For a filter on set, ultrafilters are exactly prime filters.

Question

I'm following through a book, and I'm understanding the proof of the proposition that ultrafilters are exactly prime filters.

The proof that ultrafilters are prime goes as the following.

Suppose $F$ is ultrafilter on a set $X$ that is not prime. Then there are $A, B \subseteq X$ with $A \cup B \in F$ but $A, B \notin F$. We have proved before in the text that a filter $U$ on set $X$ is an ultrafilter if and only if for every $A \subseteq X$, $A \notin U$ if and only if there is $B \in U$ with $A \cap B = \emptyset$.

So we may find $A', B' \in F$ with $A \cap A' = \emptyset = B \cap B'$. The proof says that this implies $(A \cup B) \cap (A' \cup B') = \emptyset$. I'm having trouble with this line as $A \cap A' = \emptyset = B \cap B'$ does not imply $(A \cup B) \cap (A' \cup B') = \emptyset$. For example, take $A = \{0,1\}$ $A' = \{2,3\}$, $B = \{2,4\}$, $B' = \{1,5\}$.

So with what condition do we know that $(A \cup B) \cap (A' \cup B') = \emptyset$?

After this line, the proof is easy as $(A \cup B) \cap (A' \cup B') = \emptyset$ if and only if $A \cup B \notin F$.

Thanks!

Answer 1

It appears to be a typo. $A',B'\in\mathscr{F}$, so $A'\cap B'\in\mathscr{F}$, and clearly

$$(A\cup B)\cap(A'\cap B')=\varnothing\,,$$

implying that $A\cup B\notin\mathscr{F}$.

Answer 2

$$ (A\cup B)\cap (A'\cup B') =(A\cap A')\cup (A\cap B')\cup (B\cap A')\cup (B\cap B') =(A\cap B')\cup (B\cap A')$$

Answer 3

Another approach: Let $F$ be an ultrafilter on $X$ and let $A\cup B=C\in F.$

We have

$(i).\;\, \forall E\in F\,(A\cap E\ne \emptyset),$ or

$(ii).\, \forall E\in F\,(B\cap E\ne \emptyset).$

Proof: If $\neg (i),$ let $D\in F$ with $A\cap D =\emptyset.$ So $C\cap D=B\cap D.$ So for any $E\in F$ we have $$\emptyset \ne \;C\cap D\cap E\;=\; B\cap D\cap E\;\subseteq B\cap E.$$ Now if $(i)$ holds: Then $F_A=\{(A\cap E)\cup G: E\in F\land G\subseteq X\}$ is a filter on $X$ with $A\in F_A\supseteq F.$ But $F\supseteq F_A$ by the maximality of $F,$ so $A\in F_A=F.$

Or if $(ii)$ holds, replace $A$ with $B$ in the above paragraph to conclude $B\in F.$

Remarks: $A=(A\cap X)\cup \emptyset\in F_A.$ And if $E\in F$ then $E= (A\cap E)\cup (E\setminus A)\in F_A.$ I leave it to the reader to confirm that if $(i)$ holds then $F_A$ is a filter. A further remark: $(i)\lor (ii)$ depends only on $F$ being a filter.