Intuition of Łoś's theorem [closed]

Łoś's theorem:

If $F$ is an ultrafilter on $I$, $M_i$ is a model with domain $A_i$, then for any formula $\phi$ of L and any sequence $f/F \in (\prod A_i / F)^\omega$ $$\prod M_i/F \vDash_{f/F} \phi \qquad \text{iff} \qquad \{i \in I: M_i \vDash_{f(i)} \phi \} \in F$$

Bell and Slomson's book Models and Ultraproducts: An Introduction suggests to interpret it as: a formula holds in $\prod M_i / F$ iff it holds in 'almost all' it's factors. However it seems clear that this can't be right in some cases. For take $I$ to be the set of natural numbers and let $F$ be the ultrafilter generated by $1$.Then, let the sequence $f$ be defined as: $(f_1, f_2, f_3,...)$ where each $f_i=(1, i+1, i+2, i+3, ...)$. And finally $\phi(v_0, v_1)$ is $v_0=v_1$. In this case $\{i \in I: M_i \vDash_{f(i)} \phi \}$ is just the set $\{1\}$, and so the intuition doesn't hold.

Sure, in general if the set $I$ is of cardinal $k$ the proposed intuition will hold in $2^k$ cases and won't in only $k$ cases; but I was wondering if anyone had a better intuition of how to understand this result.

closed as off-topic by Andrés E. Caicedo, Monroe Eskew, Noah Schweber, Nik Weaver, Will BrianAug 10 at 18:54

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• The "almost all" terminology is perfectly right. If you take a principal ultrafilter, that something holds at almost all points means precisely that it holds at the (one) point that generates the ultrafilter. If you want to develop an intuition, though, looking at nonprincipal ultrafilters is more interesting (and useful). – Andrés E. Caicedo Aug 10 at 17:59
• This question is good, but not appropriate for this site (which is for research mathematics). It would fit well at at math.stackexchange. – Noah Schweber Aug 10 at 18:17
• Perhaps the OP would benefit from realizing that there are different but connected usages of the "almost all" vocabulary. In some contexts, "almost all" natural numbers means all but finitely many natural numbers. In other words, the set of natural numbers for which the statement is true is in element of the co-finite filter. Such usage is generalized to the context of an ultrafilter $U$, where one says "almost all" to mean that the statement is true for a set of indices in $U$. In particular, if $U$ is principal, then $U$-almost all means true on the base of $U$. – Joel David Hamkins Aug 10 at 21:17