(Blackburn's book) Question regarding a lemma about ultrafilter extension

Question

I am reading the proof of:

Let $\tau$ be a modal similarity type, and M a $\tau$-model. Then, for any formula $\phi$ and any ultrafilter $u$ over $W$, $V (\phi)\in u$ iff $ue(M),u\Vdash \phi$.

$ue(M)$ is the ultrafilter extension of $M$, $V$ is the evaluation which takes a formula and returns the set of worlds that this formula holds.

The first lines of the proof goes as:

The proof of claim is by induction on $\phi$. The basic case is immediate from the definition of $V^{ue}$. The proofs of the boolean cases are straightforward consequences of the defining properties of ultrafilters.

The nagation case is indeed obvious. But I have trouble on disjunction case: I have $V(\phi\lor \psi)=V(\phi)\cup V(\psi)\in u$, and my inductive hypothesis is about $V(\phi),V(\psi)$ individually, so could some one please tell me what is going on here and how to use "consequences of the defining properties of ultrafilters" to prove it?

Also could someone explicitly points out which "consequences of the defining properties of ultrafilters" do we use together with some proof?

Answer 1

Basically, what you're trying to prove is:

Whenever $A\cup B$ is in the ultrafilter $u$, either $A$ is in $u$ or $B$ is in $u$.

Let's recall the definition of "ultrafilter" (this is what's meant by "defining properties"):

• $u$ must contain the whole "base set," and must not contain $\emptyset$.

• If $X\in u$ and $Y\supseteq X$, then $Y\in u$.

• If $X,Y\in u$ then $X\cap Y\in u$.

• For every $X$, either $X$ or its complement is in $u$.

(The first three conditions form the definition of "filter;" the fourth condition gives the "ultra" part. If we add "$u$ contains no singletons," we get the definition of "non-principal, or free, ultrafilter.")

None of these explicitly mention unions of two sets, but that third clause looks intriguing since it also mentions the combination of two sets. It turns out that the key here is to combine that third clause with the fourth clause (one reason to guess that we should do this is that unions and intersections are closely related via complements, and the fourth clause has something to say about complements). Specifically:

Suppose neither $A$ nor $B$ is in the ultrafilter. Can you show that $A\cup B$ is not in the ultrafilter either?

This is the contrapositive of the statement up top, so if you can show this you'll be done.

HINT: Since neither $A$ nor $B$ are in $u$, we know that $\overline{A}$ and $\overline{B}$ are in $u$. Knowing this, what third set can you conclude must be in $u$? How is this related to the set we care about ($A\cup B$)?