n \in L}{lm\vee ln = l}$$

But none of this is quite right. Observe that for no point $p$ is $p\vee p$ is ever a well-defined line, and for no line $l$ is $ll$ ever a well-defined point. So those aren't functions we've just described, they're merely partial functions. More fundamentally, the above identities are wrong. For what if the line $p\vee r$ equals the line $p\vee s$? Then their intersection $(p\vee r)(p\vee s)$ isn't even well-defined, so it surely would not be correct to say that it equals $p.$

The point, though, is that there is only a "trivial subspace" of counterexamples to the totality of our two would-be functions, and only a trivial subspace of counterexamples to our identities.

Question. Suppose we're interested in alternative semantics for first-order logic (and/or fragments thereof) in which functions are "almost-everywhere total," and universal quantifiers are interpreted "almost-everywhere." Alternatively, suppose we're interested in adjoining a third quantifier $\forall^*$ to first-order logic, such that $\forall^*x\varphi$ is intended to assert that $\varphi$ is true for almost-all $x$. Where can we learn about these ideas?

goblin GONE
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  • This really is a universal statement, you're just making it about a subset with a specific property. – Adam Hughes Jul 16 '14 at 09:30
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    You haven't really defined what "almost everywhere" means. If you want it to be "except at most five examples" then you don't get anything stronger; if you want it to be "all but finitely many", then you can give a categorical characterization to $\Bbb N$ thus destroying the compactness theorem; and if you just want an arbitrary ideal of subsets of the model to be considered "nowhere" then you have another parameter in the semantics, and I'm not quite sure how this plays out, but since this can be the finite sets, it probably usually ends with being a non-compact logic. – Asaf Karagila Jul 16 '14 at 09:34
  • @AsafKaragila, I know, I was hoping "almost everywhere" would be a free parameter of the framework. I.e. We input what we mean by "almost everywhere" and out pops a corresponding satisfaction relation $\models$. How does the categorical characterization of $\mathbb{N}$ work? – goblin GONE Jul 16 '14 at 09:51
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    It's the unique linear order without a maximal element, that for every $x$, all but finitely many $y$'s are above $x$. – Asaf Karagila Jul 16 '14 at 09:54
  • @AsafKaragila, nice. So I think that if we add to the linear order axioms the sentence $\forall x(\forall^* y(y \geq x)),$ its sole model ought to be $\mathbb{N}$ up to isomorphism, so long as $\forall^*$ is interpreted to mean "for all but finitely many." – goblin GONE Jul 16 '14 at 09:59
  • Yes. And if you interpret this as "all but countably many" you get $\omega_1$; or generally, "all but $<\kappa$" gets you $\kappa$. – Asaf Karagila Jul 16 '14 at 10:10
  • The Stanford Encyclopedia of Philosophy has some information about generalised quantifiers (and a fairly extensive bibliography, although not all are from a mathematical point of view). – user642796 Jul 16 '14 at 10:20
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    Perhaps related: filter quantifiers, which I think allow one to say "for all large"; see http://math.stackexchange.com/q/1045827/11994. – MarnixKlooster ReinstateMonica Feb 10 '15 at 07:28
  • @MarnixKlooster, thanks, that is exactly what I'm looking for. – goblin GONE Feb 10 '15 at 10:33

2 Answers2

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For generalized quantifiers, see Jouko Vaananen, Models and Games (2011), page 283-on.

On page 288, see :

Example 10.6 The dual of $\exists$ is $\forall$ and vice versa: the dual of $\forall$ is $\exists$.

The dual of $\exists^{\ge \omega}$ is the quantifier all-but-finite

$\forall^{< \omega} = \{ X : |A - X|$ is finite $\}$

and vice versa: the dual of $\forall^{< \omega}$ is the quantifier $\exists^{\ge \omega}$.

Where [page 285] :

The infinity quantifier $\exists^{\ge \omega}$ is the mapping

$\exists^{\ge \omega}(A) = \{ X \subseteq A : X$ is infinite $\}$.

Note

See page 284 :

Definition 10.1 A weak (generalized) quantifier is a mapping $Q$ which maps every non-empty set $A$ to a subset of $\mathcal P(A)$. A weak (generalized) quantifier on a domain $A$ is any subset of $\mathcal P(A)$.

Community
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Mauro ALLEGRANZA
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  • Okay, you're right that Example 1 falls under this umbrella. But I think that Example 0 and Example 2 surely don't. The counterexamples to $∀axy(ax=ay→x=y)$ form a 2-dimensional subspace of $\mathbb{Z}^3$. How do I let the mathematics know that a $2$-dimensional space of counterexamples is negligible and irrelevant? – goblin GONE Jul 16 '14 at 10:07
  • @goblin - in the above book there are other generalized quantifier; I'm mnot sure you van find what you need. But the relevant thing is that there is a semantic for them; you can try to use that approach to "develop" the suitable ones for you. – Mauro ALLEGRANZA Jul 16 '14 at 10:11
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You seem to concentrate on algebraic properties that hold on a dense open set in the sense of Zariski topology. So the latter can be considered a valid interpretation of "almost everywhere" in this context. Then again, by the very definition of Zariski topology this just means that the set of exceptions can be desribed as the zero set of a nonzero polynomial; this might look circular (the argumentation, not the zero set).

Hagen von Eitzen
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